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Do proteins contain phosphorous? If its true then how alfred hershey and martha chase used the statement that proteins do not contain phosphorous?

Do proteins contain phosphorous? If its true then how alfred hershey and martha chase used the statement that proteins do not contain phosphorous?


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I hv heard about phosphorus that it is a constituent of certain protein, although we know that no amino acids have phosphorus… and if its true then how Alfred hershey and Martha chase experiment got acclaimed whose base of the experiment was that "protein doesn't have phosphorus"??


Yes some of the proteins contain phosphorus but in the experiment (Hershey and chase experiment) they used bacteriophage it's protein sheath contains carbon,hydrogen,nitrogen and sulphur while the DNA in the bacteriophage contains carbon,hydrogen,nitrogen and phosphorus(in the acid) so they selected this for that experiment moreover sulphur is absent in the DNA but phosphorus is not so one can conclude that DNA is the thing that was transferred Reference:NCERT


Yes, None of the amino acid contain phosphorus in its side chain, but it is introduced during Post translational modification. This helps in stability and proper functioning of proteins. (Via:https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3415839/.

As, phosphorus is main building block of nucleic acid not proteins, so its usage in production of new bacteriophage will represent utilization of phosphorus in nucleic acid synthesis.Further using the blender experiment, Hershey and Chase separated the protein coat of bacteriophage from its nuclei. And then introduced this nucleic acid into the bacteria, and found that it can reproduce to synthesize new viruses. And this proved that nucleic acid is responsible for transmission of genetic information.
(Via: https://embryo.asu.edu/pages/hershey-chase-experiments-1952-alfred-hershey-and-martha-chase)


It is true that no amino acids contain phosphorus, but due to post-translational modifications some proteins can be phosphorylated. In addition, there are enzymatic proteins such as glycogen phosphorylase b, which are relatively inactive protein, and the change to the active forms (glycogen phosphorylase a) can be carried out by the action of phosphorylase kinase (Lehninger. Principles of Biochemistry. 5th Ed.)


Samacheer Kalvi 12th Bio Zoology Molecular Genetics Text Book Back Questions and Answers

Question 1.
Hershey and Chase experiment with bacteriophage showed that
(a) Protein gets into the bacterial cells
(b) DNA is the genetic material
(c) DNA contains radioactive sulphur
(d) Viruses undergo transformation
Answer:
(b) DNA is the genetic material

Question 2.
DNA and RNA are similar with respect to
(a) Thymine as a nitrogen base
(b) A single-stranded helix shape
(c) Nucleotide containing sugars, nitrogen bases and phosphates
(d) The same sequence of nucleotides for the amino acid phenyl alanine
Answer:
(c) Nucleotide containing sugars, nitrogen bases and phosphates

Question 3.
A mRNA molecule is produced by
(a) Replication
(b) Transcription
(c) Duplication
(d) Translation
Answer:
(b) Transcription

Question 4.
The total number of nitrogenous bases in human genome is estimated to be about
(a) 3.5 million
(b) 35000
(c) 35 million
(d) 3.1 billion
Answer:
(d) 3.1 billion

Question 5.
E. coli cell grown on 15N medium are transferred to 14N medium and allowed to grow for two generations. DNA extracted from these cells is ultracentrifuged in a cesium chloride density gradient. What density distribution of DNA would you expect in this experiment?
(a) One high and one low density band
(b) One intermediate density band
(c) One high and one intermediate density band
(d) One low and one intermediate density band
Answer:
(d) One low and one intermediate density band

Question 6.
What is the basis for the difference in the synthesis of the leading and lagging strand of DNA molecules?
(a) Origin of replication occurs only at the 5’ end of the molecules
(b) DNA ligase works only in the 3’ → 5’ direction
(c) DNA polymerase can join new nucleotides only to the 3 ’ end of the growing stand
(d) Helicases and single-strand binding proteins that work at the 5’ end
Answer:
(d) Helicases and single-strand binding proteins that work at the 5’ end

Question 7.
Which of the following is the correct sequence of event with reference to the central dogma?
(a) Transcription, Translation, Replication
(b) Transcription, Replication, Translation
(c) Duplication, Translation, Transcription
(d) Replication, Transcription, Translation
Answer:
(d) Replication, Transcription, Translation

Question 8.
Which of the following statements about DNA replication is not correct?
(a) Unwinding of DNA molecule occurs as hydrogen bonds break
(b) Replication occurs as each base is paired with another exactly like it
(c) Process is known as semi – conservative replication because one old strand is conserved in the new molecule
(d) Complementary base pairs are held together with hydrogen bonds
Answer:
(b) Replication occurs as each base is paired with another exactly like it

Question 9.
Which of the following statements is not true about DNA replication in eukaryotes?
(a)) Replication begins at a single origin of replication.
(b) Replication is bidirectional from the origins.
(c) Replication occurs at about 1 million base pairs per minute.
(d) There are numerous different bacterial chromosomes, with replication occurring in each at the same time.
Answer:
(d) There are numerous different bacterial chromosomes, with replication occurring in each at the same time.

Question 10.
The first codon to be deciphered was which codes for
(a) AAA, proline
(b) GGG, alanine
(c) UUU, Phenylalanine
(d) TTT, arginine
Answer:
(c) UUU, Phenylalanine

Question 11.
Meselson and Stahl’s experiment proved __________
(a) Transduction
(b) Transformation
(c) DNA is the genetic material
(d) Semi-conservative nature of DNA replication
Answer:
(d) Semi-conservative nature of DNA replication

Question 12.
Ribosomes are composed of two subunits the smaller subunit of a ribosome has a binding site for and the larger subunit has two binding sites for two
Answer:
mRNA, tRNA

Question 13.
Anoperonisa:
(a) Protein that suppresses gene expression
(b) Protein that accelerates gene expression
(c) Cluster of structural genes with related function
(d) Gene that switched other genes on or off
Answer:
(d) Gene that switched other genes on or off

Question 14.
When lactose is present in the culture medium:
(a) Transcription of lacy, lac z, lac a genes occurs
(b) Repressor is unable to bind to the operator
(c) Repressor is able to bind to the operator
(d) Both (a) and (b) are correct
Answer:
(d) Both (a) and (b) are correct

Question 15.
Give reasons: Genetic code is ‘universal’.
Answer:
The genetic code is universal. It means that all known living systems use nucleic acids and the same three base codons (triplet codon) direct the synthesis of protein from amino acids. For example, the mRNA (UUU) codon codes for phenylalanine in all cells of all organisms. Some exceptions are reported in prokaryotic, mitochondrial and chloroplast genomes. However similarities are more common than differences.

Question 16.
Name the parts marked ‘A’ and ‘B’ in the given transcription unit:
Answer:

Question 17.
Differentiate – Leading strand and lagging strand
Answer:

  1. DNA polymerase I Involved DNA repair mechanism
  2. DNA polymerase II Involved DNA repair mechanism
  3. DNA polymerase III Involved in DNA replicaton

Question 18.
Differentiate – Template strand and coding strand.
Answer:

  1. Template Strand: During replication, DNA strand having the polarity 3’ → 5’ act as template strand.
  2. Coding Strand: During replication, DNA strand having the polarity 5’ → 3’ act as coding strand.

Question 19.
Mention any two ways in which single nucleotide polymorphism (SNPs) identified in human genome can bring revolutionary change in biological and medical science.
Answer:
Scientists have identified about 1.4 million locations, where single base DNA differences (SNPs – Single nucleotide polymorphism – pronounced as ‘snips’) occur in humans. Identification of ‘SNIPS’ is helpful in finding chromosomal locations for disease associated sequences and tracing human history.

Question 20.
State any three goals of the human genome project.
Answer:

  1. Identify all the genes (approximately 30000) in human DNA.
  2. Determine the sequence of the three billion chemical base pairs that makeup the human DNA.
  3. To store this information in databases.

Question 21.
In E.coli, three enzymes 0- galactosidase, permease and transacetylase are produced in the presence of lactose. Explain why the enzymes are not synthesized in the absence of lactose.
Answer:
In the absence of lactose, the repressor protein binds to the operator and prevents the transcription of structural gene by RNA polymerase, hence the enzymes are not produced. However, there will always be a minimal level of lac operon expression even in absence of lactose.

Question 22.
Distinguish between structural gene, regulatory gene and operator gene.
Answer:
Structure of the operon: Each operon is a unit of gene expression and regulation and consists of one or more structural genes and an adjacent operator gene that controls transcriptional, activity of the structural gene.

  1. The structural gene codes for proteins, rRNA and tRNA required by the cell.
  2. Promoters are the signal sequences in DNA that initiate RNA synthesis. RNA polymerase binds to the promoter prior to the initiation of transcription.
  3. The operators are present between the promoters and structural genes. The repressor protein binds to the operator region of the operon.

Question 23.
A low level of expression of lac operon occurs at all the time in E-coli. Justify the statement.
Answer:
One of the enzyme synthesized by lac operon is permease which is involved in the transport of lactose into the cells. If the lac operon gets inactivated, permease is not synthesized hence lactose cannot enter the cell. Lactose acts as a inducer, binding to the repressor protein and switch on the operator to initiate gene expression.

Question 24.
Why the human genome project is called a mega project?
Answer:
The international human genome project was launched in the year 1990. It was a mega project and took 13 years to complete. The human genome is about 25 times larger than the genome of any organism sequenced to date and is the first vertebrate genome to be completed. Human genome is said to have approximately 3 >109 bp. HGP was closely associated with the rapid development of a new area in biology called bioinformatics.

Question 25.
From their examination of the structure of DNA, What did Watson and Crick infer about the probable mechanism of DNA replication, coding capability and mutation?
Answer:
Inference of Watson and Crick on DNA replication: They concluded that each of the DNA strand in a helix act as template during DNA replication leading to formation of new daughter DNA molecules, which are complementary to parental strand, (i.e., Semi-conservative method of replication) Inference on coding capability: During transcription, the genetic information in the DNA strand is coded to mRNA as complementary bases, (except for uracil in place of thymine in RNA) Inference on mutation: Any changes in the nucleotide sequence of DNA leads to corresponding alteration in aminoacid sequence of specific protein thus confirming the validity of genetic code.

Question 26.
Why tRNA is called an adapter molecule?
Answer:
The transfer RNA, (tRNA) molecule of a cell acts as a vehicle that picks up the amino acids scattered through the cytoplasm and also reads specific codes of mRNA molecules. Hence it is called an adapter molecule. This term was postulated by Francis Crick.

Question 27.
What are the three structural differences between RNA and DNA?
Answer:
DNA:

  1. Sugar is deoxyribose sugar.
  2. Double stranded structure.
  3. Nitrogen bases are Adenine, Guanine, Cytosine and Thymine.
  1. Sugar is ribose sugar.
  2. Single stranded molecule.
  3. Nitrogen bases are Adenine, Guanine, Cytosine and Uracil.

Question 28.
Name the anticodon required to recognize the following codons:
AAU, CGA, UAU, and GCA.
Answer:
UUA, GCU, AUA and CGU.

Question 29.
(a) Identify the figure given below
(b) Redraw the structure as a replicating fork and label the parts
(c) Write the source of energy for this replication and name the enzyme involved in this process.
(d) Mention the differences in the synthesis of protein, based on the polarity of the two template strands.
Answer:
(a) Replication fork
(b)

(c) Deoxy nucleotide, triphosphate acts as a energy source for replication. DNA polymerase is used for replication
(d) mRNA contacting information for protein synthesis will developed from DNA strand having polariy 5’ → 3’

Question 30.
If the coding sequence in a transcription unit is written as follows:
5’ TGCATGCATGCATGCATGCATGCATGC 3’
Write down the sequence of mRNA.
Answer:
mRNA sequence is 3’ACGUACGUACGUUCGUACGUACGUACG5’

Question 31.
How is the two stage process of protein synthesis advantageous?
Answer:
The split gene feature of eukaryotic genes is almost entirely absent in prokaryotes. Originally each exon may have coded for a single polypeptide chain with a specific function. Since exon arrangement and intron removal are flexible, the exon coding for these polypeptide subunits act as domains combining in various ways to form new genes. Single genes can produce different functional proteins by arranging their exons in several different ways through alternate splicing patterns, a mechanism known to play an important role in generating both protein and functional diversity in animals. Introns would have arosen before or after the evolution of eukaryotic gene.

If introns arose late how did they enter eukaryotic gene? Introns are mobile DNA sequences that can splice themselves out of, as well as into, specific ‘target sites’ acting like mobile transposon-like elements (that mediate transfer of genes between organisms – Horizontal Gene Transfer – HGT). HGT occurs between lineages of prokaryotic cells, or from prokaryotic to eukaryotic cells and between eukaryotic cells. HGT is now hypothesized to have played a major role in the evolution of life on Earth.

Question 32.
Why did Hershey and Chase use radioactively labelled phosphorous and sulphur only? Would they have got the same result if they use radiolabelled carbon and nitrogen?
Answer:
Generally proteins contain sulphur but not phosphorous and nucleic acid (DNA) contains , phosphorous but not sulphur. Hence Hershey – Chase used radioactive isotopes of sulphur ( 3 5 S) and phosphorus ( 32 P) to keep separate track of viral protein and nucleic acid in culture medium. The expected result cannot be achieved, if radioactive carbon and nitrogen is used, since these molecules are present in both DNA and proteins.

Question 33.
Explain the formation of a nucleosome.
Answer:
Komberg proposed a model for the nucleosome, in which 2 molecules of the four histone proteins H2A, H2B, H3 and H4 are organized to form a unit of eight molecules called histone octamere.

The negatively charged DNA is wrapped around the positively charged histone octamere to form a structure called nucleosome. A typical nucleosome contains 200 bp of DNA helix. The histone octameres are in close contact and DNA is coiled on the outside of nucleosome.

Question 34.
It is established that RNA is the first genetic material. Justify giving reasons.
Answer:
Three molecular biologists in the early 1980’s (Leslie Orgel, Francis Brick and Carl Woese) independently proposed the ‘RNA world’ as the first stage in the evolution of life, a stage when RNA catalysed all molecules necessary for survival and replication. The term ‘RNA world’ first used by Walter Gilbert in 1986, hypothesizes RNA as the first genetic on Earth. There is now enough evidence to suggest that essential life processes (such as metabolism, translation and splicing etc.,) evolved around RNA. RNA has the ability to act as both genetic material and catalyst. There are several biochemical reactions in living systems that are catalysed by RNA. This catalytic RNA is known as ribozyme. But, RNA being a catalyst was reactive and hence unstable.

This led to evolution of a more stable form of DNA, with certain chemical modifications. Since DNA is a double stranded molecule having complementary strand, it has resisted changes by evolving a process of repair. Some RNA molecules function as gene regulators by binding to DNA and affect gene expression. Some viruses use RNA as the genetic material. Andrew Fire and Craig Mellow (recipients of Nobel Prize in 2006) were of the opinion that RNA is an active ingredient in the chemistry of life.

Samacheer Kalvi 12th Bio Zoology Molecular Genetics Additional Questions and Answers

Question 1.
The term‘gene’was coined by ___________
Answer:
Wilhelm Johannsen

Question 2.
Whose experiment finally provided convincing evidence that DNA is the genetic material?
(a) Griffith experiment
(b) Avery, Macleod and McCarty’s experiment
(c) Hershey-Chase experiment
(d) Urey-Miller’s experiment
Answer:
(c) Hershey-Chase experiment

Question 3.
In Hershey – Chase experiment, the DNA of T 2 phase was made radioactive by using ___________
(a) 32 P
(b) 32 S
(c) 35 P
(d) 32 S
Answer:
(a) 32 P

Question 4.
A nucleoside is composed of ___________
(a) Sugar and Phosphate
(b) Nitrogen base and Phosphate
(c) Sugar and Nitrogen base
(d) Sugar, Phosphate and Nitrogenous base
Answer:
(c) Sugar and Nitrogen base

Question 5.
Identify the incorrect statement
(a) a base is a substance that accepts H+ ion
(b) Both DNA and RNA have four bases
(c) Purines have single carbon-nitrogen ring
(d) Thymine is unique for DNA
Answer:
(c) Purines have single carbon-nitrogen ring

Question 6.
Watson and Crick proposed their double helical DNA model based on the X-ray diffraction analysis o f ___________
(a) Erwin Chargaff
(b) Meselson and Stahl
(c) Wilkins and Franklin
(d) Griffith
Answer:
(c) Wilkins and Franklin

Question 7.
The term ‘RNA world’ was first used by ___________
Answer:
Walter Gilbert

Question 8.
The distance between two consecutive base pairs in DNA is ___________
(a) 0.34 nm
(b) 3.4 nm
(c) 0.034 nm
(d) 34 nm
Answer:
(a) 0.34 nm

Question 9.
If the length of E. coli DNA is 1.36 mm, the number of base pairs is ___________
(a) 0.36 × 10 6 m
(b) 4 × 10 6 m
(c) 0.34 × 10 -9 nm
(d) 4 × 10 -9 m
Answer:
(b) 4 × 10 6 m

Question 10.
Identify the proper sequence in the organisation of eukaryotic chromosome.
(a) Nucleosome – Solenoid – Chromatid
(b) Chromatid – Nucleosome – Solenoid
(c) Solenoid – chromatin – DNA
(d) Nucleosome – solenoid – genophore
Answer:
(a) Nucleosome – Solenoid – Chromatid

Question 11.
Assertion (A) : Genophore is noticed in prokaryotes.
Reason (R) : Bacteria possess circular DNA without chromatin organisation.
(a) Both A and R are correct
(b) A is correct R is incorrect
(c) R explains A
(d) A is incorrect R is correct
Answer:
(c) R explains A

Question 12.
Assertion (A): Heterochromatin is transcriptionally active.
Reason (R): Tightly packed chromatin which stains dark.
(a) Both A and R are correct
(b) A is correct R is incorrect
(c) R explains A
(d) A is incorrect R is correct
Answer:
(d) A is incorrect R is correct

Question 13.
Assertion (A) : Semi-conservative model was proposed by Hershey and Chase.
Reason (R) : The daughter DNA contains only new strands.
(a) Both A and R are incorrect
(b) A is correct R is incorrect
(c) R explains A
(d) A is incorrect R is correct
Answer:
(a) Both A and R are incorrect

Question 14.
Komberg enzyme is called as _____
Answer:
DNA polymerase I

Question 15.
Replication of DNA occurs at __________ phase of cell cycle.
(a) M
(b) S
(c) G1
(d) G2
Answer:
(b) S

Question 16.
Semi-conservative model of replication was proved by __________
(a) Hershey and Chase
(b) Griffith
(c) Meselson and Stahl
(d) Macleod and McCarty
Answer:
(c) Meselson and Stahl

Question 17.
How many types of DNA polymerases does an eukaryotic cell possess?
(a) two
(b) three
(c) four
(d) five
Answer:
(d) Five

Question 18.
Identify the incorrect statement
(a) Replication occurs at ori – site of DNA
(b) Deoxy nucleotide triphosphate acts as a substrate
(c) Unwinding of DNA strand is carried out by topoisomerase
(d) DNA polymerase catalyses the polymerization at 3-OH
Answer:
(c) Unwinding of DNA strand is carried out by topoisomerase

Question 19.
The discontinuously synthesized fragments of lagging strand are called ________
Answer:
Okazaki fragments

Question 20.
Retroviruses possess ________ as genetic material.
Answer:
RNA

Question 21.
Which is NOT a part of transcription unit?
(a) Promoter
(b) Operator
(c) Structural gene
(d) Terminator
Answer:
(b) Operator

Question 22.
Goldberg – Hogness box of eukaryotes is equivalent to ________ of prokaryotes.
Answer:
Pribnow box

Question 23.
Okazaki fragments are joined by the enzyme ________ during DNA replication.
Answer:
DNA ligase

Question 24.

Answer:
(a) A – iv, B – i, C – ii, D – iii

Question 25.
The RNA polymerase of prokaryotes binds with factor to initiate polymerization.
(a) rho
(b) theta
(c) sigma
(d) psi
Answer:
(c) sigma

Question 26.

(a) Capping
(b) Tailing
(c) Splicing
(d) Transcribing
Answer:
(c) Splicing

Question 27.
Which of the following feature is absent in prokaryotes?
(a) Prokaryotes possess three major types of RNAs
(b) Structural genes are polycistronic
(c) Initiation process of transcription requires ‘P’ factor
(d) Split gene feature
Answer:
(d) Split gene feature

Question 28.
Which of the following sequence has completely translated?
(i) AGA, UUU, UGU, AGU, UAG
(ii) AUG, UUU, AGA, UAC, UAA
(iii) AAA, UUU, UUG, UGU, UGA
(iv) AUG,AAU,AAC,UAU,UAG
(a) i and ii
(b) ii only
(c) i and iii
(d) ii and iv
Answer:
(d) ii and iv

Question 29.
Capping of mRNA occurs using __________
(a) Poly A residues
(b) Methyl guanosine triphosphate
(c) Deoxy ribonucleotide triphosphate
(d) Ribonucleotide triphosphate
Answer:
(b) Methyl guanosine triphosphate

Question 30.
One of the aspect is not a feature of genetic code?
(a) Specific
(b) Degenerate
(c) Universal
(d) Ambiguous
Answer:
(d) Ambiguous

Question 31.
Which of the triplet codon is not a code of proline?
(i) CCU
(ii) CAU
(iii) CCG
(iv) CAA
(a) i only
(b) ii and iv
(c) iii only
(d) all the above
Answer:
(b) ii and iv

Question 32.
Coding sequences found in split genes are called.
(a) Operons
(b) Introns
(c) Exons
(d) Cistron
Answer:
(c) Exons

Question 33.
Which of the following mRNA yields 6 aminoacids after translation?
(i) UCU UAU AGU CGA UGC AGU UGA AAA UUU
(ii) UGA AGA UAG GAG CAU CCC UAC UAU GAU
(iii) GUC UGC UGG GCU GAU UAA AGG AGC AUU
(iv) AUG UAC CAU UGC UGA UGC AGG AGC CCG
Answer:
(i) UCU UAU AGU CGA UGC AGU UGA AAA UUU

Question 34.
The transcription termination factor associated with RNA polymerase in prokaryotes is
(a) θ
(b) σ
(c) ρ
(d) ∑
Answer:
(c) ρ

Question 35.
In a DNA double strand, if guanine is of 30%, what will be the percentage of thymine?
(a) 100%
(b) 20%
(c) 10%
(d) 70%
Answer:
(b) 20%

Question 36.
Identify the triplet pairs that code for Tyrosine
(a) UUU,UUC
(b) UAU, UAU
(c) UGC, UGU
(d) CAU, CAC
Answer:
(b) UAU, UAU

Question 37.

Answer:
A – ii B – i C – iv D – iii

Question 38.
AUG code is for __________
(a) Arginine
(b) Tyrosine
(c) Tryptophan
(d) Methionine
Answer:
(d) Methionine

Question 39.
The sequence of bases in coding strand of DNA is G A G T T A G C A G G C, then the sequence of codons in primary transcript is
(a) C U C A U A C G C C C G
(b) C U C A A U C G U C C G
(c) U C A G A U C U G C G C
(d) U U C A A U C G U G C G
Answer:
(b) C U C A A U C G U C C G

Question 40.
The promoter region of eukaryote is __________
(a) TATAA
(b) AUGUT
(c) UUUGA
(d) AAAAU
Answer:
(a) TATAA

Question 41.
Match the following:
(A) AUG – (i) Tyrosine
(B) UGA – (ii) Glycine
(C) UUU – (iii) Methionine
(D) GGG – (iv) Phenylalanine
(a) A – iii B – i C – iv D – ii
(b) A – iii B – ii C – i D – iv
(c) A – iv B – i C – iii D – ii
(d) A – ii B – i C – iv D – iii
Answer:
(d) A – ii B – i C – iv D – iii

Question 42.
__________ number of codons, codes for cystine.
Answer:
Two

Question 43.
In sickle cell anaemia, the __________ codon of β – globin gene is modified.
(a) Eighth
(b) Seventh
(c) Sixth
(d) Nineth
Answer:
(c) Sixth

Question 44.
Pick out the incorrect statement.
(a) tRNA acts as a adapter molecule
(b) Stop codons donot have tRNA’s
(c) Addition of aminoacid leads to hydrolysis of tRNA
(d) tRNA has four major loops
Answer:
(c) Addition of aminoacid leads to hydrolysis of tRNA

Question 45.
Which of the following antibiotic inhibits the interaction between tRNA and mRNA?
(a) Neomycin
(b) Streptomycin
(c) Tetracycline
(d) Chloramphenicol
Answer:
(a) Neomycin

Question 47.
The cluster of genes with related function is called _________
(a) Cistron
(b) Operon
(c) Muton
(d) Recon
Answer:
(b) Operon

Question 48.
Repressor protein of Lac operon binds to __________ of operon.
(a) Promoter region
(b) Operator region
(c) terminator region
(d) inducer region
Answer:
(b) Operator region

Question 49.
Lac Z gene codes for __________
(a) Permease
(b) transacetylase
(c) β -galactosidase
(d) Aminoacyl transferase
Answer:
(c) β -galactosidase

Question 50.
Lac operon model was proposed by __________
Answer:
Jacob and Monod

Question 51.
Approximate count of base pair in human genome is __________
Answer:
3 × 10 9 bp

Question 52.
Automated DNA sequences are developed by.
Answer:
Frederick Sanger

Question 53.
Which of the chromosome has higher gene density?
(a) Chromosome 20
(b) Chromosome 19
(c) Chromosome 13
(d) Chromosome Y
Answer:
(b) Chromosome 19

Question 54.
Number of genes located in chromosome Y is __________
(a) 2968
(b) 213
(c) 2869
(d) 231
Answer:
(d) 231

Question 55.
How many structural genes are located in lac operon of E.Coli?
(a) 4
(b) 3
(c) 2
(d) 1
Answer:
(b) 3

Question 56.
DNA finger printing technique was developed by
(a) Jacob and Monod
(b) Alec Jeffreys
(c) Frederick Sanger
Answer:
(b) Alec Jeffreys

Question 57.
In DNA fingerprinting, separation of DNA fragments is done by __________
(a) Centrifugation
(b) Electrophoresis
(c) X-ray diffraction
(d) denaturation
Answer:
(b) Electrophoresis

Question 58.
SNP stands for
(a) Single nucleotide Polymorphism
(b) Single Nucleoside Polypeptide
(c) Single nucleotide Polymorphism
(d) Single nucleotide polymer
Answer:
(a) Single nucleotide Polymorphism

Question 59.
Specific sequences of mRNA that are not translated are __________
Answer:
UnTranslated Regions (UTR)

Question 60.
Non-coding or intervening DNA sequence is called __________

Question 61.
_______ Intron is the monomer of DNA.
Answer:
Nucleotide

Question 62.
Which one of the following is wrongly matched?
(a) Transcription – Copying information from DNA to RNA
(b) Translation – Decoding information from mRNA to protein
(c) Replication – Making of DNA copies
(d) Splicing – Joining of exons with introns
Answer:
(d) Splicing – Joining of exons with introns

Question 1.
Who proposed One gene – One enzyme hypothesis? Define it.
Answer:
George Beadle and Edward Tatum proposed One gene – One enzyme hypothesis which states that one gene controls the production of one enzyme.

Question 2.
Differentiate nucleoside from nucleotide.
Answer:

  1. Nucleoside: Nucleoside subunit is composed of nitrogenous bases linked to a pentose sugar molecule.
  2. Nucleotide: Nucleotide subunit is composed of nitrogenous bases, a pentose sugar and a phosphate group.

Question 3.
State the key differences between DNA and RNA.
Answer:
DNA:

  1. DNA is made of deoxyribose sugar.
  2. Nitrogenous bases of DNA are Adenine, Guanine, Cytosine and Thymine.
  1. RNA is made of ribose sugar.
  2. Nitrogenous bases of RNA are Adenine, Guanine, Cytosine and Uracil.

Question 4.
Point out the nitrogenous bases of RNA.
Answer:
Adenine, Guanine, Cytosine and Uracil.

Question 5.
What makes the DNA and RNA as acidic molecules?
Answer:
The phosphate functional group (PO4) gives DNA and RNA the property of an acid at physiological pH, hence the name nucleic acid.

Question 6.
Which type of bond is formed

  1. between a purine and pyrimidine base?
  2. between the pentose sugar and adjacent nucleotide?
  1. Purine and pyrimidine bases are linked by hydrogen bonds.
  2. Pentose sugar is linked to adjacent nucleotide by phosphodiester bonds.

Question 7.
DNA acts as genetic material for majority of living organisms and not the RNA. Give reasons to support the statement.
Answer:

  1. RNA was reactive and hence highly unstable.
  2. Some RNA molecules acts as gene regulators by binding to DNA and affect gene expression.
  3. Uracil of RNA is less stable than thymine of DNA.

Question 8.
Name any two viruses whose genetic material is RNA.
Answer:

Question 9.
What are the properties that a molecule must possess to act as genetic material?
Answer:

  1. Self replication
  2. Information storage
  3. Stability
  4. Variation through mutation

Question 10.
How many base pairs are present in one complete turn of DNA helix? What is the distance between two consecutive base pairs?
Answer:
There are ten base pairs in each turn with a distance of 0.34 x 109m between two adjacent base pairs.

Question 11.
What is a genophore?
Answer:
In prokaryotes such as E. coli though they do not have defined nucleus, the DNA is not scattered throughout the cell. DNA (being negatively charged) is held with some proteins (that have positive charges) in a region called the nucleoid. The DNA as a nucleoid is organized into large loops held by protein. DNA of prokaryotes is almost circular and lacks chromatin organization, hence termed genophore.

Question 12.
Whqt is nucleosome? How many base pairs are there in a typical nucleosome?
Answer:
The negatively charged DNA is wrapped around the positively charged histone octamere to form a structure called nucleosome. A typical nucleosome contains 200 bp of DNA helix.

Question 13.
Expand and define NHC
Answer:

  1. NHC : Non-histone Chromosomal protein.
  2. In eukaryotes, apart from histone proteins, additional set of proteins are required for packing of chromatin at higher level and are referred as non – histone chromosomal proteins.

Question 14.
Differentiate between Heterochromatin and Euchromatin.
Answer:
Heterochromatin:

  1. Region of nucleus where the chromatin are loosely packed and stains light are called Heterochromatin.
  2. Transcriptionally inactive.
  1. Region of nucleus where the chromatin are tightly packed and stains dark are called Euchromatin.
  2. Transcriptionally active.

Question 15.
Which is the widely accepted model of DNA replication? Who has proved it?
Answer:
Semi-conservative replication model. It was proved by Meselson and Stahl in 1958.

Question 16.
Name the chemical substance which is called by the name

  1. DNA polymerase I is also known as Komberg enzyme.
  2. Polynucleotide phosphorylase is also known as Ochoa’s enzyme.

Question 17.
Name the various types of prokaryotic DNA polymerase. State their role in replication process.
Answer:

  1. DNA Polymerase I Involver in DNA repair mechanism
  2. DNA Polymerase II Involver in DNA repair mechanism
  3. DNA Polymerase III Involver in DNA replication

Question 18.
What is the function of Deoxy nucleotide triphosphate in replication?
Answer:
Deoxy nucleotide triphosphate acts as substrate and also provides energy for polymerization reaction.

Question 19.
Given below are some events of eukaryotic replication. Name the enzymes involved in the process.

  1. Unwinding of DNA
  2. Joining of Okazaki fragments
  3. Addition of nucleotides to new strand
  4. Correcting the repair

Question 20.
Differentiate leading strand from lagging strand
Answer:
Leading strand:

Question 21.
What are Okazaki fragments?
Answer:
The discontinuously synthesized fragments of the lagging strand are called the Okazaki fragments are joined by the enzyme DNA ligase.

Question 22.
What is a replication fork?
Answer:
At the point of origin of replication, the helicases and topoisomerases (DNA gyrase) unwind and pull apart the strands, forming a Y-Shaped structure called the replication fork. There are two replication forks at each origin.

Question 23.
Apart from DNA polymerase, name any other four enzymes which were involved in DNA replication of eukaryotic cell.
Answer:
DNA ligase, Topoisomerase (DNA gyrase), Helicase and Nuclease.

Question 24.
Who proposed the central dogma? Write its concept.
Answer:
Francis Crick proposed the Central dogma in molecular biology which states that genetic information flows as follows:

Question 25.
Define transcription and name the enzyme involved in this process.
Answer:
The process of copying genetic information from one strand of DNA into RNA is termed transcription. This process takes place in presence of DNA dependent RNA polymerase.

Question 26.
What is TATA box? State its function.
Answer:
In eukaryotes, the promoter has AT rich regions called TATA box or Goldberg-Hogness box. It acts as a binding site for RNA polymerase.

Question 27.
Structural gene of eukaryotes differ from prokaryotes. How?
Answer:
In eukaryotes, the structural gene is monocistronic coding for only one protein whereas in prokaryotes the structural gene is polycistronic coding for many proteins.

Question 28.
What are the two major components of prokaryotic RNA polymerase? How do they act?
Answer:
Bacterial (prokaryotic) RNA polymerase consists of two major components, the core enzyme and the sigma subunit. The core enzyme (β1, β, and α) is responsible for RNA synthesis ” whereas a sigma subunit is responsible for recognition of the promoter.

Question 29.
Distinguish between exons and introns.
Answer:

  1. Exons: Expressed sequences (Coding sequences) of an eukaryotic gene
  2. Introns: Interveining sequences (non-coding sequences) of an eukaryotic gene

Question 30.
Define splicing.
Answer:
The process of removing introns from hnRNA is called splicing.

Question 31.
What is capping and tailing?
Answer:
In capping an unusual nucleotide, methyl guanosine triphosphate is added at the 5’ end of hnRNA, whereas adenylate residues (200-300) (Poly A) are added at the 3’ end in tailing.

Question 32.
If a double stranded DNA has 20% of cytosine, calculate the percentage of adenine in DNA.
Answer:
Cytdsine = 20, hence Guanine = 20
As per ChargafFs rule (A+T) = (G+C) =100
Percent of Thymine + Adenine = 20 + 20 = 100
(T + A) = (20 + 20) =100
(T + A)=100-(20 + 20)
T +A = 100 – 40
T + A = 60
Therefore the percent of Adenine will be 60/2 = 30%.

Question 33.
Mention the dual functions of AUG.
Answer:
AUG has dual functions. It acts as a initiator codon and also codes for the amino acid methionine.

Question 34.
How many codons are involved in termination of translation. Name them.
Answer:
Three codons terminate translation process. They are UAA, UAG and UGA.

Question 35.
Degeneracy of codon – comment.
Answer:
A degenerate code means that more than one triplet codon could code for a specific amino acid. For example, codons GUU, GUC, GUA and GUG code for valine.

Question 36.
Point out the exceptional categories to universality of genetic code.
Answer:
Exceptions to universal nature of genetic code is noticed in prokaryotic mitochondrial and chloroplast genomes.

Question 37.
What are non-sense codons?
Answer:
UGA, UAA and UAG are the non-sense codons, which terminates translation.

Question 38.
Name the triplet codons that code for

Question 39.
Why hnRNA has to undergo splicing?
Answer:
Since hnRNA contains both coding sequences (exons) and non-coding sequences (introns) it has to undergo splicing to remove introns.

Question 40.
State the role of following codons in translation process

Question 41.
Given below is mRNA sequence. Mention the aminoacids sequence that is formed after its translation.
Answer:
3’AUGAAAGAUGGGUAA5’
Methionine – Lysine – Aspartic acid – Glycine

Question 42.
Name the four codons that codes valine.
Answer:
GUU, GUC, GUA and GUG.

Question 43.
The base sequence in one of the DNA strand is TAGC ATGAT. Mention the base sequence in its complementary strand.
Answer:
The complementary strand has ATCGTACTA.

Question 44.
Why t-RNA is called as adapter molecule?
Answer:
The transfer RNA (tRNA) molecule of a cell acts as a vehicle that picks up the amino acids scattered through the cytoplasm and also reads specific codes of mRNA molecules. Hence it is called as adapter molecule.

Question 45.
What do you mean by charging of tRNA? Name the enzyme involved in this process.
Answer:
The process of addition of amino acid to tRNA is known as aminoacylation or charging and the resultant product is called aminoacyl- tRNA (charged tRNA). Aminoacylation is catalyzed by an enzyme aminoacyl – tRNA synthetase.

Question 46.
What are UTR’s?
Answer:
mRNA also have some additional sequences that are not translated and are referred to as Untranslated Regions (UTR). UTRs are present at both 5’ end (before start codon) and at 3’ end (after stop codon).

Question 47.
What is S – D sequence?
Answer:
The 5’ end of the mRNA of prokaryotes has a special sequence which precedes the initial AUG start codon of mRNA. This ribosome binding site is called the Shine – Dalgamo sequence or S-D sequence. This sequences base-pairs with a region of the 16Sr RNA of the small ribosomal subunit facilitating initiation.

Question 48.
Define translation unit.
Answer:
A translation unit in mRNA is the sequence of RNA that is flanked by the start codon on 5’ end and stop codon on 3’ end and codes of polypeptide.

Question 49.
Mention the inhibitory role of tetracycline and streptomycin in bacterial translation.
Answer:
Tetracycline inhibits binding between aminoacyl tRNA and mRNA.Streptomycin inhibits initiation of translation and causes misreading.

Question 50.
At what stage, does the gene expression is regulated?
Answer:
Gene expression can be controlled or regulated at transcriptional or translational levels.

Question 51.
What is a operon? Give example.
Answer:
The cluster of genes with related functions is called operon.
E.g: lac operon in E.coli.

Question 52.
Considering the lac operon of E.coli, name the products of the following genes.

  1. i gene – repressor protein
  2. lac Z gene – fS-galactosidase
  3. Lac Y gene – Permease
  4. lac a gene – transacetylase

Question 54.
Name the human chromosome that has

  1. Chromosome 1 has maximum number of genes (2968 genes)
  2. Chromosome Y has least genes (231 genes)

Question 55.
What are SNPs? Mention its uses.
Answer:
SNPs : Single nucleotide polymorphism. It helps to find chromosomal locations for disease associated sequences and tracing human history.

Question 56.
Mention any four areas where DNA fingerprinting can be used.
Answer:

  1. Forensic analysis
  2. Pedigree analysis
  3. Conservation of wild life
  4. Anthropological studies

Question 57.
Classify nucleic acid based on sugar molecules.
Answer:
There are two types of nucleic acids depending on the type of pentose sugar. Those containing deoxyribose sugar are called Deoxyribo Nucleic Acid (DNA) and those with ribose sugar are known as Ribonucleic Acid (RNA). The only difference between these two sugars is that there is one oxygen atom less in deoxyribose.

Question 58.
Both purines and pyrimidines are nitrogen bases yet they differ. How?
Answer:
Both purines and pyrimidines are nitrogen bases. The purine bases Adenine and Guanine have double carbon – nitrogen ring, whereas cytosine and thymine bases have single carbon nitrogen ring.

Question 59.
How 5’ of DNA differ from its 3’?
Answer:
The 5’ of DNA refers to the carbon in the sugar to which phosphate (P04V) functional group is attached. The 3’ of DNA refers to the carbon in the sugar to which a hydroxyl (OH) group is attached.

Question 60.
State Chargaff’s rule.
Answer:
According to Erwin Chargaff,

  1. Adenine pairs with Thymine with two hydrogen bonds.
  2. Guanine pairs with Cytosine with three hydrogen bonds.

Question 61.
Chemically DNA is more stable than RNA – Justify.
Answer:
In DNA, the two strands being complementary, if separated (denatured) by heating can come together (renaturation) when appropriate condition is provided. Further 2 OH group present at every nucleotide in RNA is a reactive group that makes RNA liable and easily degradable. RNA is also known to be catalytic and reactive. Hence, DNA is chemically more stable and chemically less reactive when compared to RNA. Presence of thymine instead of uracil in DNA confers additional stability to DNA.

Question 62.
Write in simple about semi-conservative mode of DNA replication.
Answer:
Semi-conservative replication was proposed by Watson and Crick in 1953. This mechanism of replication is based on the DNA model. They suggested that the two polynucleotide strands of DNA molecule unwind and start separating at one end. During this process, covalent hydrogen bonds are broken. The separated single strand then acts as template for the synthesis of a new strand. Subsequently, each daughter double helix carries one polynucleotide strand from the parent molecule that acts as a template and the other strand is newly synthesised and complementary to the parent strand.

Question 63.
Draw a simplified diagram of nucleosome and label it.
Answer:

Question 64.
What is a primer?
Answer:
A primer is a short stretch of RNA. It initiates the formation of new strand. The primer produces 3’-OH end on the sequence of ribonucleotides, to which deoxyribonucleotides are added to form a new strand.

Question 65.
Both strands of DNA are not copied during transcription. Give reason.
Answer:
Both the strands of DNA are not copied during transcription for two reasons.

  1. If both the strands act as a template, they would code for RNA with different sequences. This in turn would code for proteins with different amino acid sequences. This would result in one segment of DNA coding for two different proteins, hence complicate the genetic information transfer machinery.
  2. If two RNA molecules were produced simultaneously, double stranded RNA complementary to each other would be formed. This would prevent RNA from being translated into proteins.

Question 66.
What do you mean by a template strand and coding strand?
Answer:
DNA dependent RNA polymerase catalyses the polymerization in only one direction, the strand that has the polarity 3’→ 5’ acts as a template, and is called the template strand. The other strand which has the polarity 5’→ 3’ has a sequence same as RNA (except thymine instead of uracil) and is displaced during transcription. This strand is called coding strand.

Question 67.
Name the factors that are responsible for initiation and termination of transcription in prokaryotes.
Answer:

  1. Sigma factor is responsible for initiation of transcription.
  2. Rho factor is responsible for termination of transcription.

Question 68.
Name the major RNA types of prokaryotes and mention their role.
Answer:
In prokaryotes, there are three major types of RNAs: mRNA, tRNA, and rRNA. All three RNAs are needed to synthesize a protein in a cell. The mRNA provides the template, tRNA brings amino acids and reads the genetic code, and rRNAs play structural and catalytic role
during translation.

Question 69.
Define genetic code.
Answer:
The order of base pairs along DNA molecule controls the kind and order of amino acids found in the proteins of an organism. This specific order of base pairs is called genetic code.

Question 70.
Explain Wobble hypothesis.
Answer:
Wobble Hypothesis is proposed by Crick (1966) which states that tRNA anticodon has the ability to wobble at its 5’ end by pairing with even non-complementary base of mRNA codon.’ According to this hypothesis, in codon-anticodon pairing the third base may not be complementary.

The third base of the codon is called wobble base and this position is called wobble position. The actual base pairing occurs at first two positions only. The importance of Wobbling hypothesis is that it reduces the number of tRNAs required for polypeptide synthesis and it overcomes the effect of code degeneracy.

Question 71.
Explain the nature of eukaryotic ribosome.
Answer:
The ribosomes of eukaryotes (80 S) are larger, consisting of 60 S and 40 S sub units. ‘S’ denotes the sedimentation efficient which is expressed as Svedberg unit (S). The larger subunit in eukaryotes consist of a 23 S RNA and 5Sr RNA molecule and 31 ribosomal proteins. The smaller eukaryotic subunit consist of 18Sr RNA component and about 33 proteins.

Question 72.
Expand and define ORF.
Answer:
Any sequence of DNA or RNA, beginning with a start codon and which can be translated into a protein is known as an Open Reading Frame (ORF).

Question 73.
What are the components of initiation complex of prokaryotic translation?
Answer:
Initiation of translation in E. coli begins with the formation of an initiation complex, consisting of the 30S subunits of the ribosome, a messenger RNA and the charged N-formyl methionine tRNA (f met -1 RNA f met ), three proteinaceous initiation factors (IF 1, IF2, IF3), GTP (Guaniner Tri Phosphate) and Mg 2+ .

Question 74.
Explain the components of operon.
Answer:
Structure of the operon: Each operon is a unit of gene expression and regulation and consists
of one or more structural genes and an adjacent operator gene that controls transcriptional
activity of the structural gene.

  1. The structural gene codes for proteins, rRNA and tRNA required by the cell.
  2. Promoters are the signal sequences in DNA that initiate RNA synthesis. RNA polymerase I binds to the promoter prior to the initiation of transcription.
  3. The operators are present between the promoters and structural genes. The repressor protein binds to the operator region of the operon.

Question 75.
Describe Hershey and Chase experiment. What is concluded by their experiment?
Answer:
Alfred Hershey and Martha Chase (1952) conducted experiments on bacteriophages that infect bacteria. Phage T2 is a virus that infects the bacterium Escherichia coli. When phages (virus) are added to bacteria, they adsorb to the outer surface, some material enters the bacterium, and then later each bacterium lyses to release a large number of progeny phage. Hershey and Chase wanted to observe whether it was DNA or protein that entered the bacteria. All nucleic acids contain phosphorus, and contain sulphur (in the amino acid cysteine and methionine). Hershey and Chase designed an experiment using radioactive isotopes of Sulphur ( 35 S) and phosphorus ( 32 P) to keep separate track of the viral protein and nucleic acids during the infection process.

The phages were allowed to infect bacteria in culture medium which containing the radioactive isotopes 35 S or 32 P. The bacteriophage that grew in the presence of 35 S had labelled proteins and bacteriophages grown in the presence of 32 P had labelled DNA. The differential labelling thus enabled them to identity DNA and proteins of the phage. Hershey and Chase mixed the labelled phages with unlabeled E. coli and allowed bacteriophages to attack and inject their genetic material. Soon after infection (before lysis of bacteria), the bacterial cells were gently agitated in a blender to loosen the adhering phase particles.

It was observed that only 32 P was found associated with bacterial cells and 35 S was in the surrounding medium and not in the bacterial cells. When phage progeny was studied for radioactivity, it was found that it carried only 32 P and not 35 S. These results clearly indicate that only DNA and not protein coat entered the bacterial cells. Hershey and Chase thus conclusively proved that it was DNA, not protein, which carries the hereditary information from virus to bacteria.

Question 76.
Explain the properties of DNA that makes it an ideal genetic material.
Answer:
1. Self Replication: It should be able to replicate. According to the rule of base pairing and complementarity, both nucleic acids (DNA and RNA) have the ability to direct duplications. Proteins fail to fulfill this criteria.

2. Stability: It should he stable structurally and chemically. The genetic material should be stable enough not to change with different stages of life cycle, age or with change in physiology of the organism. Stability as one of property of genetic material was clearly evident in Griffith’s transforming principle. Heat which killed the bacteria did not destroy some of the properties of genetic material. In DNA the two strands being complementary.

if separated (denatured) by heating can come together (renaturation) when appropriate condition is provided. Further 2’ OH group present at every nucleotide in RNA is a reactive group that makes RNA liable and easily degradable. RNA is also known to be catalytic and reactive. Hence, DNA is chemically more stable and chemically less reactive when compared to RNA. Presence of thymine instead of uracil in DNA confers additional stability to DNA.

3. Information storage: It should be able to express itself in the form of ‘Mendelian characters’. RNA can directly code for protein synthesis and can easily express the characters. DNA, however depends on RNA for synthesis of proteins. Both DNA and RNA can act as a genetic material, but DNA being more stable stores the genetic information and RNA transfers the genetic information.

4. Variation through mutation: It should be able to mutate. Both DNA and RNA are able to mutate. RNA being unstable, mutates at a faster rate. Thus viruses having RNA genome with shorter life span can mutate and evolve faster. The above discussion indicates that both RNA and DNA can function as a genetic material. DNA is more stable, and is preferred for storage of genetic information.

Question 77.
How the DNA is packed in an eukaryotic cell? ft
Answer:
In eukaryotes, organization is more complex. Chromatin is formed by a series of repeating units called nucleosomes. Komberg proposed a model for the nucleosome, in which 2 molecules of the four histone proteins H2A, H2B, H3 and H4 are organized to form a unit of eight molecules called histone octamere. The negatively charged DNA is wrapped around the positively charged histone octamere to form a structure called nucleosome. A typical nucleosome contains 200 bp of DNA helix. The histone octameres are in close contact and DNA is coiled on the outside of nucleosome. Neighbouring nucleosomes are connected by linker DNA (HI) that is exposed to enzymes.

The DNA makes two complete turns around the histone octameres and the two turns are sealed off by an HI molecule. Chromatin lacking HI has a beads-on-a-string appearance in which DNA inters and leaves the nucleosomes at random places. HI of one nucleosome can interact with 33l of the neighbouring nucleosomes resulting in the further folding of the fibre.

The chrof&atin fiber in interphase nuclei and mitotic chromosomes have a diameter that vary between 200-300 nm and represents inactive chromatin. 30 nm fibre arises from the folding of nucfeosbme, chains into a solenoid structure having six nucleosomes per turn. This structure is stabilized by interaction between different HI molecules. DNA is a solenoid and packed about,%)_folds. The hierarchical nature of chromosome structure is illustrated.

Additional set of pteins are required for packing of chromatin at higher level and are referred to as non-histone chromosomal proteins (NHC). In*,a typical nucleus, some regions of chromatin are Ibosely packed (lightly stained) and are referred to as euchromatin. The chromatin that is,-tightly packed (stained darkly) is called heterochromatin. Euchromatin is transcriptionally active and heterochromatin is transcriptionally inactive.

Question 78.
Meselson and Stahl’s experiment proved the semi-coflBptervation mode of DNA replication. Explain.
Answer:
The mode of DNA replication was determined in 1958 by Meselson and Stahl. They designed an experiment to distinguish between semi-conservative, conservative and dispersive replications. In their experiment, they grew two cultures of E.coli for many generations in separate media. The ‘heavy’ culture was grown in a medium in which the nitrogen source (NH4CI) contained the heavy isotope 15 N and the ‘ light’ culture was grown in a medium in which the nitrogen source contained light isotope 14 H for many generations. At the end of growth, they observed that the bacterial DNA in the heavy culture contained only 15 N and in the light culture only 14 N. The heavy DNA could be distinguished from light DNA ( 15 N from 14 N) with a technique called Cesium Chloride (CsCl) density gradient centrifugation. In this process, heavy and light DNA extracted from cells in thtytwo cultures settled into two distinct and separate bands (hybrid DNA).

The heavy culture ( 15 N) was then transferred into a medium that had only NH4CI, and took samples at various definite time intervals (20 minutes duration). After the first replication, they extracted DNA and subjected it to density gradient centrifugation. The DNA settled into a band that was intermediate in position between the previously determined heavy and light bands. After the second replication (40 minutes duration), they again extracted DNA samples,and this time found the DNA settling into two bands, one at the light band position and one at intermediate position. These results confirm Watson and Crick’s semi – conservative replication hypothesis.

Question 79.
Give a detailed account of a transcription unit.
Answer:
A transcriptional unit in DNA is defined by three regions, a promoter, the structural gene and a terminator. The promoter is located towards the 5 ’ end. It is a DNA sequence that provides binding site for RNA polymerase. The presence of promoter in a transcription unit, defines the template and coding strands. The terminator region located towards the 3’ end of the coding strand contains a DNA sequence that causes the RNA polymerase to stop transcribing. In eukaryotes the promoter has AT rich regions called TATA box (Goldberg- Hogness box) ‘ and in prokaryotes this region is called Pribnow box.

Besides promoter, eukaryotes also require an enhancer. The two strands of the DNA in the structural gene of a transcription unit have opposite polarity. DNA dependent RNA polymerase catalyses the polymerization in only one direction, the strand that has the polarity 3’→5’ acts as a template, and is called the template strand. The other strand which has the polarity 5’→ 3’ has a sequence same as RNA (except thymine instead of uracil) and is displaced during transcription. This strand is called coding strand

The structural gene may be monocistronic (eukaryotes) or polycistronic (prokaryotes). In eukaryotes, each mRNA carries only a single gene and encodes information for only a single protein and is called monocistronic mRNA. In prokaryotes, clusters of related genes, known as operon, often found next to each other on the chromosome are transcribed together to give a single mRNA and hence are polycistronic.

Question 80.
Explain the transcription process in prokaryotes with needed diagram.
Answer:
In prokaryotes, there are three major types of RNAs:
mRNA, tRNA, and rRNA. All three RNAs are needed to synthesize a protein in a cell. The mRNA provides the template, tRNA brings amino acids and reads the genetic code, and rRNAs play structural and catalytic role during translation. There is a single DNA-dependent RNA polymerase that catalyses transcription of all types of RNA. It binds to the promoter and initiates transcription (Initiation).

The polymerases binding sites are called promoters. It uses nucleoside triphosphate as substrate and polymerases in a template depended fashion following the rule of complementarity. After the initiation of transcription, the polymerase continues to elongate the RNA, adding one nucleotide after another to the growing RNA chain. Only a short stretch of RNA remains bound to the enzyme, when the polymerase reaches a terminator at the end of a gene, the, nascent RNA falls off, so also the RNA polymerase. The RNA polymerase is only capable of catalyzing the process of elongation. The RNA polymerase associates transiently with initiation factor sigma (a) and termination factor rho (p) to initiate and terminate the transcription, respectively.

Association of RNA with these factors instructs the RNA polymerase either to initiate or terminate the process of transcription. In bacteria, since the mRNA does not require any processing to become active and also since transcription and translation take place simultaneously in the same compartment sincethere is no separation of cytosol and nucleus in bacteria), many times the translation can begin much before the mRNA is fully transcribed. This is because the genetic material is not separated from other cell organelles by a nuclear membrane consequently transcription and translation can be coupled in bacteria.

Question 81.
Write the salient features of genetic code.
Answer:
The salient features of genetic code are as follows:

  1. The genetic codon is a triplet code and 61 codons code for amino acids and 3 codons do not code for any amino acid and function as stop codon (Termination).
  2. The genetic code is universal. It means that all known living systems use nucleic acids and the same three base codons (triplet codon) direct the synthesis of protein from amino acids. For example, the mRNA (UUU) codon codes for phenylalanine in all cells of all organisms. Some exceptions are reported in prokaryotic, mitochondrial and chloroplast genomes. However similarities are more common than differences.
  3. A non-overlapping codon means that the same letter is not used for two different codons. For instance, the nucleotide sequence GUTJ and GUC represents only two codons.
  4. It is comma less, which means that the message would be read directly from one end to the other i.e., no punctuation are needed between two codes.
  5. A degenerate code means that more than one triplet codon could code for a specific amino acid. For example, codons GUU, GUC, GUA and GUG code for valine.
  6. Non-ambiguous code means that one codon will code for one amino acid.
  7. The code is always read in a fixed direction i.e. from 5’→3’ direction called polarity.
  8. AUG has dual functions. It acts as a initiator codon and also codes for the amino acid methionine.
  9. UAA, UAG (tyrosine) and UGA (tryptophan) codons are designated as termination (stop) codons and also are known as “non-sense” codons.

Question 82.
Mutations on genetic code affects the phenotype. Describe with example.
Answer:
The simplest type of mutation at the molecular level is a change in nucleotide that substitutes one base for another. Such changes are known as base substitutions which may occur spontaneously or due to the action of mutagens. A well studied example is sickle cell anaemia in humans which results from a point mutation of an allele of β-haemoglobin gene (βHb).

A haemoglobin molecule consists of four polypeptide chains of two types, two a chains and two P-chains. Each chain has a heme group on its surface. The heme groups are involved in the binding of oxygen. The jruman blood disease, sickle cell anaemia is due to abnormal haemoglobin. This abnormality in haemoglobin is due to a single base substitution at the sixth codon of the beta globingene from GAG to GTG in p -chain of haemoglobin.

It results in a change of amino acid glufeniic acid to valine at the 6th position of the p -chain. This is the classical example of point mutation that results in the change of amino acids residue glutamic acid to valine. The mutant haemoglobin undergoes polymerisation under oxygen tension causing the change in the shape of the RBC from biconcave to a sickle shaped structure.

Question 83.
Explain the mechanism of AteArperon of the E-coli.
Answer:
The Lac (Lactose) operon: The metabolism of lactose in E.coli requires three enzymes – permease, P-galactosidase (P-gat) and transacetylase. The enzyme permease is needed for entry of lactose into the cell, Pjgglactosidase brings about hydrolysis of lactose to glucose and galactose, while transacety gtransfers acetyl group from acetyl Co A to P-galactosidase. The lac operon consists of one-regulator gene (T gene refers to inhibitor) promoter sites (p), and operator site (o). Besides these, it has three structural genes namely lac z, y and lac a. The lac ‘z’ gene codes for P-gaiaqtttsidase, lac ‘y’ gene codes for permease and ‘a’ gene codes for transacetylase.

Jacob and Monod proposed the classical model of Lac operon to explain gene expression and regulation in E.coli. In lac a polycistronic structural gene is regulated by a common promoter and regulatory genfc When the cell is using its normal energy source as glucose, the ‘i’ gene transcribes a repressor mRNA and after its translation, a repressor protein is produced. It binds to the operator region of the operon and prevents translation, as a result, P-galactosidase is not produced. In the absence of preferred carbon source such as glucose, if lactose is available as an energy source for the bacteria then lactose enters the cell as a result of permease enzyme. Lactose acts as an inducer and interacts with the repressor to inactivate it.

The repressor protein binds to the operator of the operon and prevents RNA polymerase from transcribing the operon. In the presence of inducer, such as lactose or allolactose, the repressor is inactivated by interaction with the inducer. This allows RNA polymerase to bind to the promotor site and transcribe the operon to produce lac mRNA which enables formation of all the required enzymes needed for lactose metabolism. This regulation of lac operon by the repressor is an example of negative control of transcription initiation.

Question 84.
What are the objectives of Human Genome project?
Answer:
The main goals of Human Genome Project are as follows:

  1. Identify all the genes (approximately 30000) in human DNA.
  2. Determine the sequence of the three billion chemical base pairs that makeup the human DNA.
  3. To store this information in databases.
  4. Improve tools for data analysis.
  5. Transfer related technologies to other sectors, such as industries.
  6. Address the ethical, legal and social issues (ELSI) that may arise from the project.

Question 85.
Write the salient features of Human Genome Project.
Answer:

  1. Although human genome contains 3 billion nucleotide bases, the DNA sequences that encode proteins make up only about 5% of the genome.
  2. An average gene consists of 3000 bases, the largest known human gene being dystrophin with 2.4 million bases.
  3. The function of 50% of the genome is derived from transposable elements such as LINE and ALU sequence.
  4. Genes are distributed over 24 chromosomes. Chromosome 19 has the highest gene density. Chromosome 13 and Y chromosome have lowest gene densities.
  5. The chromosomal organization of human genes shows diversity.
  6. There may be 35000-40000 genes in the genome and almost 99.9 nucleotide bases are exactly the same in all people.
  7. Functions for over 50 percent of the discovered genes are unknown.
  8. Less than 2 percent of the genome codes for proteins.
  9. Repeated sequences make up very large portion of the human genome. Repetitive sequences have no direct coding functions but they shed light on chromosome structure, dynamics and evolution (genetic diversity).
  10. Chromosome 1 has 2968 genes, whereas chromosome ’Y’ has 231 genes.
  11. Scientists have identified about 1.4 million locations, where single base DNA differences (SNPs – Single nucleotidepolymorphism – pronounce as ‘snips’) occur in humans. Identification of ‘SNIPS’ is helpful in finding chromosomal locations for disease associated sequences and tracing human history.

Question 86.
Describe the principle involved in DNA fingerprinting technique.
Answer:
The DNA fingerprinting technique was first developed by Alec Jeffreys in 1985. The DNA of a person and finger prints are unique. There are 23 pairs of human chromosomes with 1.5 million pairs of genes. It is a well known fact that genes are segments of DNA which differ in the sequence of their nucleotides. Not all segments of DNA code for proteins, some DNA segments have a regulatory function, while others are intervening sequences (introns) and still others are repeated DNA sequences. In DNA fingerprinting, short repetitive nucleotide sequences are specific for a person. These nucleotide sequences are called as variable number tandem repeats (VNTR). The VNTRs of two persons generally show variations and are useful as genetic markers.

DNA finger printing involves identifying differences in some specific regions in DNA sequence called repetitive DNA, because in these sequences, a small stretch of DNA is repeated many times. These repetitive DNA are separated from bulk genomic DNA as different peaks during density gradient centrifugation. The bulk DNA forms a major peak and the other small peaks are referred to as satellite DNA. Depending on base composition (A: T rich or G : C rich), length of segment and number of repetitive units, the satellite DNA is classified into many sub – categories such as micro-satellites and mini satellites, etc.

These sequences do not code for any proteins, but they form a large portion of human genome. These sequences show high degree of polymorphism and form the basis of DNA fingerprinting. DNA isolated from blood, hair, skin cells, or other genetic evidences left at the scene of a crime can be compared through VNTR patterns, with the DNA of a criminal suspect to determine guilt or innocence. VNTR patterns are also useful in establishing the identity of a homicide victim, either from DNA found as evidence or from the body itself.

Question 87.
Draw a flow chart depicting the steps of DNA finger printings technique
Answer:

Higher Order Thinking Skills (HOTs) Questions

Question 1.
A mRNA strand has a series of triplet codons of which the first three codons are given below
(a) AUG
(b) UUU
(c) UGC
(i) Name the amino acid encoded by these triplet codons.
(ii) Mention the DNA sequence from which these triplet codons would have transcribed?
Answer:
(i) AUG codes for Methionine
UUU codes for Phenylalanine
UGC codes for Cysteine
(ii) TAC sequence of DNA is transcribed to AUG
AAA sequence of DNA is transcribed to UUU
ACG sequence of DNA is transcribed to UGC

Question 2.
Given below are the structures of tRNA molecules which are involved in translation process. In one tRNA, codon is mentioned but not the amino acid. In another tRNA molecule, amino acid is named and not the codon. Complete the figure by mentioning the respective amino acids and codons.
Answer:

Question 3.
A DNA fragment possesses 32 adenine bases and 32 cytosine bases. How many total number of nucleotides does that DNA fragment contains? Explain.
Answer:
128 nucleotides. Adenine always pair Thymine base. If there are 32 adenine bases then there must be 32 Thymine bases. Similarly cytosine pairs with guanine. If cytosine bases are 32 in number the guanine bases will be equal to cytosine. So it make a total of 128 nucleotides.

Question 4.
Following is a DNA sequence representing a part of gene TAC TCG CCC TAT UAA CCC AAA ACC TCT using this derive A.


66 Historical Basis of Modern Understanding

By the end of this section, you will be able to do the following:

  • Explain transformation of DNA
  • Describe the key experiments that helped identify that DNA is the genetic material
  • State and explain Chargaff’s rules

Our current understanding of DNA began with the discovery of nucleic acids followed by the development of the double-helix model. In the 1860s, Friedrich Miescher ((Figure)), a physician by profession, isolated phosphate-rich chemicals from white blood cells (leukocytes). He named these chemicals (which would eventually be known as DNA) nuclein because they were isolated from the nuclei of the cells.


To see Miescher conduct his experiment that led to his discovery of DNA and associated proteins in the nucleus, click through this review.

A half century later, in 1928, British bacteriologist Frederick Griffith reported the first demonstration of bacterial transformation —a process in which external DNA is taken up by a cell, thereby changing its morphology and physiology. Griffith conducted his experiments with Streptococcus pneumoniae, a bacterium that causes pneumonia. Griffith worked with two strains of this bacterium called rough (R) and smooth (S). (The two cell types were called “rough” and “smooth” after the appearance of their colonies grown on a nutrient agar plate.)

The R strain is non-pathogenic (does not cause disease). The S strain is pathogenic (disease-causing), and has a capsule outside its cell wall. The capsule allows the cell to escape the immune responses of the host mouse.

When Griffith injected the living S strain into mice, they died from pneumonia. In contrast, when Griffith injected the live R strain into mice, they survived. In another experiment, when he injected mice with the heat-killed S strain, they also survived. This experiment showed that the capsule alone was not the cause of death. In a third set of experiments, a mixture of live R strain and heat-killed S strain were injected into mice, and—to his surprise—the mice died. Upon isolating the live bacteria from the dead mouse, only the S strain of bacteria was recovered. When this isolated S strain was injected into fresh mice, the mice died. Griffith concluded that something had passed from the heat-killed S strain into the live R strain and transformed it into the pathogenic S strain. He called this the transforming principle ((Figure)). These experiments are now known as Griffith’s transformation experiments.


Scientists Oswald Avery, Colin MacLeod, and Maclyn McCarty (1944) were interested in exploring this transforming principle further. They isolated the S strain from the dead mice and isolated the proteins and nucleic acids (RNA and DNA) as these were possible candidates for the molecule of heredity. They used enzymes that specifically degraded each component and then used each mixture separately to transform the R strain. They found that when DNA was degraded, the resulting mixture was no longer able to transform the bacteria, whereas all of the other combinations were able to transform the bacteria. This led them to conclude that DNA was the transforming principle.

Forensic Scientist Forensic Scientists used DNA analysis evidence for the first time to solve an immigration case. The story started with a teenage boy returning to London from Ghana to be with his mother. Immigration authorities at the airport were suspicious of him, thinking that he was traveling on a forged passport. After much persuasion, he was allowed to go live with his mother, but the immigration authorities did not drop the case against him. All types of evidence, including photographs, were provided to the authorities, but deportation proceedings were started nevertheless. Around the same time, Dr. Alec Jeffreys of Leicester University in the United Kingdom had invented a technique known as DNA fingerprinting . The immigration authorities approached Dr. Jeffreys for help. He took DNA samples from the mother and three of her children, as well as an unrelated mother, and compared the samples with the boy’s DNA. Because the biological father was not in the picture, DNA from the three children was compared with the boy’s DNA. He found a match in the boy’s DNA for both the mother and his three siblings. He concluded that the boy was indeed the mother’s son.

Forensic scientists analyze many items, including documents, handwriting, firearms, and biological samples. They analyze the DNA content of hair, semen, saliva, and blood, and compare it with a database of DNA profiles of known criminals. Analysis includes DNA isolation, sequencing, and sequence analysis. Forensic scientists are expected to appear at court hearings to present their findings. They are usually employed in crime labs of city and state government agencies. Geneticists experimenting with DNA techniques also work for scientific and research organizations, pharmaceutical industries, and college and university labs. Students wishing to pursue a career as a forensic scientist should have at least a bachelor’s degree in chemistry, biology, or physics, and preferably some experience working in a laboratory.

Although the experiments of Avery, McCarty and McLeod had demonstrated that DNA was the informational component transferred during transformation, DNA was still considered to be too simple a molecule to carry biological information. Proteins, with their 20 different amino acids, were regarded as more likely candidates. The decisive experiment, conducted by Martha Chase and Alfred Hershey in 1952, provided confirmatory evidence that DNA was indeed the genetic material and not proteins. Chase and Hershey were studying a bacteriophage —a virus that infects bacteria. Viruses typically have a simple structure: a protein coat, called the capsid, and a nucleic acid core that contains the genetic material (either DNA or RNA). The bacteriophage infects the host bacterial cell by attaching to its surface, and then it injects its nucleic acids inside the cell. The phage DNA makes multiple copies of itself using the host machinery, and eventually the host cell bursts, releasing a large number of bacteriophages. Hershey and Chase selected radioactive elements that would specifically distinguish the protein from the DNA in infected cells. They labeled one batch of phage with radioactive sulfur, 35 S, to label the protein coat. Another batch of phage were labeled with radioactive phosphorus, 32 P. Because phosphorous is found in DNA, but not protein, the DNA and not the protein would be tagged with radioactive phosphorus. Likewise, sulfur is absent from DNA, but present in several amino acids such as methionine and cysteine.

Each batch of phage was allowed to infect the cells separately. After infection, the phage bacterial suspension was put in a blender, which caused the phage coat to detach from the host cell. Cells exposed long enough for infection to occur were then examined to see which of the two radioactive molecules had entered the cell. The phage and bacterial suspension was spun down in a centrifuge. The heavier bacterial cells settled down and formed a pellet, whereas the lighter phage particles stayed in the supernatant. In the tube that contained phage labeled with 35 S, the supernatant contained the radioactively labeled phage, whereas no radioactivity was detected in the pellet. In the tube that contained the phage labeled with 32 P, the radioactivity was detected in the pellet that contained the heavier bacterial cells, and no radioactivity was detected in the supernatant. Hershey and Chase concluded that it was the phage DNA that was injected
into the cell and carried information to produce more phage particles, thus providing evidence that DNA was the genetic material and not proteins ((Figure)).


Around this same time, Austrian biochemist Erwin Chargaff examined the content of DNA in different species and found that the amounts of adenine, thymine, guanine, and cytosine were not found in equal quantities, and that relative concentrations of the four nucleotide bases varied from species to species, but not within tissues of the same individual or between individuals of the same species. He also discovered something unexpected: That the amount of adenine equaled the amount of thymine, and the amount of cytosine equaled the amount of guanine (that is, A = T and G = C). Different species had equal amounts of purines (A+G) and pyrimidines (T + C), but different ratios of A+T to G+C. These observations became known as Chargaff’s rules . Chargaff’s findings proved immensely useful when Watson and Crick were getting ready to propose their DNA double helix model! You can see after reading the past few pages how science builds upon previous discoveries, sometimes in a slow and laborious process.

Section Summary

DNA was first isolated from white blood cells by Friedrich Miescher, who called it nuclein because it was isolated from nuclei. Frederick Griffith’s experiments with strains of Streptococcus pneumoniae provided the first hint that DNA may be the transforming principle. Avery, MacLeod, and McCarty showed that DNA is required for the transformation of bacteria. Later experiments by Hershey and Chase using bacteriophage T2 proved that DNA is the genetic material. Chargaff found that the ratio of A = T and C = G, and that the percentage content of A, T, G, and C is different for different species.

Review Questions

If DNA of a particular species was analyzed and it was found that it contains 27 percent A, what would be the percentage of C?

The experiments by Hershey and Chase helped confirm that DNA was the hereditary material on the basis of the finding that:

  1. radioactive phage were found in the pellet
  2. radioactive cells were found in the supernatant
  3. radioactive sulfur was found inside the cell
  4. radioactive phosphorus was found in the cell

Bacterial transformation is a major concern in many medical settings. Why might health care providers be concerned?

  1. Pathogenic bacteria could introduce disease-causing genes in non-pathogenic bacteria.
  2. Antibiotic resistance genes could be introduced to new bacteria to create “superbugs.”
  3. Bacteriophages could spread DNA encoding toxins to new bacteria.
  4. All of the above.

Critical Thinking Questions

Explain Griffith’s transformation experiments. What did he conclude from them?

Live R cells acquired genetic information from the heat-killed S cells that “transformed” the R cells into S cells.

Why were radioactive sulfur and phosphorous used to label bacteriophage in Hershey and Chase’s experiments?

Sulfur is an element found in proteins and phosphorus is a component of nucleic acids.

When Chargaff was performing his experiments, the tetranucleotide hypothesis, which stated that DNA was composed of GACT nucleotide repeats, was the most widely accepted view of DNA’s composition. How did Chargaff disprove this hypothesis?

If the tetranucleotide hypothesis were true, then DNA would have to contain equal amounts of all four nucleotides (A=T=G=C). However, Chargaff demonstrated that A=T and G=C, but that the four nucleotides are not present in equal amounts.


Martha Chase: For the Books

The Hershey-Chase experiment is the historic study that solidified the function of DNA as the carrier of genetic information over protein.

Martha Chase

Martha Chase was an American geneticist who was part of the duo with Alfred Hershey who conducted the groundbreaking work. The work garnered Hershey the Nobel Prize in Physiology or Medicine in 1969, shared with Max Delbrück and Salvador Luria, for their “discoveries concerning the genetic structure of viruses.” Chase’s exclusion from the prize remains a mystery. Some question Chase’s intellectual contributions to the work, as well as the fact that she was a research assistant on the team and not a lead investigator. Nevertheless, her contribution is apparent given that the historic Hershey-Chase experiment includes her name. It could thus be speculated that the Nobel Prize snub may have to do with patriarchal attitudes at the time.

Chase was born in Cleveland, Ohio in 1927 and received a bachelor’s degree from the College of Wooster in 1950. She worked as a research assistant at Cold Spring Harbor Laboratory in the laboratory of bacteriologist and geneticist Alfred Hershey before going back to school in 1959, earning a PhD in Microbiology from the University of Southern California in 1964.

In California, Chase met and married fellow scientist Richard Epstein in the late 1950s and changed her name to Martha C. Epstein. The marriage ended in divorce after a few short years. After that, some personal setbacks in the 1960s led her to end her scientific career. She moved back to Ohio to live with family, and in her later years, developed dementia that led to short-term memory loss. She died of pneumonia at the age of 75 on August 8, 2003.

It was during her time in Hershey’s lab in the 1950s, pre-PhD, when she and Hershey performed the Hershey-Chase experiment, which helped confirm that DNA is the carrier and transmitter of genetic information.

The Hershey-Chase experiment was actually a set of experiments conducted by Hershey and Chase that helped them conclude that DNA served as the genetic material of a cell. While the structure of DNA had been elucidated a few short years earlier by the group of Rosalind Franklin, James Watson, Francis Crick and Maurice Wilkins, its function remained largely unknown. At the same time, it was thought that protein was the carrier of genetic information as DNA was too inert of a substance — and was also shielded away in the nucleus — to be able to harbor such a role.

In the experiments, Hershey and Chase infected bacteria with radioactively labeled bacteriophages (viruses that specifically infect bacteria), which consist of both DNA and protein. Bacteriophages inject their DNA into cells but not protein, which is mainly a constituent of their outer envelope, or capsid. Through the radioactive labels, they were able to track which of the two, DNA or protein, was injected into the bacteria.

They conducted two experiments, one with radioactive phosphorous to label DNA, and one with radioactive sulfur to tag protein (DNA does not contain sulfur and protein does not contain phosphate groups). In each of the experiments, they looked at whether the radioactive signal was found in the cell or outside the cell.

The experiment is also known as the “Waring blender experiment” as a Waring kitchen blender was used to separate the bacteria from the viruses after infection. After detachment, the mixture was centrifuged this was done to destroy the outer membrane integrity of bacterial cells, and hence remove any phages attached to the outside of the bacteria. Meanwhile, any radioactive material that had entered the cells would be intact and detectable.

The experiment revealed that in the radioactive phosphorus sample, 40 percent of the labeled particles were removed, while in the sulfur-labeled sample 80 percent of the labeled particles were removed. Since more protein from the phages was removed off of the bacteria than DNA, the results demonstrated that DNA was able to readily enter cells and not protein. And since DNA made its way into the bacteria, it was the carrier of genetic information, and played a role in reproduction.

Despite not winning formal accolades for her work, namely the Nobel Prize, Chase’s work is certainly recognized each time a molecular biology or genetics student picks up a textbook and reads about the foundational Hershey-Chase experiment, which has her name stamped on it, quite literally.


Tamilnadu Samacheer Kalvi 12th Bio Zoology Solutions Chapter 5 Molecular Genetics

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Samacheer Kalvi 12th Bio Zoology Molecular Genetics Text Book Back Questions and Answers

Question 1.
Hershey and Chase experiment with bacteriophage showed that
(a) Protein gets into the bacterial cells
(b) DNA is the genetic material
(c) DNA contains radioactive sulphur
(d) Viruses undergo transformation
Answer:
(b) DNA is the genetic material

Question 2.
DNA and RNA are similar with respect to
(a) Thymine as a nitrogen base
(b) A single-stranded helix shape
(c) Nucleotide containing sugars, nitrogen bases and phosphates
(d) The same sequence of nucleotides for the amino acid phenyl alanine
Answer:
(c) Nucleotide containing sugars, nitrogen bases and phosphates

Question 3.
A mRNA molecule is produced by
(a) Replication
(b) Transcription
(c) Duplication
(d) Translation
Answer:
(b) Transcription

Question 4.
The total number of nitrogenous bases in human genome is estimated to be about
(a) 3.5 million
(b) 35000
(c) 35 million
(d) 3.1 billion
Answer:
(d) 3.1 billion

Question 5.
E. coli cell grown on 15N medium are transferred to 14N medium and allowed to grow for two generations. DNA extracted from these cells is ultracentrifuged in a cesium chloride density gradient. What density distribution of DNA would you expect in this experiment?
(a) One high and one low density band
(b) One intermediate density band
(c) One high and one intermediate density band
(d) One low and one intermediate density band
Answer:
(d) One low and one intermediate density band

Question 6.
What is the basis for the difference in the synthesis of the leading and lagging strand of DNA molecules?
(a) Origin of replication occurs only at the 5’ end of the molecules
(b) DNA ligase works only in the 3’ → 5’ direction
(c) DNA polymerase can join new nucleotides only to the 3 ’ end of the growing stand
(d) Helicases and single-strand binding proteins that work at the 5’ end
Answer:
(d) Helicases and single-strand binding proteins that work at the 5’ end

Question 7.
Which of the following is the correct sequence of event with reference to the central dogma?
(a) Transcription, Translation, Replication
(b) Transcription, Replication, Translation
(c) Duplication, Translation, Transcription
(d) Replication, Transcription, Translation
Answer:
(d) Replication, Transcription, Translation

Question 8.
Which of the following statements about DNA replication is not correct?
(a) Unwinding of DNA molecule occurs as hydrogen bonds break
(b) Replication occurs as each base is paired with another exactly like it
(c) Process is known as semi – conservative replication because one old strand is conserved in the new molecule
(d) Complementary base pairs are held together with hydrogen bonds
Answer:
(b) Replication occurs as each base is paired with another exactly like it

Question 9.
Which of the following statements is not true about DNA replication in eukaryotes?
(a)) Replication begins at a single origin of replication.
(b) Replication is bidirectional from the origins.
(c) Replication occurs at about 1 million base pairs per minute.
(d) There are numerous different bacterial chromosomes, with replication occurring in each at the same time.
Answer:
(d) There are numerous different bacterial chromosomes, with replication occurring in each at the same time.

Question 10.
The first codon to be deciphered was which codes for
(a) AAA, proline
(b) GGG, alanine
(c) UUU, Phenylalanine
(d) TTT, arginine
Answer:
(c) UUU, Phenylalanine

Question 11.
Meselson and Stahl’s experiment proved __________
(a) Transduction
(b) Transformation
(c) DNA is the genetic material
(d) Semi-conservative nature of DNA replication
Answer:
(d) Semi-conservative nature of DNA replication

Question 12.
Ribosomes are composed of two subunits the smaller subunit of a ribosome has a binding site for and the larger subunit has two binding sites for two
Answer:
mRNA, tRNA

Question 13.
Anoperonisa:
(a) Protein that suppresses gene expression
(b) Protein that accelerates gene expression
(c) Cluster of structural genes with related function
(d) Gene that switched other genes on or off
Answer:
(d) Gene that switched other genes on or off

Question 14.
When lactose is present in the culture medium:
(a) Transcription of lacy, lac z, lac a genes occurs
(b) Repressor is unable to bind to the operator
(c) Repressor is able to bind to the operator
(d) Both (a) and (b) are correct
Answer:
(d) Both (a) and (b) are correct

Question 15.
Give reasons: Genetic code is ‘universal’.
Answer:
The genetic code is universal. It means that all known living systems use nucleic acids and the same three base codons (triplet codon) direct the synthesis of protein from amino acids. For example, the mRNA (UUU) codon codes for phenylalanine in all cells of all organisms. Some exceptions are reported in prokaryotic, mitochondrial and chloroplast genomes. However similarities are more common than differences.

Question 16.
Name the parts marked ‘A’ and ‘B’ in the given transcription unit:
Answer:

Question 17.
Differentiate – Leading strand and lagging strand
Answer:

  1. DNA polymerase I Involved DNA repair mechanism
  2. DNA polymerase II Involved DNA repair mechanism
  3. DNA polymerase III Involved in DNA replicaton

Question 18.
Differentiate – Template strand and coding strand.
Answer:

  1. Template Strand: During replication, DNA strand having the polarity 3’ → 5’ act as template strand.
  2. Coding Strand: During replication, DNA strand having the polarity 5’ → 3’ act as coding strand.

Question 19.
Mention any two ways in which single nucleotide polymorphism (SNPs) identified in human genome can bring revolutionary change in biological and medical science.
Answer:
Scientists have identified about 1.4 million locations, where single base DNA differences (SNPs – Single nucleotide polymorphism – pronounced as ‘snips’) occur in humans. Identification of ‘SNIPS’ is helpful in finding chromosomal locations for disease associated sequences and tracing human history.

Question 20.
State any three goals of the human genome project.
Answer:

  1. Identify all the genes (approximately 30000) in human DNA.
  2. Determine the sequence of the three billion chemical base pairs that makeup the human DNA.
  3. To store this information in databases.

Question 21.
In E.coli, three enzymes 0- galactosidase, permease and transacetylase are produced in the presence of lactose. Explain why the enzymes are not synthesized in the absence of lactose.
Answer:
In the absence of lactose, the repressor protein binds to the operator and prevents the transcription of structural gene by RNA polymerase, hence the enzymes are not produced. However, there will always be a minimal level of lac operon expression even in absence of lactose.

Question 22.
Distinguish between structural gene, regulatory gene and operator gene.
Answer:
Structure of the operon: Each operon is a unit of gene expression and regulation and consists of one or more structural genes and an adjacent operator gene that controls transcriptional, activity of the structural gene.

  1. The structural gene codes for proteins, rRNA and tRNA required by the cell.
  2. Promoters are the signal sequences in DNA that initiate RNA synthesis. RNA polymerase binds to the promoter prior to the initiation of transcription.
  3. The operators are present between the promoters and structural genes. The repressor protein binds to the operator region of the operon.

Question 23.
A low level of expression of lac operon occurs at all the time in E-coli. Justify the statement.
Answer:
One of the enzyme synthesized by lac operon is permease which is involved in the transport of lactose into the cells. If the lac operon gets inactivated, permease is not synthesized hence lactose cannot enter the cell. Lactose acts as a inducer, binding to the repressor protein and switch on the operator to initiate gene expression.

Question 24.
Why the human genome project is called a mega project?
Answer:
The international human genome project was launched in the year 1990. It was a mega project and took 13 years to complete. The human genome is about 25 times larger than the genome of any organism sequenced to date and is the first vertebrate genome to be completed. Human genome is said to have approximately 3 >109 bp. HGP was closely associated with the rapid development of a new area in biology called bioinformatics.

Question 25.
From their examination of the structure of DNA, What did Watson and Crick infer about the probable mechanism of DNA replication, coding capability and mutation?
Answer:
Inference of Watson and Crick on DNA replication: They concluded that each of the DNA strand in a helix act as template during DNA replication leading to formation of new daughter DNA molecules, which are complementary to parental strand, (i.e., Semi-conservative method of replication) Inference on coding capability: During transcription, the genetic information in the DNA strand is coded to mRNA as complementary bases, (except for uracil in place of thymine in RNA) Inference on mutation: Any changes in the nucleotide sequence of DNA leads to corresponding alteration in aminoacid sequence of specific protein thus confirming the validity of genetic code.

Question 26.
Why tRNA is called an adapter molecule?
Answer:
The transfer RNA, (tRNA) molecule of a cell acts as a vehicle that picks up the amino acids scattered through the cytoplasm and also reads specific codes of mRNA molecules. Hence it is called an adapter molecule. This term was postulated by Francis Crick.

Question 27.
What are the three structural differences between RNA and DNA?
Answer:
DNA:

  1. Sugar is deoxyribose sugar.
  2. Double stranded structure.
  3. Nitrogen bases are Adenine, Guanine, Cytosine and Thymine.
  1. Sugar is ribose sugar.
  2. Single stranded molecule.
  3. Nitrogen bases are Adenine, Guanine, Cytosine and Uracil.

Question 28.
Name the anticodon required to recognize the following codons:
AAU, CGA, UAU, and GCA.
Answer:
UUA, GCU, AUA and CGU.

Question 29.
(a) Identify the figure given below
(b) Redraw the structure as a replicating fork and label the parts
(c) Write the source of energy for this replication and name the enzyme involved in this process.
(d) Mention the differences in the synthesis of protein, based on the polarity of the two template strands.
Answer:
(a) Replication fork
(b)

(c) Deoxy nucleotide, triphosphate acts as a energy source for replication. DNA polymerase is used for replication
(d) mRNA contacting information for protein synthesis will developed from DNA strand having polariy 5’ → 3’

Question 30.
If the coding sequence in a transcription unit is written as follows:
5’ TGCATGCATGCATGCATGCATGCATGC 3’
Write down the sequence of mRNA.
Answer:
mRNA sequence is 3’ACGUACGUACGUUCGUACGUACGUACG5’

Question 31.
How is the two stage process of protein synthesis advantageous?
Answer:
The split gene feature of eukaryotic genes is almost entirely absent in prokaryotes. Originally each exon may have coded for a single polypeptide chain with a specific function. Since exon arrangement and intron removal are flexible, the exon coding for these polypeptide subunits act as domains combining in various ways to form new genes. Single genes can produce different functional proteins by arranging their exons in several different ways through alternate splicing patterns, a mechanism known to play an important role in generating both protein and functional diversity in animals. Introns would have arosen before or after the evolution of eukaryotic gene.

If introns arose late how did they enter eukaryotic gene? Introns are mobile DNA sequences that can splice themselves out of, as well as into, specific ‘target sites’ acting like mobile transposon-like elements (that mediate transfer of genes between organisms – Horizontal Gene Transfer – HGT). HGT occurs between lineages of prokaryotic cells, or from prokaryotic to eukaryotic cells and between eukaryotic cells. HGT is now hypothesized to have played a major role in the evolution of life on Earth.

Question 32.
Why did Hershey and Chase use radioactively labelled phosphorous and sulphur only? Would they have got the same result if they use radiolabelled carbon and nitrogen?
Answer:
Generally proteins contain sulphur but not phosphorous and nucleic acid (DNA) contains , phosphorous but not sulphur. Hence Hershey – Chase used radioactive isotopes of sulphur ( 3 5 S) and phosphorus ( 32 P) to keep separate track of viral protein and nucleic acid in culture medium. The expected result cannot be achieved, if radioactive carbon and nitrogen is used, since these molecules are present in both DNA and proteins.

Question 33.
Explain the formation of a nucleosome.
Answer:
Komberg proposed a model for the nucleosome, in which 2 molecules of the four histone proteins H2A, H2B, H3 and H4 are organized to form a unit of eight molecules called histone octamere.

The negatively charged DNA is wrapped around the positively charged histone octamere to form a structure called nucleosome. A typical nucleosome contains 200 bp of DNA helix. The histone octameres are in close contact and DNA is coiled on the outside of nucleosome.

Question 34.
It is established that RNA is the first genetic material. Justify giving reasons.
Answer:
Three molecular biologists in the early 1980’s (Leslie Orgel, Francis Brick and Carl Woese) independently proposed the ‘RNA world’ as the first stage in the evolution of life, a stage when RNA catalysed all molecules necessary for survival and replication. The term ‘RNA world’ first used by Walter Gilbert in 1986, hypothesizes RNA as the first genetic on Earth. There is now enough evidence to suggest that essential life processes (such as metabolism, translation and splicing etc.,) evolved around RNA. RNA has the ability to act as both genetic material and catalyst. There are several biochemical reactions in living systems that are catalysed by RNA. This catalytic RNA is known as ribozyme. But, RNA being a catalyst was reactive and hence unstable.

This led to evolution of a more stable form of DNA, with certain chemical modifications. Since DNA is a double stranded molecule having complementary strand, it has resisted changes by evolving a process of repair. Some RNA molecules function as gene regulators by binding to DNA and affect gene expression. Some viruses use RNA as the genetic material. Andrew Fire and Craig Mellow (recipients of Nobel Prize in 2006) were of the opinion that RNA is an active ingredient in the chemistry of life.

Samacheer Kalvi 12th Bio Zoology Molecular Genetics Additional Questions and Answers

Question 1.
The term‘gene’was coined by ___________
Answer:
Wilhelm Johannsen

Question 2.
Whose experiment finally provided convincing evidence that DNA is the genetic material?
(a) Griffith experiment
(b) Avery, Macleod and McCarty’s experiment
(c) Hershey-Chase experiment
(d) Urey-Miller’s experiment
Answer:
(c) Hershey-Chase experiment

Question 3.
In Hershey – Chase experiment, the DNA of T 2 phase was made radioactive by using ___________
(a) 32 P
(b) 32 S
(c) 35 P
(d) 32 S
Answer:
(a) 32 P

Question 4.
A nucleoside is composed of ___________
(a) Sugar and Phosphate
(b) Nitrogen base and Phosphate
(c) Sugar and Nitrogen base
(d) Sugar, Phosphate and Nitrogenous base
Answer:
(c) Sugar and Nitrogen base

Question 5.
Identify the incorrect statement
(a) a base is a substance that accepts H+ ion
(b) Both DNA and RNA have four bases
(c) Purines have single carbon-nitrogen ring
(d) Thymine is unique for DNA
Answer:
(c) Purines have single carbon-nitrogen ring

Question 6.
Watson and Crick proposed their double helical DNA model based on the X-ray diffraction analysis o f ___________
(a) Erwin Chargaff
(b) Meselson and Stahl
(c) Wilkins and Franklin
(d) Griffith
Answer:
(c) Wilkins and Franklin

Question 7.
The term ‘RNA world’ was first used by ___________
Answer:
Walter Gilbert

Question 8.
The distance between two consecutive base pairs in DNA is ___________
(a) 0.34 nm
(b) 3.4 nm
(c) 0.034 nm
(d) 34 nm
Answer:
(a) 0.34 nm

Question 9.
If the length of E. coli DNA is 1.36 mm, the number of base pairs is ___________
(a) 0.36 × 10 6 m
(b) 4 × 10 6 m
(c) 0.34 × 10 -9 nm
(d) 4 × 10 -9 m
Answer:
(b) 4 × 10 6 m

Question 10.
Identify the proper sequence in the organisation of eukaryotic chromosome.
(a) Nucleosome – Solenoid – Chromatid
(b) Chromatid – Nucleosome – Solenoid
(c) Solenoid – chromatin – DNA
(d) Nucleosome – solenoid – genophore
Answer:
(a) Nucleosome – Solenoid – Chromatid

Question 11.
Assertion (A) : Genophore is noticed in prokaryotes.
Reason (R) : Bacteria possess circular DNA without chromatin organisation.
(a) Both A and R are correct
(b) A is correct R is incorrect
(c) R explains A
(d) A is incorrect R is correct
Answer:
(c) R explains A

Question 12.
Assertion (A): Heterochromatin is transcriptionally active.
Reason (R): Tightly packed chromatin which stains dark.
(a) Both A and R are correct
(b) A is correct R is incorrect
(c) R explains A
(d) A is incorrect R is correct
Answer:
(d) A is incorrect R is correct

Question 13.
Assertion (A) : Semi-conservative model was proposed by Hershey and Chase.
Reason (R) : The daughter DNA contains only new strands.
(a) Both A and R are incorrect
(b) A is correct R is incorrect
(c) R explains A
(d) A is incorrect R is correct
Answer:
(a) Both A and R are incorrect

Question 14.
Komberg enzyme is called as _____
Answer:
DNA polymerase I

Question 15.
Replication of DNA occurs at __________ phase of cell cycle.
(a) M
(b) S
(c) G1
(d) G2
Answer:
(b) S

Question 16.
Semi-conservative model of replication was proved by __________
(a) Hershey and Chase
(b) Griffith
(c) Meselson and Stahl
(d) Macleod and McCarty
Answer:
(c) Meselson and Stahl

Question 17.
How many types of DNA polymerases does an eukaryotic cell possess?
(a) two
(b) three
(c) four
(d) five
Answer:
(d) Five

Question 18.
Identify the incorrect statement
(a) Replication occurs at ori – site of DNA
(b) Deoxy nucleotide triphosphate acts as a substrate
(c) Unwinding of DNA strand is carried out by topoisomerase
(d) DNA polymerase catalyses the polymerization at 3-OH
Answer:
(c) Unwinding of DNA strand is carried out by topoisomerase

Question 19.
The discontinuously synthesized fragments of lagging strand are called ________
Answer:
Okazaki fragments

Question 20.
Retroviruses possess ________ as genetic material.
Answer:
RNA

Question 21.
Which is NOT a part of transcription unit?
(a) Promoter
(b) Operator
(c) Structural gene
(d) Terminator
Answer:
(b) Operator

Question 22.
Goldberg – Hogness box of eukaryotes is equivalent to ________ of prokaryotes.
Answer:
Pribnow box

Question 23.
Okazaki fragments are joined by the enzyme ________ during DNA replication.
Answer:
DNA ligase

Question 24.

Answer:
(a) A – iv, B – i, C – ii, D – iii

Question 25.
The RNA polymerase of prokaryotes binds with factor to initiate polymerization.
(a) rho
(b) theta
(c) sigma
(d) psi
Answer:
(c) sigma

Question 26.

(a) Capping
(b) Tailing
(c) Splicing
(d) Transcribing
Answer:
(c) Splicing

Question 27.
Which of the following feature is absent in prokaryotes?
(a) Prokaryotes possess three major types of RNAs
(b) Structural genes are polycistronic
(c) Initiation process of transcription requires ‘P’ factor
(d) Split gene feature
Answer:
(d) Split gene feature

Question 28.
Which of the following sequence has completely translated?
(i) AGA, UUU, UGU, AGU, UAG
(ii) AUG, UUU, AGA, UAC, UAA
(iii) AAA, UUU, UUG, UGU, UGA
(iv) AUG,AAU,AAC,UAU,UAG
(a) i and ii
(b) ii only
(c) i and iii
(d) ii and iv
Answer:
(d) ii and iv

Question 29.
Capping of mRNA occurs using __________
(a) Poly A residues
(b) Methyl guanosine triphosphate
(c) Deoxy ribonucleotide triphosphate
(d) Ribonucleotide triphosphate
Answer:
(b) Methyl guanosine triphosphate

Question 30.
One of the aspect is not a feature of genetic code?
(a) Specific
(b) Degenerate
(c) Universal
(d) Ambiguous
Answer:
(d) Ambiguous

Question 31.
Which of the triplet codon is not a code of proline?
(i) CCU
(ii) CAU
(iii) CCG
(iv) CAA
(a) i only
(b) ii and iv
(c) iii only
(d) all the above
Answer:
(b) ii and iv

Question 32.
Coding sequences found in split genes are called.
(a) Operons
(b) Introns
(c) Exons
(d) Cistron
Answer:
(c) Exons

Question 33.
Which of the following mRNA yields 6 aminoacids after translation?
(i) UCU UAU AGU CGA UGC AGU UGA AAA UUU
(ii) UGA AGA UAG GAG CAU CCC UAC UAU GAU
(iii) GUC UGC UGG GCU GAU UAA AGG AGC AUU
(iv) AUG UAC CAU UGC UGA UGC AGG AGC CCG
Answer:
(i) UCU UAU AGU CGA UGC AGU UGA AAA UUU

Question 34.
The transcription termination factor associated with RNA polymerase in prokaryotes is
(a) θ
(b) σ
(c) ρ
(d) ∑
Answer:
(c) ρ

Question 35.
In a DNA double strand, if guanine is of 30%, what will be the percentage of thymine?
(a) 100%
(b) 20%
(c) 10%
(d) 70%
Answer:
(b) 20%

Question 36.
Identify the triplet pairs that code for Tyrosine
(a) UUU,UUC
(b) UAU, UAU
(c) UGC, UGU
(d) CAU, CAC
Answer:
(b) UAU, UAU

Question 37.

Answer:
A – ii B – i C – iv D – iii

Question 38.
AUG code is for __________
(a) Arginine
(b) Tyrosine
(c) Tryptophan
(d) Methionine
Answer:
(d) Methionine

Question 39.
The sequence of bases in coding strand of DNA is G A G T T A G C A G G C, then the sequence of codons in primary transcript is
(a) C U C A U A C G C C C G
(b) C U C A A U C G U C C G
(c) U C A G A U C U G C G C
(d) U U C A A U C G U G C G
Answer:
(b) C U C A A U C G U C C G

Question 40.
The promoter region of eukaryote is __________
(a) TATAA
(b) AUGUT
(c) UUUGA
(d) AAAAU
Answer:
(a) TATAA

Question 41.
Match the following:
(A) AUG – (i) Tyrosine
(B) UGA – (ii) Glycine
(C) UUU – (iii) Methionine
(D) GGG – (iv) Phenylalanine
(a) A – iii B – i C – iv D – ii
(b) A – iii B – ii C – i D – iv
(c) A – iv B – i C – iii D – ii
(d) A – ii B – i C – iv D – iii
Answer:
(d) A – ii B – i C – iv D – iii

Question 42.
__________ number of codons, codes for cystine.
Answer:
Two

Question 43.
In sickle cell anaemia, the __________ codon of β – globin gene is modified.
(a) Eighth
(b) Seventh
(c) Sixth
(d) Nineth
Answer:
(c) Sixth

Question 44.
Pick out the incorrect statement.
(a) tRNA acts as a adapter molecule
(b) Stop codons donot have tRNA’s
(c) Addition of aminoacid leads to hydrolysis of tRNA
(d) tRNA has four major loops
Answer:
(c) Addition of aminoacid leads to hydrolysis of tRNA

Question 45.
Which of the following antibiotic inhibits the interaction between tRNA and mRNA?
(a) Neomycin
(b) Streptomycin
(c) Tetracycline
(d) Chloramphenicol
Answer:
(a) Neomycin

Question 47.
The cluster of genes with related function is called _________
(a) Cistron
(b) Operon
(c) Muton
(d) Recon
Answer:
(b) Operon

Question 48.
Repressor protein of Lac operon binds to __________ of operon.
(a) Promoter region
(b) Operator region
(c) terminator region
(d) inducer region
Answer:
(b) Operator region

Question 49.
Lac Z gene codes for __________
(a) Permease
(b) transacetylase
(c) β -galactosidase
(d) Aminoacyl transferase
Answer:
(c) β -galactosidase

Question 50.
Lac operon model was proposed by __________
Answer:
Jacob and Monod

Question 51.
Approximate count of base pair in human genome is __________
Answer:
3 × 10 9 bp

Question 52.
Automated DNA sequences are developed by.
Answer:
Frederick Sanger

Question 53.
Which of the chromosome has higher gene density?
(a) Chromosome 20
(b) Chromosome 19
(c) Chromosome 13
(d) Chromosome Y
Answer:
(b) Chromosome 19

Question 54.
Number of genes located in chromosome Y is __________
(a) 2968
(b) 213
(c) 2869
(d) 231
Answer:
(d) 231

Question 55.
How many structural genes are located in lac operon of E.Coli?
(a) 4
(b) 3
(c) 2
(d) 1
Answer:
(b) 3

Question 56.
DNA finger printing technique was developed by
(a) Jacob and Monod
(b) Alec Jeffreys
(c) Frederick Sanger
Answer:
(b) Alec Jeffreys

Question 57.
In DNA fingerprinting, separation of DNA fragments is done by __________
(a) Centrifugation
(b) Electrophoresis
(c) X-ray diffraction
(d) denaturation
Answer:
(b) Electrophoresis

Question 58.
SNP stands for
(a) Single nucleotide Polymorphism
(b) Single Nucleoside Polypeptide
(c) Single nucleotide Polymorphism
(d) Single nucleotide polymer
Answer:
(a) Single nucleotide Polymorphism

Question 59.
Specific sequences of mRNA that are not translated are __________
Answer:
UnTranslated Regions (UTR)

Question 60.
Non-coding or intervening DNA sequence is called __________

Question 61.
_______ Intron is the monomer of DNA.
Answer:
Nucleotide

Question 62.
Which one of the following is wrongly matched?
(a) Transcription – Copying information from DNA to RNA
(b) Translation – Decoding information from mRNA to protein
(c) Replication – Making of DNA copies
(d) Splicing – Joining of exons with introns
Answer:
(d) Splicing – Joining of exons with introns

Question 1.
Who proposed One gene – One enzyme hypothesis? Define it.
Answer:
George Beadle and Edward Tatum proposed One gene – One enzyme hypothesis which states that one gene controls the production of one enzyme.

Question 2.
Differentiate nucleoside from nucleotide.
Answer:

  1. Nucleoside: Nucleoside subunit is composed of nitrogenous bases linked to a pentose sugar molecule.
  2. Nucleotide: Nucleotide subunit is composed of nitrogenous bases, a pentose sugar and a phosphate group.

Question 3.
State the key differences between DNA and RNA.
Answer:
DNA:

  1. DNA is made of deoxyribose sugar.
  2. Nitrogenous bases of DNA are Adenine, Guanine, Cytosine and Thymine.
  1. RNA is made of ribose sugar.
  2. Nitrogenous bases of RNA are Adenine, Guanine, Cytosine and Uracil.

Question 4.
Point out the nitrogenous bases of RNA.
Answer:
Adenine, Guanine, Cytosine and Uracil.

Question 5.
What makes the DNA and RNA as acidic molecules?
Answer:
The phosphate functional group (PO4) gives DNA and RNA the property of an acid at physiological pH, hence the name nucleic acid.

Question 6.
Which type of bond is formed

  1. between a purine and pyrimidine base?
  2. between the pentose sugar and adjacent nucleotide?
  1. Purine and pyrimidine bases are linked by hydrogen bonds.
  2. Pentose sugar is linked to adjacent nucleotide by phosphodiester bonds.

Question 7.
DNA acts as genetic material for majority of living organisms and not the RNA. Give reasons to support the statement.
Answer:

  1. RNA was reactive and hence highly unstable.
  2. Some RNA molecules acts as gene regulators by binding to DNA and affect gene expression.
  3. Uracil of RNA is less stable than thymine of DNA.

Question 8.
Name any two viruses whose genetic material is RNA.
Answer:

Question 9.
What are the properties that a molecule must possess to act as genetic material?
Answer:

  1. Self replication
  2. Information storage
  3. Stability
  4. Variation through mutation

Question 10.
How many base pairs are present in one complete turn of DNA helix? What is the distance between two consecutive base pairs?
Answer:
There are ten base pairs in each turn with a distance of 0.34 x 109m between two adjacent base pairs.

Question 11.
What is a genophore?
Answer:
In prokaryotes such as E. coli though they do not have defined nucleus, the DNA is not scattered throughout the cell. DNA (being negatively charged) is held with some proteins (that have positive charges) in a region called the nucleoid. The DNA as a nucleoid is organized into large loops held by protein. DNA of prokaryotes is almost circular and lacks chromatin organization, hence termed genophore.

Question 12.
Whqt is nucleosome? How many base pairs are there in a typical nucleosome?
Answer:
The negatively charged DNA is wrapped around the positively charged histone octamere to form a structure called nucleosome. A typical nucleosome contains 200 bp of DNA helix.

Question 13.
Expand and define NHC
Answer:

  1. NHC : Non-histone Chromosomal protein.
  2. In eukaryotes, apart from histone proteins, additional set of proteins are required for packing of chromatin at higher level and are referred as non – histone chromosomal proteins.

Question 14.
Differentiate between Heterochromatin and Euchromatin.
Answer:
Heterochromatin:

  1. Region of nucleus where the chromatin are loosely packed and stains light are called Heterochromatin.
  2. Transcriptionally inactive.
  1. Region of nucleus where the chromatin are tightly packed and stains dark are called Euchromatin.
  2. Transcriptionally active.

Question 15.
Which is the widely accepted model of DNA replication? Who has proved it?
Answer:
Semi-conservative replication model. It was proved by Meselson and Stahl in 1958.

Question 16.
Name the chemical substance which is called by the name

  1. DNA polymerase I is also known as Komberg enzyme.
  2. Polynucleotide phosphorylase is also known as Ochoa’s enzyme.

Question 17.
Name the various types of prokaryotic DNA polymerase. State their role in replication process.
Answer:

  1. DNA Polymerase I Involver in DNA repair mechanism
  2. DNA Polymerase II Involver in DNA repair mechanism
  3. DNA Polymerase III Involver in DNA replication

Question 18.
What is the function of Deoxy nucleotide triphosphate in replication?
Answer:
Deoxy nucleotide triphosphate acts as substrate and also provides energy for polymerization reaction.

Question 19.
Given below are some events of eukaryotic replication. Name the enzymes involved in the process.

  1. Unwinding of DNA
  2. Joining of Okazaki fragments
  3. Addition of nucleotides to new strand
  4. Correcting the repair

Question 20.
Differentiate leading strand from lagging strand
Answer:
Leading strand:

Question 21.
What are Okazaki fragments?
Answer:
The discontinuously synthesized fragments of the lagging strand are called the Okazaki fragments are joined by the enzyme DNA ligase.

Question 22.
What is a replication fork?
Answer:
At the point of origin of replication, the helicases and topoisomerases (DNA gyrase) unwind and pull apart the strands, forming a Y-Shaped structure called the replication fork. There are two replication forks at each origin.

Question 23.
Apart from DNA polymerase, name any other four enzymes which were involved in DNA replication of eukaryotic cell.
Answer:
DNA ligase, Topoisomerase (DNA gyrase), Helicase and Nuclease.

Question 24.
Who proposed the central dogma? Write its concept.
Answer:
Francis Crick proposed the Central dogma in molecular biology which states that genetic information flows as follows:

Question 25.
Define transcription and name the enzyme involved in this process.
Answer:
The process of copying genetic information from one strand of DNA into RNA is termed transcription. This process takes place in presence of DNA dependent RNA polymerase.

Question 26.
What is TATA box? State its function.
Answer:
In eukaryotes, the promoter has AT rich regions called TATA box or Goldberg-Hogness box. It acts as a binding site for RNA polymerase.

Question 27.
Structural gene of eukaryotes differ from prokaryotes. How?
Answer:
In eukaryotes, the structural gene is monocistronic coding for only one protein whereas in prokaryotes the structural gene is polycistronic coding for many proteins.

Question 28.
What are the two major components of prokaryotic RNA polymerase? How do they act?
Answer:
Bacterial (prokaryotic) RNA polymerase consists of two major components, the core enzyme and the sigma subunit. The core enzyme (β1, β, and α) is responsible for RNA synthesis ” whereas a sigma subunit is responsible for recognition of the promoter.

Question 29.
Distinguish between exons and introns.
Answer:

  1. Exons: Expressed sequences (Coding sequences) of an eukaryotic gene
  2. Introns: Interveining sequences (non-coding sequences) of an eukaryotic gene

Question 30.
Define splicing.
Answer:
The process of removing introns from hnRNA is called splicing.

Question 31.
What is capping and tailing?
Answer:
In capping an unusual nucleotide, methyl guanosine triphosphate is added at the 5’ end of hnRNA, whereas adenylate residues (200-300) (Poly A) are added at the 3’ end in tailing.

Question 32.
If a double stranded DNA has 20% of cytosine, calculate the percentage of adenine in DNA.
Answer:
Cytdsine = 20, hence Guanine = 20
As per ChargafFs rule (A+T) = (G+C) =100
Percent of Thymine + Adenine = 20 + 20 = 100
(T + A) = (20 + 20) =100
(T + A)=100-(20 + 20)
T +A = 100 – 40
T + A = 60
Therefore the percent of Adenine will be 60/2 = 30%.

Question 33.
Mention the dual functions of AUG.
Answer:
AUG has dual functions. It acts as a initiator codon and also codes for the amino acid methionine.

Question 34.
How many codons are involved in termination of translation. Name them.
Answer:
Three codons terminate translation process. They are UAA, UAG and UGA.

Question 35.
Degeneracy of codon – comment.
Answer:
A degenerate code means that more than one triplet codon could code for a specific amino acid. For example, codons GUU, GUC, GUA and GUG code for valine.

Question 36.
Point out the exceptional categories to universality of genetic code.
Answer:
Exceptions to universal nature of genetic code is noticed in prokaryotic mitochondrial and chloroplast genomes.

Question 37.
What are non-sense codons?
Answer:
UGA, UAA and UAG are the non-sense codons, which terminates translation.

Question 38.
Name the triplet codons that code for

Question 39.
Why hnRNA has to undergo splicing?
Answer:
Since hnRNA contains both coding sequences (exons) and non-coding sequences (introns) it has to undergo splicing to remove introns.

Question 40.
State the role of following codons in translation process

Question 41.
Given below is mRNA sequence. Mention the aminoacids sequence that is formed after its translation.
Answer:
3’AUGAAAGAUGGGUAA5’
Methionine – Lysine – Aspartic acid – Glycine

Question 42.
Name the four codons that codes valine.
Answer:
GUU, GUC, GUA and GUG.

Question 43.
The base sequence in one of the DNA strand is TAGC ATGAT. Mention the base sequence in its complementary strand.
Answer:
The complementary strand has ATCGTACTA.

Question 44.
Why t-RNA is called as adapter molecule?
Answer:
The transfer RNA (tRNA) molecule of a cell acts as a vehicle that picks up the amino acids scattered through the cytoplasm and also reads specific codes of mRNA molecules. Hence it is called as adapter molecule.

Question 45.
What do you mean by charging of tRNA? Name the enzyme involved in this process.
Answer:
The process of addition of amino acid to tRNA is known as aminoacylation or charging and the resultant product is called aminoacyl- tRNA (charged tRNA). Aminoacylation is catalyzed by an enzyme aminoacyl – tRNA synthetase.

Question 46.
What are UTR’s?
Answer:
mRNA also have some additional sequences that are not translated and are referred to as Untranslated Regions (UTR). UTRs are present at both 5’ end (before start codon) and at 3’ end (after stop codon).

Question 47.
What is S – D sequence?
Answer:
The 5’ end of the mRNA of prokaryotes has a special sequence which precedes the initial AUG start codon of mRNA. This ribosome binding site is called the Shine – Dalgamo sequence or S-D sequence. This sequences base-pairs with a region of the 16Sr RNA of the small ribosomal subunit facilitating initiation.

Question 48.
Define translation unit.
Answer:
A translation unit in mRNA is the sequence of RNA that is flanked by the start codon on 5’ end and stop codon on 3’ end and codes of polypeptide.

Question 49.
Mention the inhibitory role of tetracycline and streptomycin in bacterial translation.
Answer:
Tetracycline inhibits binding between aminoacyl tRNA and mRNA.Streptomycin inhibits initiation of translation and causes misreading.

Question 50.
At what stage, does the gene expression is regulated?
Answer:
Gene expression can be controlled or regulated at transcriptional or translational levels.

Question 51.
What is a operon? Give example.
Answer:
The cluster of genes with related functions is called operon.
E.g: lac operon in E.coli.

Question 52.
Considering the lac operon of E.coli, name the products of the following genes.

  1. i gene – repressor protein
  2. lac Z gene – fS-galactosidase
  3. Lac Y gene – Permease
  4. lac a gene – transacetylase

Question 54.
Name the human chromosome that has

  1. Chromosome 1 has maximum number of genes (2968 genes)
  2. Chromosome Y has least genes (231 genes)

Question 55.
What are SNPs? Mention its uses.
Answer:
SNPs : Single nucleotide polymorphism. It helps to find chromosomal locations for disease associated sequences and tracing human history.

Question 56.
Mention any four areas where DNA fingerprinting can be used.
Answer:

  1. Forensic analysis
  2. Pedigree analysis
  3. Conservation of wild life
  4. Anthropological studies

Question 57.
Classify nucleic acid based on sugar molecules.
Answer:
There are two types of nucleic acids depending on the type of pentose sugar. Those containing deoxyribose sugar are called Deoxyribo Nucleic Acid (DNA) and those with ribose sugar are known as Ribonucleic Acid (RNA). The only difference between these two sugars is that there is one oxygen atom less in deoxyribose.

Question 58.
Both purines and pyrimidines are nitrogen bases yet they differ. How?
Answer:
Both purines and pyrimidines are nitrogen bases. The purine bases Adenine and Guanine have double carbon – nitrogen ring, whereas cytosine and thymine bases have single carbon nitrogen ring.

Question 59.
How 5’ of DNA differ from its 3’?
Answer:
The 5’ of DNA refers to the carbon in the sugar to which phosphate (P04V) functional group is attached. The 3’ of DNA refers to the carbon in the sugar to which a hydroxyl (OH) group is attached.

Question 60.
State Chargaff’s rule.
Answer:
According to Erwin Chargaff,

  1. Adenine pairs with Thymine with two hydrogen bonds.
  2. Guanine pairs with Cytosine with three hydrogen bonds.

Question 61.
Chemically DNA is more stable than RNA – Justify.
Answer:
In DNA, the two strands being complementary, if separated (denatured) by heating can come together (renaturation) when appropriate condition is provided. Further 2 OH group present at every nucleotide in RNA is a reactive group that makes RNA liable and easily degradable. RNA is also known to be catalytic and reactive. Hence, DNA is chemically more stable and chemically less reactive when compared to RNA. Presence of thymine instead of uracil in DNA confers additional stability to DNA.

Question 62.
Write in simple about semi-conservative mode of DNA replication.
Answer:
Semi-conservative replication was proposed by Watson and Crick in 1953. This mechanism of replication is based on the DNA model. They suggested that the two polynucleotide strands of DNA molecule unwind and start separating at one end. During this process, covalent hydrogen bonds are broken. The separated single strand then acts as template for the synthesis of a new strand. Subsequently, each daughter double helix carries one polynucleotide strand from the parent molecule that acts as a template and the other strand is newly synthesised and complementary to the parent strand.

Question 63.
Draw a simplified diagram of nucleosome and label it.
Answer:

Question 64.
What is a primer?
Answer:
A primer is a short stretch of RNA. It initiates the formation of new strand. The primer produces 3’-OH end on the sequence of ribonucleotides, to which deoxyribonucleotides are added to form a new strand.

Question 65.
Both strands of DNA are not copied during transcription. Give reason.
Answer:
Both the strands of DNA are not copied during transcription for two reasons.

  1. If both the strands act as a template, they would code for RNA with different sequences. This in turn would code for proteins with different amino acid sequences. This would result in one segment of DNA coding for two different proteins, hence complicate the genetic information transfer machinery.
  2. If two RNA molecules were produced simultaneously, double stranded RNA complementary to each other would be formed. This would prevent RNA from being translated into proteins.

Question 66.
What do you mean by a template strand and coding strand?
Answer:
DNA dependent RNA polymerase catalyses the polymerization in only one direction, the strand that has the polarity 3’→ 5’ acts as a template, and is called the template strand. The other strand which has the polarity 5’→ 3’ has a sequence same as RNA (except thymine instead of uracil) and is displaced during transcription. This strand is called coding strand.

Question 67.
Name the factors that are responsible for initiation and termination of transcription in prokaryotes.
Answer:

  1. Sigma factor is responsible for initiation of transcription.
  2. Rho factor is responsible for termination of transcription.

Question 68.
Name the major RNA types of prokaryotes and mention their role.
Answer:
In prokaryotes, there are three major types of RNAs: mRNA, tRNA, and rRNA. All three RNAs are needed to synthesize a protein in a cell. The mRNA provides the template, tRNA brings amino acids and reads the genetic code, and rRNAs play structural and catalytic role
during translation.

Question 69.
Define genetic code.
Answer:
The order of base pairs along DNA molecule controls the kind and order of amino acids found in the proteins of an organism. This specific order of base pairs is called genetic code.

Question 70.
Explain Wobble hypothesis.
Answer:
Wobble Hypothesis is proposed by Crick (1966) which states that tRNA anticodon has the ability to wobble at its 5’ end by pairing with even non-complementary base of mRNA codon.’ According to this hypothesis, in codon-anticodon pairing the third base may not be complementary.

The third base of the codon is called wobble base and this position is called wobble position. The actual base pairing occurs at first two positions only. The importance of Wobbling hypothesis is that it reduces the number of tRNAs required for polypeptide synthesis and it overcomes the effect of code degeneracy.

Question 71.
Explain the nature of eukaryotic ribosome.
Answer:
The ribosomes of eukaryotes (80 S) are larger, consisting of 60 S and 40 S sub units. ‘S’ denotes the sedimentation efficient which is expressed as Svedberg unit (S). The larger subunit in eukaryotes consist of a 23 S RNA and 5Sr RNA molecule and 31 ribosomal proteins. The smaller eukaryotic subunit consist of 18Sr RNA component and about 33 proteins.

Question 72.
Expand and define ORF.
Answer:
Any sequence of DNA or RNA, beginning with a start codon and which can be translated into a protein is known as an Open Reading Frame (ORF).

Question 73.
What are the components of initiation complex of prokaryotic translation?
Answer:
Initiation of translation in E. coli begins with the formation of an initiation complex, consisting of the 30S subunits of the ribosome, a messenger RNA and the charged N-formyl methionine tRNA (f met -1 RNA f met ), three proteinaceous initiation factors (IF 1, IF2, IF3), GTP (Guaniner Tri Phosphate) and Mg 2+ .

Question 74.
Explain the components of operon.
Answer:
Structure of the operon: Each operon is a unit of gene expression and regulation and consists
of one or more structural genes and an adjacent operator gene that controls transcriptional
activity of the structural gene.

  1. The structural gene codes for proteins, rRNA and tRNA required by the cell.
  2. Promoters are the signal sequences in DNA that initiate RNA synthesis. RNA polymerase I binds to the promoter prior to the initiation of transcription.
  3. The operators are present between the promoters and structural genes. The repressor protein binds to the operator region of the operon.

Question 75.
Describe Hershey and Chase experiment. What is concluded by their experiment?
Answer:
Alfred Hershey and Martha Chase (1952) conducted experiments on bacteriophages that infect bacteria. Phage T2 is a virus that infects the bacterium Escherichia coli. When phages (virus) are added to bacteria, they adsorb to the outer surface, some material enters the bacterium, and then later each bacterium lyses to release a large number of progeny phage. Hershey and Chase wanted to observe whether it was DNA or protein that entered the bacteria. All nucleic acids contain phosphorus, and contain sulphur (in the amino acid cysteine and methionine). Hershey and Chase designed an experiment using radioactive isotopes of Sulphur ( 35 S) and phosphorus ( 32 P) to keep separate track of the viral protein and nucleic acids during the infection process.

The phages were allowed to infect bacteria in culture medium which containing the radioactive isotopes 35 S or 32 P. The bacteriophage that grew in the presence of 35 S had labelled proteins and bacteriophages grown in the presence of 32 P had labelled DNA. The differential labelling thus enabled them to identity DNA and proteins of the phage. Hershey and Chase mixed the labelled phages with unlabeled E. coli and allowed bacteriophages to attack and inject their genetic material. Soon after infection (before lysis of bacteria), the bacterial cells were gently agitated in a blender to loosen the adhering phase particles.

It was observed that only 32 P was found associated with bacterial cells and 35 S was in the surrounding medium and not in the bacterial cells. When phage progeny was studied for radioactivity, it was found that it carried only 32 P and not 35 S. These results clearly indicate that only DNA and not protein coat entered the bacterial cells. Hershey and Chase thus conclusively proved that it was DNA, not protein, which carries the hereditary information from virus to bacteria.

Question 76.
Explain the properties of DNA that makes it an ideal genetic material.
Answer:
1. Self Replication: It should be able to replicate. According to the rule of base pairing and complementarity, both nucleic acids (DNA and RNA) have the ability to direct duplications. Proteins fail to fulfill this criteria.

2. Stability: It should he stable structurally and chemically. The genetic material should be stable enough not to change with different stages of life cycle, age or with change in physiology of the organism. Stability as one of property of genetic material was clearly evident in Griffith’s transforming principle. Heat which killed the bacteria did not destroy some of the properties of genetic material. In DNA the two strands being complementary.

if separated (denatured) by heating can come together (renaturation) when appropriate condition is provided. Further 2’ OH group present at every nucleotide in RNA is a reactive group that makes RNA liable and easily degradable. RNA is also known to be catalytic and reactive. Hence, DNA is chemically more stable and chemically less reactive when compared to RNA. Presence of thymine instead of uracil in DNA confers additional stability to DNA.

3. Information storage: It should be able to express itself in the form of ‘Mendelian characters’. RNA can directly code for protein synthesis and can easily express the characters. DNA, however depends on RNA for synthesis of proteins. Both DNA and RNA can act as a genetic material, but DNA being more stable stores the genetic information and RNA transfers the genetic information.

4. Variation through mutation: It should be able to mutate. Both DNA and RNA are able to mutate. RNA being unstable, mutates at a faster rate. Thus viruses having RNA genome with shorter life span can mutate and evolve faster. The above discussion indicates that both RNA and DNA can function as a genetic material. DNA is more stable, and is preferred for storage of genetic information.

Question 77.
How the DNA is packed in an eukaryotic cell? ft
Answer:
In eukaryotes, organization is more complex. Chromatin is formed by a series of repeating units called nucleosomes. Komberg proposed a model for the nucleosome, in which 2 molecules of the four histone proteins H2A, H2B, H3 and H4 are organized to form a unit of eight molecules called histone octamere. The negatively charged DNA is wrapped around the positively charged histone octamere to form a structure called nucleosome. A typical nucleosome contains 200 bp of DNA helix. The histone octameres are in close contact and DNA is coiled on the outside of nucleosome. Neighbouring nucleosomes are connected by linker DNA (HI) that is exposed to enzymes.

The DNA makes two complete turns around the histone octameres and the two turns are sealed off by an HI molecule. Chromatin lacking HI has a beads-on-a-string appearance in which DNA inters and leaves the nucleosomes at random places. HI of one nucleosome can interact with 33l of the neighbouring nucleosomes resulting in the further folding of the fibre.

The chrof&atin fiber in interphase nuclei and mitotic chromosomes have a diameter that vary between 200-300 nm and represents inactive chromatin. 30 nm fibre arises from the folding of nucfeosbme, chains into a solenoid structure having six nucleosomes per turn. This structure is stabilized by interaction between different HI molecules. DNA is a solenoid and packed about,%)_folds. The hierarchical nature of chromosome structure is illustrated.

Additional set of pteins are required for packing of chromatin at higher level and are referred to as non-histone chromosomal proteins (NHC). In*,a typical nucleus, some regions of chromatin are Ibosely packed (lightly stained) and are referred to as euchromatin. The chromatin that is,-tightly packed (stained darkly) is called heterochromatin. Euchromatin is transcriptionally active and heterochromatin is transcriptionally inactive.

Question 78.
Meselson and Stahl’s experiment proved the semi-coflBptervation mode of DNA replication. Explain.
Answer:
The mode of DNA replication was determined in 1958 by Meselson and Stahl. They designed an experiment to distinguish between semi-conservative, conservative and dispersive replications. In their experiment, they grew two cultures of E.coli for many generations in separate media. The ‘heavy’ culture was grown in a medium in which the nitrogen source (NH4CI) contained the heavy isotope 15 N and the ‘ light’ culture was grown in a medium in which the nitrogen source contained light isotope 14 H for many generations. At the end of growth, they observed that the bacterial DNA in the heavy culture contained only 15 N and in the light culture only 14 N. The heavy DNA could be distinguished from light DNA ( 15 N from 14 N) with a technique called Cesium Chloride (CsCl) density gradient centrifugation. In this process, heavy and light DNA extracted from cells in thtytwo cultures settled into two distinct and separate bands (hybrid DNA).

The heavy culture ( 15 N) was then transferred into a medium that had only NH4CI, and took samples at various definite time intervals (20 minutes duration). After the first replication, they extracted DNA and subjected it to density gradient centrifugation. The DNA settled into a band that was intermediate in position between the previously determined heavy and light bands. After the second replication (40 minutes duration), they again extracted DNA samples,and this time found the DNA settling into two bands, one at the light band position and one at intermediate position. These results confirm Watson and Crick’s semi – conservative replication hypothesis.

Question 79.
Give a detailed account of a transcription unit.
Answer:
A transcriptional unit in DNA is defined by three regions, a promoter, the structural gene and a terminator. The promoter is located towards the 5 ’ end. It is a DNA sequence that provides binding site for RNA polymerase. The presence of promoter in a transcription unit, defines the template and coding strands. The terminator region located towards the 3’ end of the coding strand contains a DNA sequence that causes the RNA polymerase to stop transcribing. In eukaryotes the promoter has AT rich regions called TATA box (Goldberg- Hogness box) ‘ and in prokaryotes this region is called Pribnow box.

Besides promoter, eukaryotes also require an enhancer. The two strands of the DNA in the structural gene of a transcription unit have opposite polarity. DNA dependent RNA polymerase catalyses the polymerization in only one direction, the strand that has the polarity 3’→5’ acts as a template, and is called the template strand. The other strand which has the polarity 5’→ 3’ has a sequence same as RNA (except thymine instead of uracil) and is displaced during transcription. This strand is called coding strand

The structural gene may be monocistronic (eukaryotes) or polycistronic (prokaryotes). In eukaryotes, each mRNA carries only a single gene and encodes information for only a single protein and is called monocistronic mRNA. In prokaryotes, clusters of related genes, known as operon, often found next to each other on the chromosome are transcribed together to give a single mRNA and hence are polycistronic.

Question 80.
Explain the transcription process in prokaryotes with needed diagram.
Answer:
In prokaryotes, there are three major types of RNAs:
mRNA, tRNA, and rRNA. All three RNAs are needed to synthesize a protein in a cell. The mRNA provides the template, tRNA brings amino acids and reads the genetic code, and rRNAs play structural and catalytic role during translation. There is a single DNA-dependent RNA polymerase that catalyses transcription of all types of RNA. It binds to the promoter and initiates transcription (Initiation).

The polymerases binding sites are called promoters. It uses nucleoside triphosphate as substrate and polymerases in a template depended fashion following the rule of complementarity. After the initiation of transcription, the polymerase continues to elongate the RNA, adding one nucleotide after another to the growing RNA chain. Only a short stretch of RNA remains bound to the enzyme, when the polymerase reaches a terminator at the end of a gene, the, nascent RNA falls off, so also the RNA polymerase. The RNA polymerase is only capable of catalyzing the process of elongation. The RNA polymerase associates transiently with initiation factor sigma (a) and termination factor rho (p) to initiate and terminate the transcription, respectively.

Association of RNA with these factors instructs the RNA polymerase either to initiate or terminate the process of transcription. In bacteria, since the mRNA does not require any processing to become active and also since transcription and translation take place simultaneously in the same compartment sincethere is no separation of cytosol and nucleus in bacteria), many times the translation can begin much before the mRNA is fully transcribed. This is because the genetic material is not separated from other cell organelles by a nuclear membrane consequently transcription and translation can be coupled in bacteria.

Question 81.
Write the salient features of genetic code.
Answer:
The salient features of genetic code are as follows:

  1. The genetic codon is a triplet code and 61 codons code for amino acids and 3 codons do not code for any amino acid and function as stop codon (Termination).
  2. The genetic code is universal. It means that all known living systems use nucleic acids and the same three base codons (triplet codon) direct the synthesis of protein from amino acids. For example, the mRNA (UUU) codon codes for phenylalanine in all cells of all organisms. Some exceptions are reported in prokaryotic, mitochondrial and chloroplast genomes. However similarities are more common than differences.
  3. A non-overlapping codon means that the same letter is not used for two different codons. For instance, the nucleotide sequence GUTJ and GUC represents only two codons.
  4. It is comma less, which means that the message would be read directly from one end to the other i.e., no punctuation are needed between two codes.
  5. A degenerate code means that more than one triplet codon could code for a specific amino acid. For example, codons GUU, GUC, GUA and GUG code for valine.
  6. Non-ambiguous code means that one codon will code for one amino acid.
  7. The code is always read in a fixed direction i.e. from 5’→3’ direction called polarity.
  8. AUG has dual functions. It acts as a initiator codon and also codes for the amino acid methionine.
  9. UAA, UAG (tyrosine) and UGA (tryptophan) codons are designated as termination (stop) codons and also are known as “non-sense” codons.

Question 82.
Mutations on genetic code affects the phenotype. Describe with example.
Answer:
The simplest type of mutation at the molecular level is a change in nucleotide that substitutes one base for another. Such changes are known as base substitutions which may occur spontaneously or due to the action of mutagens. A well studied example is sickle cell anaemia in humans which results from a point mutation of an allele of β-haemoglobin gene (βHb).

A haemoglobin molecule consists of four polypeptide chains of two types, two a chains and two P-chains. Each chain has a heme group on its surface. The heme groups are involved in the binding of oxygen. The jruman blood disease, sickle cell anaemia is due to abnormal haemoglobin. This abnormality in haemoglobin is due to a single base substitution at the sixth codon of the beta globingene from GAG to GTG in p -chain of haemoglobin.

It results in a change of amino acid glufeniic acid to valine at the 6th position of the p -chain. This is the classical example of point mutation that results in the change of amino acids residue glutamic acid to valine. The mutant haemoglobin undergoes polymerisation under oxygen tension causing the change in the shape of the RBC from biconcave to a sickle shaped structure.

Question 83.
Explain the mechanism of AteArperon of the E-coli.
Answer:
The Lac (Lactose) operon: The metabolism of lactose in E.coli requires three enzymes – permease, P-galactosidase (P-gat) and transacetylase. The enzyme permease is needed for entry of lactose into the cell, Pjgglactosidase brings about hydrolysis of lactose to glucose and galactose, while transacety gtransfers acetyl group from acetyl Co A to P-galactosidase. The lac operon consists of one-regulator gene (T gene refers to inhibitor) promoter sites (p), and operator site (o). Besides these, it has three structural genes namely lac z, y and lac a. The lac ‘z’ gene codes for P-gaiaqtttsidase, lac ‘y’ gene codes for permease and ‘a’ gene codes for transacetylase.

Jacob and Monod proposed the classical model of Lac operon to explain gene expression and regulation in E.coli. In lac a polycistronic structural gene is regulated by a common promoter and regulatory genfc When the cell is using its normal energy source as glucose, the ‘i’ gene transcribes a repressor mRNA and after its translation, a repressor protein is produced. It binds to the operator region of the operon and prevents translation, as a result, P-galactosidase is not produced. In the absence of preferred carbon source such as glucose, if lactose is available as an energy source for the bacteria then lactose enters the cell as a result of permease enzyme. Lactose acts as an inducer and interacts with the repressor to inactivate it.

The repressor protein binds to the operator of the operon and prevents RNA polymerase from transcribing the operon. In the presence of inducer, such as lactose or allolactose, the repressor is inactivated by interaction with the inducer. This allows RNA polymerase to bind to the promotor site and transcribe the operon to produce lac mRNA which enables formation of all the required enzymes needed for lactose metabolism. This regulation of lac operon by the repressor is an example of negative control of transcription initiation.

Question 84.
What are the objectives of Human Genome project?
Answer:
The main goals of Human Genome Project are as follows:

  1. Identify all the genes (approximately 30000) in human DNA.
  2. Determine the sequence of the three billion chemical base pairs that makeup the human DNA.
  3. To store this information in databases.
  4. Improve tools for data analysis.
  5. Transfer related technologies to other sectors, such as industries.
  6. Address the ethical, legal and social issues (ELSI) that may arise from the project.

Question 85.
Write the salient features of Human Genome Project.
Answer:

  1. Although human genome contains 3 billion nucleotide bases, the DNA sequences that encode proteins make up only about 5% of the genome.
  2. An average gene consists of 3000 bases, the largest known human gene being dystrophin with 2.4 million bases.
  3. The function of 50% of the genome is derived from transposable elements such as LINE and ALU sequence.
  4. Genes are distributed over 24 chromosomes. Chromosome 19 has the highest gene density. Chromosome 13 and Y chromosome have lowest gene densities.
  5. The chromosomal organization of human genes shows diversity.
  6. There may be 35000-40000 genes in the genome and almost 99.9 nucleotide bases are exactly the same in all people.
  7. Functions for over 50 percent of the discovered genes are unknown.
  8. Less than 2 percent of the genome codes for proteins.
  9. Repeated sequences make up very large portion of the human genome. Repetitive sequences have no direct coding functions but they shed light on chromosome structure, dynamics and evolution (genetic diversity).
  10. Chromosome 1 has 2968 genes, whereas chromosome ’Y’ has 231 genes.
  11. Scientists have identified about 1.4 million locations, where single base DNA differences (SNPs – Single nucleotidepolymorphism – pronounce as ‘snips’) occur in humans. Identification of ‘SNIPS’ is helpful in finding chromosomal locations for disease associated sequences and tracing human history.

Question 86.
Describe the principle involved in DNA fingerprinting technique.
Answer:
The DNA fingerprinting technique was first developed by Alec Jeffreys in 1985. The DNA of a person and finger prints are unique. There are 23 pairs of human chromosomes with 1.5 million pairs of genes. It is a well known fact that genes are segments of DNA which differ in the sequence of their nucleotides. Not all segments of DNA code for proteins, some DNA segments have a regulatory function, while others are intervening sequences (introns) and still others are repeated DNA sequences. In DNA fingerprinting, short repetitive nucleotide sequences are specific for a person. These nucleotide sequences are called as variable number tandem repeats (VNTR). The VNTRs of two persons generally show variations and are useful as genetic markers.

DNA finger printing involves identifying differences in some specific regions in DNA sequence called repetitive DNA, because in these sequences, a small stretch of DNA is repeated many times. These repetitive DNA are separated from bulk genomic DNA as different peaks during density gradient centrifugation. The bulk DNA forms a major peak and the other small peaks are referred to as satellite DNA. Depending on base composition (A: T rich or G : C rich), length of segment and number of repetitive units, the satellite DNA is classified into many sub – categories such as micro-satellites and mini satellites, etc.

These sequences do not code for any proteins, but they form a large portion of human genome. These sequences show high degree of polymorphism and form the basis of DNA fingerprinting. DNA isolated from blood, hair, skin cells, or other genetic evidences left at the scene of a crime can be compared through VNTR patterns, with the DNA of a criminal suspect to determine guilt or innocence. VNTR patterns are also useful in establishing the identity of a homicide victim, either from DNA found as evidence or from the body itself.

Question 87.
Draw a flow chart depicting the steps of DNA finger printings technique
Answer:

Higher Order Thinking Skills (HOTs) Questions

Question 1.
A mRNA strand has a series of triplet codons of which the first three codons are given below
(a) AUG
(b) UUU
(c) UGC
(i) Name the amino acid encoded by these triplet codons.
(ii) Mention the DNA sequence from which these triplet codons would have transcribed?
Answer:
(i) AUG codes for Methionine
UUU codes for Phenylalanine
UGC codes for Cysteine
(ii) TAC sequence of DNA is transcribed to AUG
AAA sequence of DNA is transcribed to UUU
ACG sequence of DNA is transcribed to UGC

Question 2.
Given below are the structures of tRNA molecules which are involved in translation process. In one tRNA, codon is mentioned but not the amino acid. In another tRNA molecule, amino acid is named and not the codon. Complete the figure by mentioning the respective amino acids and codons.
Answer:

Question 3.
A DNA fragment possesses 32 adenine bases and 32 cytosine bases. How many total number of nucleotides does that DNA fragment contains? Explain.
Answer:
128 nucleotides. Adenine always pair Thymine base. If there are 32 adenine bases then there must be 32 Thymine bases. Similarly cytosine pairs with guanine. If cytosine bases are 32 in number the guanine bases will be equal to cytosine. So it make a total of 128 nucleotides.

Question 4.
Following is a DNA sequence representing a part of gene TAC TCG CCC TAT UAA CCC AAA ACC TCT using this derive A.

  1. The RNA transcript
  2. The spliced mRNA (consider all the codons with two Aderine bases are introns)
  3. The total number of aminoacids coded by the mRNA
  1. RNA transcript: AUG UGC GGG AUA GGG UUU UGG AGA
  2. Spliced mRNA : AUG UGC GGG GGG UUU UGG
  3. 6 aminoacids are coded by mRNA

Question 5.
Complete the molecular processes by naming them

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Do proteins contain phosphorous? If its true then how alfred hershey and martha chase used the statement that proteins do not contain phosphorous? - Biology

178 notecards = 45 pages ( 4 cards per page)

Sells and Jeans Final

1.) Of all nature’s molecules, nucleic acids are unique in their ability to direct their own replication. What feature(s) of DNA causes it to have this unusual characteristic?

The information contained in one strand is the same as that in the other.

2.) Which of the following is a correct statement about the two DNA strands in a typical double stranded DNA molecule.

d. The two strands will be antiparallel to each other (the 5’ end of one strand will match up with the 3’ end of the other).

3.) Each nucleotide is composed of two constant parts (a phosphate group and a pentose sugar) and one variable nitrogenous base. Which of the following is a not a difference between RNA and DNA?

e. Nucleotides are added to the 5’ end of RNA and to the 3’ end of DNA molecules.

4.) DNA polymerases are notable enzymes both for their speed and their accuracy. Consider a nucleotide that is being added by a DNA polymerase to a growing polynucleotide chain. To which position on the ribose sugar of the nucleotide already present is a phosphodiester bond made?

5.) The sum total of the DNA content within each of an organism’s cells is often referred to as the organism’s “C-value.” It is reasonable to assume that C-value increases in a way that is proportional to organismal complexity. Which of the following organisms is most likely to have the smallest C-value?

6.) Individuals with sickle cell anemia have a single difference in the amino acid sequence of their -globin protein relative to people who do not have the disease. The specific mutation, a substitution, also resulted in the failure of a restriction enzyme to generate a particular fragment of DNA in afflicted individuals. What was that mutation?

c. A change that results in the incorporation of a valine instead of a glutamic acid as the sixth amino acid in the -globin protein.

7.) Even though there are only four different nucleotides in DNA molecules, the specific order in which they are found is used by cells to spell out the order in which 20 different amino acids are used in making proteins. How many of the 64 possible triplet codes are reserved for use as stop codons?

8.) The wheat genome is about five times larger than the human genome. It may contain more protein coding regions (genes) as well. How can it be that humans are able to make many more different proteins than wheat?

b. Humans do a lot more alternative splicing than wheat does.

9.) Griffith found that Streptococcus pnemoniae that formed smooth colonies were capable of killing mice but that bacteria from rough colonies could not. What would he have found if he had infected mice with just the DNA of the smooth bacteria?

The mice would not become ill.

10.) The viruses that Alfred Hershey and Martha Chase radiolabelled with radioactive isotopes of phosphorous (32P) and of sulfur (35S) were T2 bacteriophage. T2 viruses, like the T4 viruses that destroy their host cell’s genomic DNA as the first stage of their infection cycle, have an elaborate, “lunar-lander” like exterior when viewed with an electron microscope. Many viruses have less elaborate outer coats but from what bio-polymer are they all constructed?

11.) Meselson and Stahl designed an experiment using two different isotopes of nitrogen (15N and 14N) that tested Watson and Crick’s prediction that DNA replication occurred by a semi-conservative process. What would they have found after one round of cell divisions if DNA replication was actually conservative?

d. Some double stranded DNA molecules would contain only 15N and others would contain only 14N.

12.) DNA is typically present in cells as a double helix consisting of two strands of polynucleotides. Why is the information on one strand of a DNA molecule redundant with that on its complementary strand?

c. For every A and G on one strand, there is a T and C on the other.

13.) The mapping of human chromosome 21 (one of the smallest) required 198 STSs that were used to screen more than 70,000 YAC clones and identify a total of 810 YACs that yielded an overlapping array of clones that spanned the entire chromosome. Why is it more convenient to use yeast artificial chromosome (YAC) clones for this kind of mapping than clones in bacterial vectors like plasmids?

d. The cloned regions in YACs can be many times bigger than those found in bacterial clones so fewer “steps” need to be taken in chromosome walks.

14.) As a general rule, eukaryotes are more complex than prokaryotes. Which of the following differences between prokaryotes and eukaryotes requires eukaryotic RNA polymerases and promoters to be substantially more complicated than their prokaryotic counterparts?

b. Eukaryotes have to accomplish tissue-specific regulation of gene expression.

15.) If one strand of DNA has its nitrogenous bases arranged in this order: 5’ - TACGAA - 3’, what would be the order of nucleotides in its complementary strand?

16.) It will not be long before physicians routinely use information from the complete genome sequence of their patients. That information should allow the medicines that are proscribed to be tailored to the particular metabolism so as to have the maximum effect with the smallest possible dose and side effects. What is this kind of individualized medicine called?

17.) An average of three to four chiasma can be seen along the length of every human chromosome during prophase I of meiosis. Some regions appear to be less likely to form chiasma than others though. What effect would this have on recombination-frequency based gene mapping studies in these regions?

e. Genes that seem to be relatively close to each other may really be distantly linked.

As cells translate new proteins, appropriate amino acids are carried to the ribosomes by an RNA molecule with a characteristic cruciform (clover leaf) secondary structure. What are members of that family of RNA molecules called?

Proteins are never made directly from the information stored within DNA. Rather, information is transcribed into an RNA intermediate by RNA polymerase. What is the name of those RNA intermediates?

Which of the many kinds of RNA molecules present within cells is specifically recognized by ribosomes due to the presence of a modification (the addition of a “cap” – a single guanine by an unusual 5’-to-5’ tri-phosphate linkage) at its 5’ end?

Unlike prokaryotes, eukaryotes extensively modify the RNA transcripts of their genes. What are eukaryotic RNAs destined for translation into proteins called before that processing is complete (i.e. its introns have not yet been spliced out or it has not yet been polyadenylated).

. hnRNAs (heterogeneous RNAs).

22.) Thomas Hunt Morgan in his studies with Drosophila was the first to realize that heritable traits (genes) were correlated with chromosomes. How was he able to reconcile the fact that Drosophila appeared to have a much larger number of traits than chromosomes?

c. He found that many traits were coded for on each of the four chromosomes.

23.) George Beadle and Edward Tatum used X-rays to alter the genetic material of a pink bread mold. Subsequent tests of the molds’ ability to synthesize the materials it needed to grow revealed that the alteration of specific genes also altered specific enzymes. These observations were the basis of a theory that earned them a Nobel prize in 1958. What was that theory?

c. For every one enzyme in an organism there was one gene.

24.) What is the central dogma of molecular biology?

Information passes from DNA to RNA to protein.

25.) In both prokaryotes and eukaryotes the first amino acid incorporated into all new proteins is methionine. Two researchers found that prokaryotic ribosomes also required the sequence 5’-UAAGGAGG-3’ (named the Shine-Delgarno sequence in their honor) 5’ to a start codon in order to start translation. What causes ribosomes to release an mRNA and stop adding amino acids to a growing polypeptide?

c. When ribosomes encounter a codon for which there are no charged tRNAs they stop translating mRNAs.

26.) Jacob and Monod won a Nobel prize in 1961 for their work with partial diploids and the lac operon. If a prokaryotic cell had a mutant (non-functioning) version of the lacI gene but a normal copy of the lacZ gene, how would it respond to the presence and absence of lactose?

e. It would make beta galactosidase both when lactose was present and absent.

27.) A chemical whose abbreviated name is IPTG is sufficiently similar to lactose to bind to the lac repressor bu it cannot be degraded by the enzyme -galactosidase. What effect would the presence of IPTG have on the expression of the lac operon?

a. The lac operon would be expressed at high levels even when there was no lactose.

28.) The binding site for the protein product of the lacI gene is normally found only once in the entire E. coli genome – the operator of the lac operon. What would happen if molecular biologists genetically engineered an E. coli genome in such a way as to include the nucleotide sequence of that binding site 50 nucleotides upstream of the promoter of the trp operon?

a. There would be no effect on the expression of the trp operon.

Different Hfr strains exhibit different origins and directions of transfer of their bacterial chromosome. The order of transfer of genes close to the origin (O) of five different Hfr strains is given below:

Hfr strain/ order of transfer of genes

Which of the following is a portion of the correct conjugational map of an E. coli chromosome?

30.) F-factors are one of many different kinds of plasmids (extra-chromosomal pieces of DNA) found in prokaryotes. What exterior structure that is fundamentally important to bacterial sex do most of the 25 genes of a typical F-factor code for?

31.) A protein coded for by a gene immediately upstream of the lactose operon was found to specifically bind to the promoter in front of the beta-galactosidase gene but only when no lactose was present. When this protein was bound, RNA polymerase could not bind and begin transcription of the lactose operon. What type of regulation of gene expression is that protein involved in?

32.) Most biological pathways require cells to possess two or more different enzymes in equal quantities for the efficient conversion of a starting material into an end-product. What feature of prokaryotes helps assure coordinated expression of genes in such pathways?

b. Genes with related functions are usually found in a single operon.

33.) Which of the following is not an important difference between E. coli’s single chromosome and the many chromosomes of human cells?

d. E. coli’s chromosome has long regions of non-coding sequence interrupting the coding regions of its genes.

34.) When DNA is isolated from eukaryotic cells it is usually found to be tightly associated with a large amount of highly positively charged proteins called histones. The fact that histones are virtually identical across a wide variety of eukaryotic organisms implies that their function must be critically important. What is the complex of histones and DNA in eukaryotic nuclei called?

35.) Humans who are lactose intolerant no longer make the equivalent enzyme to beta galactosidase in the lac operon. Which of the following mutations to the lac operon of a prokaryotic cell might make it unable to synthesize beta galactosidase?

A deletion of the start codon.

36.) By convention, the first nucleotide incorporated into an mRNA by an RNA polymerase is said to be the “+1” position for that gene. In prokaryotes, two relatively well-conserved sets of nucleotide sequences 35 and 10 base pairs (bp) upstream of that position are required for RNA polymerase to bind and be correctly oriented. How large are those Pribnow and TATA boxes?

37.) Four different but complementary approaches were employed by the molecular biologists that obtained the complete nucleotide sequence of the human genome. Those approaches are: linkage mapping, physical mapping, sequencing and comparative sequence analysis. How do linkage and physical mapping differ?

a. They use analyses of families and pieces of clones, respectively.

38.) Sir Alec Jefferies was the first person to actually see that the length of eukaryotic mRNAs was actually much shorter than the genes that coded for them when he hybridized mRNAs to genomic DNA and looked at the products with an electron microscope. What are the coding portions that are spliced out of the primary transcripts of eukaryotic genes?

39.) The chicken ovalbumin gene is expressed exclusively in hen oviduct cells. Which of the following is likely to be true?

a. It would be in heterochromatin in most tissues but in euchromatin in oviduct cells.

40.) Many eukaryotic genomes contain vast tracts of non-coding or redundant DNA. Plants seem to have frequently accumulated numerous duplications of their whole genomes over the course of their evolutionary histories. What happens to a gene whose function is no longer needed because its copy serves its original purpose?

c. Some acquire new functions that confer a selective advantage and come under functional constraint

41.) Edward Southern was the first person to show that DNA molecules attached to paper-like membranes could specifically hybridize with other, free-floating DNA molecules that had been radio-labeled. Upon which of the following important features of DNA does Southern’s technique rely?

c. Complementary strands of single stranded DNA have an affinity for each other.

42.) Once translation starts, ribosomes are careful to move along mRNA molecules in three nucleotide steps. What would happen if ribosomes were to disregard that reading frame?

d. The amino acids downstream of the frameshift would be incorrect.

43.) The ability to determine the order in which nucleotides appear within the DNA of our chromosomes was obviously a major advance in the study of biology since it allowed us to directly examine our genetic code. The first DNA sequencing method to be used won Maxam and Gilbert a Nobel Prize and was widely used until Sanger developed a new method roughly ten years later. How do Maxam and Gilbert’s and Sanger’s methods differ?

c. The first involves chemical degradation of DNA while Sanger’s is an enzymatic process.

44.) Agarose gel electrophoresis of DNA involves the movement of DNA in an electric field through the pores of a rectangular slab of agarose (a firm, gelatin-like material). Since DNA has a pronounced negative charge, it naturally moves toward positive electrodes when an electric current is applied to the gel. How do large pieces of DNA move relative to small ones in such a system?

a. Slower because they become more tangled with the gel matrix.

45.) The discovery in 1974 of a group of enzymes in prokaryotes that break DNA chains at very specific sequences of nucleotides proved to be invaluable for researchers interested in both cloning and sequencing DNA. What are enzymes such as BamHI which cleaves DNA at the sequence 5’-GGATCC-3’ called?

46.) The cloning of a sheep, Dolly, from one of her own mammary gland cells in 1997 made possible many new opportunities for molecular biology. Which of the following would not be true of a human clone?

b. It would have the same memories as its “parent.”

47.) Which of the following lists things in increasing order of size?

d. Amino acids, tRNA, mRNA, ribosomes.

48.) Diabetics generally need regular doses of the hormone insulin to survive. Which of the following represents a reasonable way of making a transgenic organism that could provide insulin for such treatments?

e. Putting a bacterial promoter in front of the DNA sequence for the human insulin protein and inserting that construct as part of a plasmid into a bacterial cell.

49.) Keri Mullis won a Nobel Prize for his description of what is now a very popular methodology for making billions of copies of a specific portion of a genome through repeated cycles of heating and cooling of the genome with a DNA polyermase, primers that flank the region of interest, and just a few additional reagents. What is this amplification process called?

. Polymerase chain reaction (PCR).

50.) The burning of glucose releases 684 kcal per mole. The synthesis of glucose is just as energetically unfavorable as its combustion is favorable. How are living things able to make glucose synthesis a spontaneous reaction?

e. By coupling the unfavorable energetics to chemical reactions that are energetically favorable like the conversion of ATP to ADP and Pi.

51.) Complementary DNA sequences have a natural affinity for each other primarily because of their ability to form hydrogen between their nitrogenous bases. The three hydrogen bonds that form between guanines and cytosines and the two that form between adenines and thymines are known as “base pairs.” How do these hydrogen bonds differ from the covalent bonds that connect a nitrogenous base to a sugar like deoxyribose in DNA?

d. They do not take as much energy to break because no sharing of electrons actually takes place.

52.) Most of the energy extracted from glucose during glycolysis is initially harvested in the citric acid cycle as NADH. The high-energy electrons of NADH are used to create a proton gradient within mitochondria. Which of the following energy-rich compounds is the one that is synthesized as a direct result of that proton gradient?

53.) When populations of prokaryotic organisms coordinate their activities they can form communities that begin to rival the complexity of some eukaryotic organisms. How do such populations of bacteria determine how many bacteria are nearby?

54.) pH and pOH are both related to each other and are both logarithmic scales. What would be the concentration of hydrogen ions in a solution whose pOH is 12?

55.) In the first step of glycolysis a six-carbon sugar, glucose, interacts with ATP and, as a result, ADP and a six-carbon sugar with one phosphate group attached to it is formed. What is the name of the enzyme that catalyzes that reaction?

56.) All living things on Earth use the same set of 20 amino acids to build proteins. How do those 20 amino acids differ from each other?

b. Each has its own R-group.

57.) Carbon (6C) has only one electron in each of the four orbitals of its second level. Which of the atoms listed below would you expect to be much more electronegative than carbon?

58.) The components of the nuclear membranes that are taken apart close to the end of prophase are re-used to assemble the nuclear envelope of daughter cells. At what stage of mitosis do those new nuclear membranes form?

59.) The different building blocks used to make each of the four kinds of biopolymers cause each kind of macromolecule to have a different function within living cells. What are the primary roles of carbohydrates, lipids, proteins and nucleic acids?

e. Energy transfer, energy storage, catalysis and information storage, respectively.

60.) One of the many fruit fly genetics experiments performed by Thomas Hunt Morgan involved mating a homozygous white-eyed and yellow bodied fly (yw/yw) with another fly that was heterozygous for those traits (yw/++). Morgan found that 487 progeny were white-eyed and yellow bodied, 491 had normal eye and body color, 10 were white-eyed and normal body colored and 12 were yellow-bodied and normal eye colored. How many map units apart are the y and w genes?

1.) In a series of mating experiments Thomas Hunt Morgan found that the genes responsible for black body color and cinnabar eyes in fruit flies recombined with a frequency of 9%. He also found that the gene for cinnabar eyes was 9.5 map units from the gene for vestigial wings. What can you conclude about the genes for black body color and vestigial wings?

b. They recombine with a frequency of either 0.5% or 18.5%.

2.) Birds determine sex differently than mammals – roosters are XX and hens are XY. Chicken farmers often find it useful to be able to determine the sex of newborn chicks from the color of their plumage. How can this be done by using an X-linked dominant allele (S) for silver plumage and a recessive allele (s) for gold plumage?

c. Golden roosters mated with silver hens will always produce golden hens and silver roosters.

3.) DNA is “de-oxy” nucleic acid because the hydroxyl group on the 2’ carbon of ribose is replaced with a hydrogen atom. Which of the following carbons is also “de-oxy” in the dide-oxy nucleotides used for the Sanger method of DNA sequencing?

4.) The further apart two genes are upon a single chromosome, the more likely they are to have a synaptonemal complex form between them during meiosis. Why does the resulting recombination provide an advantage to sexually reproducing organisms?

d. It allows advantageous genes to lose their association with disadvantageous ones.

5.) Arthur Kornberg was the first to isolate from E. coli the enzyme that is responsible for replicating DNA in living cells: DNA polymerase. Which of the following is a correct description of the 5’ end of a DNA strand?

b. The 5’ hydroxyl (-OH) group is not attached to anything but a phosphate group.

6.) What other somewhat surprising feature(s) of DNA polymerases did Arthur Kornberg discover in the course of his experimentation?

e. They must start with at least a short stretch of double stranded DNA as they add nucleotides in a 5’ to 3’ direction.

7.) Nucleic acids are unusual in their ability to direct their own replication. What feature(s) of DNA causes it to have this unusual characteristic?

b. The information contained in one strand is the same as that in the other.

8.) Which of the following is a correct statement about the two DNA strands in a typical double stranded DNA molecule.

d. The two strands will be antiparallel to each other (the 5’ end of one strand will match up with the 3’ end of the other).

9.) In 1978 researchers confirmed Vernon Ingram’s hypothesis that the single amino acid difference in the -globin protein found in individuals with sickle cell anemia relative to those without the disease was due to a single difference in the nucleotide sequence of that gene. The specific mutation, a substitution, also resulted in the failure of a restriction enzyme to generate a particular fragment of DNA in afflicted individuals. What was that mutation?

c. The presence of an A at a particular position within healthy individual’s -globin gene versus the presence of a T in afflicted individuals at the same position.

10.) Each nucleotide is composed of two constant parts (a phosphate group and a pentose sugar) and one variable nitrogenous base. Which of the following is a difference between RNA and DNA?

c. RNA is used to convey information to ribosomes and DNA is used to store information.

11.) The sum total of the DNA content within each of an organism’s cells is often referred to as the organism’s “C-value.” As a general rule, C-values are correlated with organismal complexity but there are striking exceptions that, taken together, are known as the C-value paradox. What is a reasonable conclusion that can be drawn from the observation that the frog genome is hundreds of times larger than the human genome?

b. Much of the frog genome must be “junk.”

12.) Griffith found that Streptococcus pneumoniae that formed smooth colonies were capable of killing mice but that bacteria from rough colonies could not. A mixture of heat-killed smooth bacteria and live rough bacteria killed mice too. What kind of bacteria did he find in the mice after they died?

13.) Even though there are only four different nucleotides in DNA molecules, the specific order in which they are found is used by cells to spell out the order in which 20 different amino acids are used in making proteins. There are 64 possible triplet codons with four different nucleotides. How many different possible triplet codons would there be if there were six different nucleotides?

14.) As result of Griffith’s work, Alfred Hershey and Martha Chase radio-labeled viruses with a radioactive isotopes of phosphorous (32P) and of sulfur (35S) in 1952. After allowing the viruses to interact with living bacterial cells for about half an hour, where did they find the 32P and 35S?

. Inside and outside the cells, respectively.

15.) The viruses that Alfred Hershey and Martha Chase radiolabelled with radioactive isotopes of phosphorous (32P) and of sulfur (35S) were T2 bacteriophage. T2 viruses, like the T4 viruses that destroy their host cell’s genomic DNA as the first stage of their infection cycle, have an elaborate, “lunar-lander” like exterior when viewed with an electron microscope. The outer coat of all viruses are made of proteins. Which of the following molecules might be used to store their genetic information?

16.) Meselson and Stahl designed an experiment using two different isotopes of nitrogen (15N and 14N) that tested Watson and Crick’s prediction that DNA replication occurred by a semi-conservative process. What would they have found after one round of cell divisions if DNA replication was actually conservative?

d. Some double stranded DNA molecules would contain only 15N and others would contain only 14N.

17.) The mapping of human chromosome 21 (one of the smallest) required 198 sequence tagged sites (STSs) that were used to screen more than 70,000 yeast artificial chromosome (YAC) clones and identify a total of 810 YACs that yielded an overlapping array of clones that spanned the entire chromosome. Why is it more convenient to use the larger yeast artificial chromosome (YAC) clones for this kind of mapping than smaller clones in bacterial vectors like plasmids?

d. The larger the cloned portions of a chromosome, the easier it is to make a physical map of the chromosome.

18.) As a general rule, eukaryotes are more complex than prokaryotes. Which of the following differences between prokaryotes and eukaryotes requires eukaryotic RNA polymerases and promoters to be substantially more complicated than their prokaryotic counterparts?

b. Eukaryotes have to accomplish tissue-specific regulation of gene expression.

19.) If one strand of DNA has its nitrogenous bases arranged in this order: 5’ - TACGAA - 3’, what would be the order of nucleotides in its complementary strand?

20.) In human DNA, the four nucleotides are present in the following percentages: A = 30.2% T = 30.2% G = 19.8% and C = 19.8%. How are these numbers consistent with Chargaff’s rule?

d. The molar ratio of A’s and T’s in DNA are the same.

Which of the many kinds of RNA molecules present within cells is specifically recognized by ribosomes due to the presence of a modification (the addition of a “cap” – a single guanine by an unusual 5’-to-5’ tri-phosphate linkage) at its 5’ end?

Unlike prokaryotes, eukaryotes extensively modify the RNA transcripts of their genes. What are eukaryotic RNAs destined for translation into proteins called before that processing is complete (i.e. its introns have not yet been spliced out or it has not yet been polyadenylated).

hnRNAs (heterogeneous RNAs).

As cells translate new proteins, appropriate amino acids are carried to the ribosomes by an RNA molecule with a characteristic cruciform (clover leaf) secondary structure. What are members of that family of RNA molecules called?

Proteins are never made directly from the information stored within DNA. Rather, information is transcribed into an RNA intermediate by RNA polymerase. What is the name of those RNA intermediates?

25.) It will not be long before physicians routinely use information from the complete genome sequence of their patients. That information should allow the medicines that are proscribed to be tailored to the particular metabolism so as to have the maximum effect with the smallest possible dose and side effects. What is this kind of individualized medicine called?

26.) George Beadle and Edward Tatum shared a Nobel prize in 1958 for elucidating the precise relationship between genes and enzymes. The bread mold that they experimented with usually grew well on a media containing only simple compounds such as sugar, salts and nitrogen. Why did some of the auxotrophic mutants they found require additional compounds like arginine in their media?

e. One or more genes involved in the synthesis of arginine had undergone a mutation.

27.) In both prokaryotes and eukaryotes the first amino acid incorporated into all new proteins is methionine. Two researchers found that prokaryotic ribosomes also required the sequence 5’-UAAGGAGG-3’ (named the Shine-Delgarno sequence in their honor) 5’ to a start codon in order to start translation. What causes ribosomes to release an mRNA and stop adding amino acids to a growing polypeptide?

c. When ribosomes encounter a codon for which there are no charged tRNAs they stop translating mRNAs.

28.) What is the central dogma of molecular biology?

a. Information passes from DNA to RNA to protein.

29.) During the 1960’s, the genetic code was deciphered by two different research groups – one headed by Nirenberg and the other by Ochoa. They found that ribosomes translated strings of poly-uracil (e.g. 5’-UUUUUUUUU-3’) into polypeptides containing only the amino acid phenylalanine. What is the nucleotide sequence of the anticodon of a phenylalanine tRNA?

30.) Genetic analyses of lactose utilization by E. coli, carried out by François Jacob and Jacques Monod, provided the first major insights into bacterial gene regulation and resulted in the coining of the word “operon.” Jacob and Monod found that the genes essential to lactose metabolism were in very close proximity and arranged on E. coli’s chromosome in this order: beta-galactosidase, lactose permease, and transacetylase. What did these genes of the first studied operon have in common?

31.) A chemical whose abbreviated name is IPTG is sufficiently similar to lactose to bind to the lac repressor but it cannot be degraded by the enzyme -galactosidase. What effect would the presence of IPTG have on the expression of the lac operon?

b. The lac operon would be expressed at high levels even when there was no lactose.

32.) The binding site for the protein product of the lacI gene is normally found only once in the entire E. coli genome – the operator of the lac operon. What would happen if molecular biologists genetically engineered an E. coli genome in such a way as to include the nucleotide sequence of that binding site 50 nucleotides downstream of the promoter of the trp operon?

c. The trp operon’s expression would also be associated with lactose levels in the cell.

33.) Different Hfr strains exhibit different origins and directions of transfer of their bacterial chromosome. The order of transfer of genes close to the origin (O) of five different Hfr strains is given below.

HFR Strain/ order of ttransfer of genes

Which of the following is a portion of the correct conjugational map of an E. coli chromosome?

34.) A protein (cyclic AMP receptor protein) has been found to specifically bind to the promoter in front of the beta-galactosidase gene but only when glucose levels are low. Only when this protein is bound can RNA polymerase begin transcription of the lactose operon. In what type of regulation of gene expression is this protein involved?

35.) Most biological pathways require cells to possess two or more different enzymes in equal quantities for the efficient conversion of a starting material into an end-product. What feature of prokaryotes helps assure coordinated expression of genes in such pathways?

b. Genes with related functions are usually found in a single operon.

36.) Which of the following is an important difference between E. coli’s single chromosome and the many chromosomes of human cells?

a. E. coli’s chromosome is circular while human chromosomes are linear.

37.) Humans who are lactose intolerant no longer make the equivalent enzyme to beta galactosidase in the lac operon. Which of the following mutations to the lac operon of a prokaryotic cell might make it unable to synthesize beta galactosidase?

A deletion of the promoter

38.) 70 is an important component of E. coli’s RNA polymerase that best recognizes promoters that contain Pribnow and TATA-boxes whose sequences are “TTGACA” and “TATAAT,” respectively. How can genes whose promoter sequences are significantly different ever be recognized by RNA polymerases and transcribed?

e. They must be transcribed by RNA polymerases with different factors.

39.) The Human Genome Project was an ambitious attempt to compile a kind of address list of where all the genes reside along human chromosomes. The completion of the DNA sequencing involved in this “molecular biology moon shot” came as a result of previous genetic mapping and physical mapping phases. What was (and still is) the fourth and final phase of this endeavor?

e. Comparing homologous sequences found in the genomes of other organisms to determine which are functionally important and what that function might be.

40.) Sir Alec Jefferies was the first person to actually see that the length of eukaryotic mRNAs was actually much shorter than the genes that coded for them when he hybridized mRNAs to genomic DNA and looked at the products with an electron microscope. What are the non-coding portions that are spliced out of the primary transcripts of eukaryotic genes?

41.) Many eukaryotic genomes contain vast tracts of non-coding or redundant DNA. Plants seem to have frequently accumulated numerous duplications of their whole genomes over the course of their evolutionary histories. What happens to a gene whose function is no longer needed because its copy serves its original purpose?

c. Some acquire new functions that confer a selective advantage and come under functional constraint

42.) In 1997 scientists in Edinburgh, Scotland cloned an adult Finn Dorset lamb. The researchers involved with this work were careful to use mammary cells as the source of the nuclear DNA from the lamb to be cloned because they are known to be less differentiated than most other mammalian cells. Why would it be necessary to utilize a nucleus from an undifferentiated cell line as opposed to a differentiated one?

e. As cells become differentiated they convert some portions of their chromatin to heterochromatin and that state is faithfully passed on to subsequent generations.

43.) Edward Southern was the first person to show that DNA molecules attached to paper-like membranes could specifically hybridize with other, free-floating DNA molecules that had been radio-labeled. Upon which of the following important features of DNA does Southern’s technique rely?

c. Complementary strands of single stranded DNA have an affinity for each other.

44.) The ability to determine the order in which nucleotides appear within the DNA of our chromosomes was obviously a major advance in the study of biology since it allowed us to directly examine our genetic code. The first DNA sequencing method to be used won Maxam and Gilbert a Nobel Prize and was widely used until Sanger developed a new method roughly ten years later. How do Maxam and Gilbert’s and Sanger’s methods differ?

e. The first involves chemical degradation of DNA while Sanger’s is an enzymatic process.

45.) Diabetics generally need regular doses of the hormone insulin to survive. Which of the following represents a reasonable way of making a transgenic organism that could provide insulin for such treatments?

e. Putting a bacterial promoter in front of the DNA sequence for the human insulin protein and inserting that construct as part of a plasmid into a bacterial cell.

46.) Which of the following lists things in increasing order of size?

e. Nitrogenous base, nucleotide, gene, chromosome.

47.) The restriction enzyme HpaII specifically makes a double stranded break in DNA whenever it encounters the nucleotide sequence 5’-CCGG-3’. If the human genome is roughly 3,000,000,000 nucleotides long, how many pieces would you expect it to be cut into when HpaII cuts at every one of its restriction sites?

Roughly 11,700,000 (3,000,000,000/256).

48.) 35,000 is a good estimate of the number of genes in the human genome. Human cells use a wide variety of mechanisms to express different genes in different tissues. Which of the following is likely to be different between brain and liver cells as a result?

e. The packaging of parts of their chromosomes.

49.) Keri Mullis won a Nobel Prize for his description of what is now a very popular methodology for making billions of copies of a specific portion of a genome through repeated cycles of heating and cooling of the genome with a DNA polyermase, primers that flank the region of interest, and just a few additional reagents. What is this amplification process called?

. Polymerase chain reaction (PCR).

50.) The burning of glucose releases 684 kcal per mole. The synthesis of glucose is just as energetically unfavorable as its combustion is favorable. How are living things able to make glucose synthesis a spontaneous reaction?

e. By coupling the unfavorable energetics to chemical reactions that are energetically favorable like the conversion of ATP to ADP and Pi.

51.) Complementary DNA sequences have a natural affinity for each other primarily because of their ability to form hydrogen between their nitrogenous bases. The three hydrogen bonds that form between guanines and cytosines and the two that form between adenines and thymines are known as “base pairs.” How do these hydrogen bonds differ from the covalent bonds that connect a nitrogenous base to a sugar like deoxyribose in DNA?

c. They do not take as much energy to break because no sharing of electrons actually takes place.

52.) Most of the energy extracted from glucose during glycolysis is initially harvested in the citric acid cycle as NADH. The high-energy electrons of NADH are used to create a proton gradient within mitochondria. Which of the following energy-rich compounds is the one that is synthesized as a direct result of that proton gradient?

53.) When populations of prokaryotic organisms coordinate their activities they can form communities that begin to rival the complexity of some eukaryotic organisms. How do such populations of bacteria determine how many bacteria are nearby?

54.) Membranes are often used by cells to create a variety of micro-environments in which otherwise incompatible reactions can take place. Which of the following intra-cellular compartments is actually enveloped by a double membrane that may reflect the fact that it was once a free-living organism of its own that is now living symbiotically in eukaryotic cells?

55.) In eukaryotes, the processes of glycolysis and the citric acid cycle are physically separated by membrane compartments. The NADH and FADH molecules generated in the mitochondria during the citric acid cycle carry with them the bulk of the harvestable energy from glucose. What is the molecule that enters into the citric acid cycle with so much energy associated with it?

c. A two carbon molecule attached to coenzyme A (acetly-CoA).

56.) The complete oxidation of glucose releases a large amount of energy in living cells: a net change of –686 kcal/mole. Cells convert as much of that energy as possible into energy-rich ATP molecules that are used to drive a large number of other chemical reactions. The hydrolysis of ATP to ADP and inorganic phosphate has a free energy change of –7.3 kcal/mole. What is the net amount of energy that cells harvest in the course of converting glucose to pyruvate at the end of glycolysis?

57.) Photosynthesis is a highly organized process that is divided into to major stages. One set of reactions produce ATP and NADPH as well as liberate oxygen gas (O2). The other set uses energy-rich ATP and NADPH molecules to synthesize carbohydrates. Which of these sets of reactions can continue to take place in the absence of light?

58.) The components of the nuclear membranes that are taken apart close to the end of prophase are re-used to assemble the nuclear envelope of daughter cells. At what stage of mitosis do those new nuclear membranes form?

59.) The different building blocks used to make each of the four kinds of biopolymers cause each kind of macromolecule to have a different function within living cells. What are the primary roles of carbohydrates, lipids, proteins and nucleic acids?

e. Energy transfer, energy storage, catalysis and information storage, respectively.

60.) Assume seed color (G/g) and seed shape (S/s) are coded for by two different but linked genes in a newly isolated species of beans. A cross between a true-breeding green (GG) smooth (SS) seeded plant and a yellow (gg) wrinkled (ss) seeded plant gives an F1 generation in which all individuals are green (Gg) smooth (Ss) seeded. In an F2 generation 50 green-smooth plants, 30 yellow-wrinkled plants, 10 green-wrinkled plants and 10 yellow-smooth plants are observed. How many map units separate the genes for seed color and shape in this species of beans?


Do proteins contain phosphorous? If its true then how alfred hershey and martha chase used the statement that proteins do not contain phosphorous? - Biology

By the end of this section, you will be able to do the following:

  • Explain transformation of DNA
  • Describe the key experiments that helped identify that DNA is the genetic material
  • State and explain Chargaff’s rules

Our current understanding of DNA began with the discovery of nucleic acids followed by the development of the double-helix model. In the 1860s, Friedrich Miescher ((Figure)), a physician by profession, isolated phosphate-rich chemicals from white blood cells (leukocytes). He named these chemicals (which would eventually be known as DNA) nuclein because they were isolated from the nuclei of the cells.

Figure 1. Friedrich Miescher (1844–1895) discovered nucleic acids.

Link to Learning

To see Miescher conduct his experiment that led to his discovery of DNA and associated proteins in the nucleus, click through this review.

A half century later, in 1928, British bacteriologist Frederick Griffith reported the first demonstration of bacterial transformation—a process in which external DNA is taken up by a cell, thereby changing its morphology and physiology. Griffith conducted his experiments with Streptococcus pneumoniae, a bacterium that causes pneumonia. Griffith worked with two strains of this bacterium called rough (R) and smooth (S). (The two cell types were called “rough” and “smooth” after the appearance of their colonies grown on a nutrient agar plate.)

The R strain is non-pathogenic (does not cause disease). The S strain is pathogenic (disease-causing), and has a capsule outside its cell wall. The capsule allows the cell to escape the immune responses of the host mouse.

When Griffith injected the living S strain into mice, they died from pneumonia. In contrast, when Griffith injected the live R strain into mice, they survived. In another experiment, when he injected mice with the heat-killed S strain, they also survived. This experiment showed that the capsule alone was not the cause of death. In a third set of experiments, a mixture of live R strain and heat-killed S strain were injected into mice, and—to his surprise—the mice died. Upon isolating the live bacteria from the dead mouse, only the S strain of bacteria was recovered. When this isolated S strain was injected into fresh mice, the mice died. Griffith concluded that something had passed from the heat-killed S strain into the live R strain and transformed it into the pathogenic S strain. He called this the transforming principle ((Figure)). These experiments are now known as Griffith’s transformation experiments.

Figure 2. Two strains of S. pneumoniae were used in Griffith’s transformation experiments. The R strain is non-pathogenic, whereas the S strain is pathogenic and causes death. When Griffith injected a mouse with the heat-killed S strain and a live R strain, the mouse died. The S strain was recovered from the dead mouse. Griffith concluded that something had passed from the heat-killed S strain to the R strain, transforming the R strain into the S strain in the process. (credit “living mouse”: modification of work by NIH credit “dead mouse”: modification of work by Sarah Marriage)

Scientists Oswald Avery, Colin MacLeod, and Maclyn McCarty (1944) were interested in exploring this transforming principle further. They isolated the S strain from the dead mice and isolated the proteins and nucleic acids (RNA and DNA) as these were possible candidates for the molecule of heredity. They used enzymes that specifically degraded each component and then used each mixture separately to transform the R strain. They found that when DNA was degraded, the resulting mixture was no longer able to transform the bacteria, whereas all of the other combinations were able to transform the bacteria. This led them to conclude that DNA was the transforming principle.

Career Connection

Forensic Scientists used DNA analysis evidence for the first time to solve an immigration case. The story started with a teenage boy returning to London from Ghana to be with his mother. Immigration authorities at the airport were suspicious of him, thinking that he was traveling on a forged passport. After much persuasion, he was allowed to go live with his mother, but the immigration authorities did not drop the case against him. All types of evidence, including photographs, were provided to the authorities, but deportation proceedings were started nevertheless. Around the same time, Dr. Alec Jeffreys of Leicester University in the United Kingdom had invented a technique known as DNA fingerprinting. The immigration authorities approached Dr. Jeffreys for help. He took DNA samples from the mother and three of her children, as well as an unrelated mother, and compared the samples with the boy’s DNA. Because the biological father was not in the picture, DNA from the three children was compared with the boy’s DNA. He found a match in the boy’s DNA for both the mother and his three siblings. He concluded that the boy was indeed the mother’s son.

Forensic scientists analyze many items, including documents, handwriting, firearms, and biological samples. They analyze the DNA content of hair, semen, saliva, and blood, and compare it with a database of DNA profiles of known criminals. Analysis includes DNA isolation, sequencing, and sequence analysis. Forensic scientists are expected to appear at court hearings to present their findings. They are usually employed in crime labs of city and state government agencies. Geneticists experimenting with DNA techniques also work for scientific and research organizations, pharmaceutical industries, and college and university labs. Students wishing to pursue a career as a forensic scientist should have at least a bachelor’s degree in chemistry, biology, or physics, and preferably some experience working in a laboratory.

Although the experiments of Avery, McCarty and McLeod had demonstrated that DNA was the informational component transferred during transformation, DNA was still considered to be too simple a molecule to carry biological information. Proteins, with their 20 different amino acids, were regarded as more likely candidates. The decisive experiment, conducted by Martha Chase and Alfred Hershey in 1952, provided confirmatory evidence that DNA was indeed the genetic material and not proteins. Chase and Hershey were studying a bacteriophage—a virus that infects bacteria. Viruses typically have a simple structure: a protein coat, called the capsid, and a nucleic acid core that contains the genetic material (either DNA or RNA). The bacteriophage infects the host bacterial cell by attaching to its surface, and then it injects its nucleic acids inside the cell. The phage DNA makes multiple copies of itself using the host machinery, and eventually the host cell bursts, releasing a large number of bacteriophages. Hershey and Chase selected radioactive elements that would specifically distinguish the protein from the DNA in infected cells. They labeled one batch of phage with radioactive sulfur, 35 S, to label the protein coat. Another batch of phage were labeled with radioactive phosphorus, 32 P. Because phosphorous is found in DNA, but not protein, the DNA and not the protein would be tagged with radioactive phosphorus. Likewise, sulfur is absent from DNA, but present in several amino acids such as methionine and cysteine.

Each batch of phage was allowed to infect the cells separately. After infection, the phage bacterial suspension was put in a blender, which caused the phage coat to detach from the host cell. Cells exposed long enough for infection to occur were then examined to see which of the two radioactive molecules had entered the cell. The phage and bacterial suspension was spun down in a centrifuge. The heavier bacterial cells settled down and formed a pellet, whereas the lighter phage particles stayed in the supernatant. In the tube that contained phage labeled with 35 S, the supernatant contained the radioactively labeled phage, whereas no radioactivity was detected in the pellet. In the tube that contained the phage labeled with 32 P, the radioactivity was detected in the pellet that contained the heavier bacterial cells, and no radioactivity was detected in the supernatant. Hershey and Chase concluded that it was the phage DNA that was injected
into the cell and carried information to produce more phage particles, thus providing evidence that DNA was the genetic material and not proteins ((Figure)).

Figure 3. In Hershey and Chase’s experiments, bacteria were infected with phage radiolabeled with either 35S, which labels protein, or 32P, which labels DNA. Only 32P entered the bacterial cells, indicating that DNA is the genetic material.

Around this same time, Austrian biochemist Erwin Chargaff examined the content of DNA in different species and found that the amounts of adenine, thymine, guanine, and cytosine were not found in equal quantities, and that relative concentrations of the four nucleotide bases varied from species to species, but not within tissues of the same individual or between individuals of the same species. He also discovered something unexpected: That the amount of adenine equaled the amount of thymine, and the amount of cytosine equaled the amount of guanine (that is, A = T and G = C). Different species had equal amounts of purines (A+G) and pyrimidines (T + C), but different ratios of A+T to G+C. These observations became known as Chargaff’s rules. Chargaff’s findings proved immensely useful when Watson and Crick were getting ready to propose their DNA double helix model! You can see after reading the past few pages how science builds upon previous discoveries, sometimes in a slow and laborious process.

Section Summary

DNA was first isolated from white blood cells by Friedrich Miescher, who called it nuclein because it was isolated from nuclei. Frederick Griffith’s experiments with strains of Streptococcus pneumoniae provided the first hint that DNA may be the transforming principle. Avery, MacLeod, and McCarty showed that DNA is required for the transformation of bacteria. Later experiments by Hershey and Chase using bacteriophage T2 proved that DNA is the genetic material. Chargaff found that the ratio of A = T and C = G, and that the percentage content of A, T, G, and C is different for different species.

Review Questions

If DNA of a particular species was analyzed and it was found that it contains 27 percent A, what would be the percentage of C?


Basics of DNA Replication

Figure 4. The three suggested models of DNA replication. Grey indicates the original DNA strands, and blue indicates newly synthesized DNA.

The elucidation of the structure of the double helix provided a hint as to how DNA divides and makes copies of itself. This model suggests that the two strands of the double helix separate during replication, and each strand serves as a template from which the new complementary strand is copied. What was not clear was how the replication took place. There were three models suggested: conservative, semi-conservative, and dispersive (see Figure 4).

In conservative replication, the parental DNA remains together, and the newly formed daughter strands are together. The semi-conservative method suggests that each of the two parental DNA strands act as a template for new DNA to be synthesized after replication, each double-stranded DNA includes one parental or “old” strand and one “new” strand. In the dispersive model, both copies of DNA have double-stranded segments of parental DNA and newly synthesized DNA interspersed.

Meselson and Stahl were interested in understanding how DNA replicates. They grew E. coli for several generations in a medium containing a “heavy” isotope of nitrogen ( 15 N) that gets incorporated into nitrogenous bases, and eventually into the DNA (Figure 5).

Figure 5. Meselson and Stahl experimented with E. coli grown first in heavy nitrogen ( 15 N) then in 14 N. DNA grown in 15 N (red band) is heavier than DNA grown in 14 N (orange band), and sediments to a lower level in cesium chloride solution in an ultracentrifuge. When DNA grown in 15 N is switched to media containing 14 N, after one round of cell division the DNA sediments halfway between the 15 N and 14 N levels, indicating that it now contains fifty percent 14 N. In subsequent cell divisions, an increasing amount of DNA contains 14 N only. This data supports the semi-conservative replication model. (credit: modification of work by Mariana Ruiz Villareal)

The E. coli culture was then shifted into medium containing 14 N and allowed to grow for one generation. The cells were harvested and the DNA was isolated. The DNA was centrifuged at high speeds in an ultracentrifuge. Some cells were allowed to grow for one more life cycle in 14 N and spun again. During the density gradient centrifugation, the DNA is loaded into a gradient (typically a salt such as cesium chloride or sucrose) and spun at high speeds of 50,000 to 60,000 rpm. Under these circumstances, the DNA will form a band according to its density in the gradient. DNA grown in 15 N will band at a higher density position than that grown in 14 N. Meselson and Stahl noted that after one generation of growth in 14 N after they had been shifted from 15 N, the single band observed was intermediate in position in between DNA of cells grown exclusively in 15 N and 14 N. This suggested either a semi-conservative or dispersive mode of replication. The DNA harvested from cells grown for two generations in 14 N formed two bands: one DNA band was at the intermediate position between 15 N and 14 N, and the other corresponded to the band of 14 N DNA. These results could only be explained if DNA replicates in a semi-conservative manner. Therefore, the other two modes were ruled out.

During DNA replication, each of the two strands that make up the double helix serves as a template from which new strands are copied. The new strand will be complementary to the parental or “old” strand. When two daughter DNA copies are formed, they have the same sequence and are divided equally into the two daughter cells.


From discovering to understanding

It might seem as though the role of DNA as the carrier of genetic information was not realized until the mid-1940s, when Oswald Avery (1877–1955) and colleagues demonstrated that DNA could transform bacteria ( Avery et al, 1944 ). Although these experiments provided direct evidence for the function of DNA, the first ideas that it might have an important role in processes such as cell proliferation, fertilization and the transmission of heritable traits had already been put forward more than half a century earlier. Friedrich Miescher (1844–1895 Fig 1), the Swiss scientist who discovered DNA in 1869 ( Miescher, 1869a ), developed surprisingly insightful theories to explain its function and how biological molecules could encode information. Although his ideas were incorrect from today's point of view, his work contains concepts that come tantalizingly close to our current understanding. But Miescher's career also holds lessons beyond his scientific insights. It is the story of a brilliant scientist well on his way to making one of the most fundamental discoveries in the history of science, who ultimately fell short of his potential because he clung to established theories and failed to follow through with the interpretation of his findings in a new light.

…a brilliant scientist well on his way to making one of the most fundamental discoveries in the history of science […] fell short of his potential because he clung to established theories…

It is a curious coincidence in the history of genetics that three of the most decisive discoveries in this field occurred within a decade: in 1859, Charles Darwin (1809–1882) published On the Origin of Species by Means of Natural Selection, in which he expounded the mechanism driving the evolution of species seven years later, Gregor Mendel's (1822–1884) paper describing the basic laws of inheritance appeared and in early 1869, Miescher discovered DNA. Yet, although the magnitude of Darwin's theory was realized almost immediately, and at least Mendel himself seems to have grasped the importance of his work, Miescher is often viewed as oblivious to the significance of his discovery. It would be another 75 years before Oswald Avery, Colin MacLeod (1909–1972) and Maclyn McCarthy (1911–2005) could convincingly show that DNA was the carrier of genetic information, and another decade before James Watson and Francis Crick (1916–2004) unravelled its structure ( Watson & Crick, 1953 ), paving the way to our understanding of how DNA encodes information and how this is translated into proteins. But Miescher already had astonishing insights into the function of DNA.

Between 1868 and 1869, Miescher worked at the University of Tübingen in Germany (Figs 2,3), where he tried to understand the chemical basis of life. A crucial difference in his approach compared with earlier attempts was that he worked with isolated cells—leukocytes that he obtained from pus—and later purified nuclei, rather than whole organs or tissues. The innovative protocols he developed allowed him to investigate the chemical composition of an isolated organelle ( Dahm, 2005 ), which significantly reduced the complexity of his starting material and enabled him to analyse its constituents.

In carefully designed experiments, Miescher discovered DNA—or “Nuclein” as he called it—and showed that it differed from the other classes of biological molecule known at that time ( Miescher, 1871a ). Most notably, nuclein's elementary composition with its high phosphorous content convinced him that he had discovered a substance sui generis, that is, of its own kind a conclusion subsequently confirmed by Miescher's mentor in Tübingen, the eminent biochemist Felix Hoppe-Seyler (1825–1895 Hoppe-Seyler, 1871 Miescher, 1871a ). After his initial analyses, Miescher was convinced that nuclein was an important molecule and suggested in his first publication that it would “merit to be considered equal to the proteins” ( Miescher, 1871a ).

Moreover, Miescher recognized immediately that nuclein could be used to define the nucleus ( Miescher, 1870 ). This was an important realization, as at the time the unequivocal identification of nuclei, and hence their study, was often difficult or even impossible to achieve because their morphology, subcellular localization and staining properties differed between tissues, cell types and states of the cells. Instead, Miescher proposed to base the characterization of nuclei on the presence of this molecule ( Miescher, 1870, 1874 ). Moreover, he held that the nucleus should be defined by properties that are related to its physiological activity, which he believed to be closely linked to nuclein. Miescher had thus made a significant first step towards defining an organelle in terms of its function rather than its appearance.

Importantly, his findings also showed that the nucleus is chemically distinct from the cytoplasm at a time when many scientists still assumed that there was nothing unique about this organelle. Miescher thus paved the way for the subsequent realization that cells are subdivided into compartments with distinct molecular composition and functions. On the basis of his observations that nuclein appeared able to separate itself from the “protoplasm” (cytoplasm), Miescher even went so far as to suggest the “possibility that [nuclein can be] distributed in the protoplasm, which could be the precursor for some of the de novo formations of nuclei” ( Miescher, 1874 ). He seemed to anticipate that the nucleus re-forms around the chromosomes after cell division, but unfortunately did not elaborate on under which conditions this might occur. It is therefore impossible to know with certainty to which circumstances he was referring.

Miescher thus paved the way for the subsequent realization that cells are subdivided into compartments with distinct molecular composition and functions

In this context, it is interesting to note that in 1872, Edmund Russow (1841–1897) observed that chromosomes appeared to dissolve in basic solutions. Intriguingly, Miescher had also found that he could precipitate nuclein by using acids and then return it to solution by increasing the pH ( Miescher, 1871a ). At the time, however, he did not make the link between nuclein and chromatin. This happened around a decade later, in 1881, when Eduard Zacharias (1852–1911) studied the nature of chromosomes by using some of the same methods Miescher had used when characterizing nuclein. Zacharias found that chromosomes, such as nuclein, were resistant to digestion by pepsin solutions and that the chromatin disappeared when he extracted the pepsin-treated cells with dilute alkaline solutions. This led Walther Flemming (1843–1905) to speculate in 1882 that nuclein and chromatin are identical ( Mayr, 1982 ).

Alas, Miescher was not convinced. His reluctance to accept these developments was at least partly based on a profound scepticism towards the methods—and hence results—of cytologists and histologists, which, according to Miescher, lacked the precision of chemical approaches as he applied them. The fact that DNA was crucially linked to the function of the nucleus was, however, firmly established in Miescher's mind and in the following years he tried to obtain additional evidence. He later wrote: “Above all, using a range of suitable plant and animal specimens, I want to prove that Nuclein really specifically belongs to the life of the nucleus” ( Miescher, 1876 ).

Although the acidic nature of DNA, its large molecular weight, elementary composition and presence in the nucleus are some of its central properties—all first determined by Miescher—they reveal nothing about its function. Having convinced himself that he had discovered a new type of molecule, Miescher rapidly set out to understand its role in different biological contexts. As a first step, he determined that nuclein occurs in a variety of cell types. Unfortunately, he did not elaborate on the types of tissue or the species his samples were derived from. The only hints as to the specimens he worked with come from letters he wrote to his uncle, the Swiss anatomist Wilhelm His (1831–1904), and his parents his father, Friedrich Miescher-His (1811–1887), was professor of anatomy in Miescher's native Basel. In his correspondence, Miescher mentioned other cell types that he had studied for the presence of nuclein, including liver, kidney, yeast cells, erythrocytes and chicken eggs, and hinted at having found nuclein in these as well ( Miescher, 1869b His, 1897 ). Moreover, Miescher had also planned to look for nuclein in plants, especially in their spores ( Miescher, 1869c ). This is an intriguing choice given his later fascination with vertebrate germ cells and his speculation on the processes of fertilization and heredity ( Miescher, 1871b, 1874 ).

Another clue to the tissues and cell types that Miescher might have examined comes from two papers published by Hoppe-Seyler, who wanted to confirm his student's results, which he initially viewed with scepticism, before their publication. In the first, another of Hoppe-Seyler's students, Pal Plósz, reported that nuclein is present in the nucleated erythrocytes of snakes and birds but not in the anuclear erythrocytes of cows ( Plósz, 1871 ). In the second paper, Hoppe-Seyler himself confirmed Miescher's findings and reported that he had detected nuclein in yeast cells ( Hoppe-Seyler, 1871 ).

In an addendum to his 1871 paper, published posthumously, Miescher stated that the apparently ubiquitous presence of nuclein meant that “a new factor has been found for the life of the most basic as well as for the most advanced organisms,” thus opening up a wide range of questions for physiology in general ( Miescher, 1870 ). To argue that Miescher understood that DNA was an essential component of all forms of life is probably an over-interpretation of his words. His statement does, however, clearly show that he believed DNA to be an important factor in the life of a wide range of species.

In addition, Miescher looked at tissues under different physiological conditions. He quickly noticed that both nuclein and nuclei were significantly more abundant in proliferating tissues for instance, he noted that in plants, large amounts of phosphorous are found predominantly in regions of growth and that these parts show the highest densities of nuclei and actively proliferating cells ( Miescher, 1871a ). Miescher had thus taken the first step towards linking the presence of phosphorous—that is, DNA in this context—to cell proliferation. Some years later, while examining changes in the bodies of salmon as they migrate upstream to their spawning grounds, he noticed that he could, with minimal effort, purify large amounts of pure nuclein from the testes, as they were at the height of cell proliferation in preparation for mating ( Miescher, 1874 ). This provided additional evidence for a link between proliferation and the presence of a high concentration of nuclein.

Miescher's most insightful comments on this issue, however, date from his time in Hoppe-Seyler's laboratory in Tübingen. He was convinced that histochemical analyses would lead to a much better understanding of certain pathological states than would microscopic studies. He also believed that physiological processes, which at the time were seen as similar, might turn out to be very different if the chemistry were better understood. As early as 1869, the year in which he discovered nuclein, he wrote in a letter to His: “Based on the relative amounts of nuclear substances [DNA], proteins and secondary degradation products, it would be possible to assess the physiological significance of changes with greater accuracy than is feasible now” ( Miescher, 1869c ).

Importantly, Miescher proposed three exemplary processes that might benefit from such analyses: “nutritive progression”, characterized by an increase in the cytoplasmic proteins and the enlargement of the cell “generative progression”, defined as an increase in “nuclear substances” (nuclein) and as a preliminary phase of cell division in proliferating cells and possibly in tumours and “regression”, an accumulation of lipids and degenerative products ( Miescher, 1869c ).

When we consider the first two categories, Miescher seems to have understood that an increase in DNA was not only associated with, but also a prerequisite for cell proliferation. Subsequently, cells that are no longer proliferating would increase in size through the synthesis of proteins and hence cytoplasm. Crucially, he believed that chemical analyses of such different states would enable him to obtain a more fundamental insight into the causes underlying these processes. These are astonishingly prescient insights. Sadly, Miescher never followed up on these ideas and, apart from the thoughts expressed in his letter, never published on the topic.

…Miescher seems to have understood that an increase in DNA was not only associated with, but also a prerequisite for cell proliferation

It is likely, however, that he had preliminary data supporting these views. Miescher was generally careful to base statements on facts rather than speculation. But, being a perfectionist who published only after extensive verification of his results, he presumably never pursued these studies to such a satisfactory point. It is possible his plans were cut short by leaving Hoppe-Seyler's laboratory to receive additional training under the supervision of Carl Ludwig (1816–1895) in Leipzig. While there, Miescher turned his attention to matters entirely unrelated to DNA and only resumed his work on nuclein after returning to his native Basel in 1871.

Crucially for these subsequent studies of nuclein, Miescher made an important choice: he turned to sperm as his main source of DNA. When analysing the sperm from different species, he noted that the spermatozoa, especially of salmon, have comparatively small tails and thus consist mainly of a nucleus ( Miescher, 1874 ). He immediately grasped that this would greatly facilitate his efforts to isolate DNA at much higher purity (Fig 4). Yet, Miescher also saw beyond the possibility of obtaining pure nuclein from salmon sperm. He realized it also indicated that the nucleus and the nuclein therein might play a crucial role in fertilization and the transmission of heritable traits. In a letter to his colleague Rudolf Boehm (1844–1926) in Würzburg, Miescher wrote: “Ultimately, I expect insights of a more fundamental importance than just for the physiology of sperm” ( Miescher, 1871c ). It was the beginning of a fascination with the phenomena of fertilization and heredity that would occupy Miescher to the end of his days.

Miescher had entered this field at a critical time. By the middle of the nineteenth century, the old view that cells arise through spontaneous generation had been challenged. Instead, it was widely recognized that cells always arise from other cells ( Mayr, 1982 ). In particular, the development and function of spermatozoa and oocytes, which in the mid-1800s had been shown to be cells, were seen in a new light. Moreover, in 1866, three years before Miescher discovered DNA, Ernst Haeckel (1834–1919) had postulated that the nucleus contained the factors that transmit heritable traits. This proposition from one of the most influential scientists of the time brought the nucleus to the centre of attention for many biologists. Having discovered nuclein as a distinctive molecule present exclusively in this organelle, Miescher realized that he was in an excellent position to make a contribution to this field. Thus, he set about trying to better characterize nuclein with the aim of correlating its chemical properties with the morphology and function of cells, especially of sperm cells.

His analyses of the chemical composition of the heads of salmon spermatozoa led Miescher to identify two principal components: in addition to the acidic nuclein, he found an alkaline protein for which he coined the term 'protamin' the name is still in use today protamines are small proteins that replace histones during spermatogenesis. He further determined that these two molecules occur in a “salt-like, not an ether-like [that is, covalent] association” ( Miescher, 1874 ). Following his meticulous analyses of the chemical composition of sperm, he concluded that, “aside from the mentioned substances [protamin and nuclein] nothing is present in significant quantity. As this is crucial for the theory of fertilization, I carry this business out as quantitatively as possible right from the beginning” ( Miescher, 1872a ). His analyses showed him that the DNA and protamines in sperm occur at constant ratios a fact that Miescher considered “is certainly of special importance,” without, however, elaborating on what might be this importance. Today, of course, we know that proteins, such as histones and protamines, bind to DNA in defined stoichiometric ratios.

Miescher went on to analyse the spermatozoa of carp, frogs (Rana esculenta) and bulls, in which he confirmed the presence of large amounts of nuclein ( Miescher, 1874 ). Importantly, he could show that nuclein is present in only the heads of sperm—the tails being composed largely of lipids and proteins—and that within the head, the nuclein is located in the nucleus ( Miescher, 1874 Schmiedeberg & Miescher, 1896 ). With this discovery, Miescher had not only demonstrated that DNA is a constant component of spermatozoa, but also directed his attention to the sperm heads. On the basis of the observations of other investigators, such as those of Albert von Kölliker (1817–1905) concerning the morphology of spermatozoa in some myriapods and arachnids, Miescher knew that the spermatozoa of some species are aflagellate, that is, lack a tail. This confirmed that the sperm head, and thus the nucleus, was the crucial component. But, the question remained: what in the sperm cells mediated fertilization and the transmission of hereditary traits from one generation to the next?

On the basis of his chemical analyses of sperm, Miescher speculated on the existence of molecules that have a crucial part in these processes. In a letter to Boehm, Miescher wrote: “If chemicals do play a role in procreation at all, then the decisive factor is now a known substance” ( Miescher, 1872b ). But Miescher was unsure as to what might be this substance. He did, however, strongly suspect the combination of nuclein and protamin was the key and that the oocyte might lack a crucial component to be able to develop: “If now the decisive difference between the oocyte and an ordinary cell would be that from the roster of factors, which account for an active arrangement, an element has been removed? For otherwise all proper cellular substances are present in the egg,” he later wrote ( Miescher, 1872b ).

Owing to his inability to detect protamin in the oocyte, Miescher initially favoured this molecule as responsible for fertilization. Later, however, when he failed to detect protamin in the sperm of other species, such as bulls, he changed his mind: “The Nuclein by contrast has proved to be constant [that is, present in the sperm cells of all species Miescher analysed] so far to it and its associations I will direct my interest from now on” ( Miescher, 1872b ). Unfortunately, however, although he came tantalizingly close, he never made a clear link between nuclein and heredity.

The final section of his 1874 paper on the occurrence and properties of nuclein in the spermatozoa of different vertebrate species is of particular interest because Miescher tried to correlate his chemical findings about nuclein with the physiological role of spermatozoa. He had realized that spermatozoa represented an ideal model system to study the role of DNA because, as he would later put it, “[f]or the actual chemical–biological problems, the great advantage of sperm [cells] is that everything is reduced to the really active substances and that they are caught just at the moment when they exert their greatest physiological function” ( Miescher, 1893a ). He appreciated that his data were still incomplete, yet wanted to make a first attempt to pull his results together and integrate them into a broader picture to explain fertilization.

At the time, Wilhelm Kühne (1837–1900), among others, was putting forward the idea that spermatozoa are the carriers of specific substances that, through their chemical properties, achieve fertilization ( Kühne, 1868 ). Miescher considered his results of the chemical composition of spermatozoa in this context. While critically considering the possibility of a chemical substance explaining fertilization, he stated that: “if we were to assume at all that a single substance, as an enzyme or in any other way, for instance as a chemical impulse, could be the specific cause of fertilization, we would without a doubt first and foremost have to consider Nuclein. Nuclein-bodies were consistently found to be the main components [of spermatozoa]” ( Miescher, 1874 ).

With hindsight, these statements seem to suggest that Miescher had identified nuclein as the molecule that mediates fertilization—a crucial assumption to follow up on its role in heredity. Unfortunately, however, Miescher himself was far from convinced that a molecule (or molecules) was responsible for this. There are several reasons for his reluctance, although the influence of his uncle was presumably a crucial factor as it was he who had been instrumental in fostering the young Miescher's interest in biochemistry and remained a strong influence throughout his life. Indeed, when Miescher came tantalizingly close to uncovering the function of DNA, His's views proved counterproductive, probably preventing him from interpreting his findings in the context of new results from other scientists at the time. Miescher thus failed to take his studies of nuclein and its function in fertilization and heredity to the next level, which might well have resulted in recognizing DNA as the central molecule in both processes.

One specific aspect that diverted Miescher from contemplating the role of nuclein in fertilization was a previous study in which he had erroneously identified the yolk platelets in chicken oocytes as a large number of nuclein-containing granules ( Miescher, 1871b ). This led him to conclude that the comparatively minimal quantitative contribution of DNA from a spermatozoon to an oocyte, which already contained so much more of the substance, could not have a significant impact on the latter's physiology. He therefore concluded that, “not in a specific substance can the mystery of fertilization be concealed. […] Not a part, but the whole must act through the cooperation of all its parts” ( Miescher, 1874 ).

It is all the more unfortunate that Miescher had identified the yolk platelets in oocytes as nuclein-containing cells because he had realized that the presumed nuclein in these granules differed from the nuclein (that is, DNA) he had isolated previously from other sources, notably by its much higher phosphorous content. But influenced by His's strong view that these structures were genuine cells, Miescher saw his results in this light. Only several years later, based on results from his contemporaries Flemming and Eduard A. Strasburger (1844–1912) on the morphological properties of nuclei and their behaviour during cell divisions, and Albrecht Kossel's (1853–1927) discoveries about the composition of DNA ( Portugal & Cohen, 1977 ), did Miescher revise his initial assumption that chicken oocytes contain a large number of nuclein-containing granules. Instead, he finally conceded that the molecules comprising these granules were different from nuclein ( Miescher, 1890 ).

Another factor that prevented Miescher from concluding that nuclein was the basis for the transmission of hereditary traits was that he could not conceive of how a single substance might explain the multitude of heritable traits. How, he wondered, could a specific molecule be responsible for the differences between species, races and individuals? Although he granted that “differences in the chemical constitution of these molecules [different types of nuclein] will occur, but only to a limited extent” ( Miescher, 1874 ).

And thus, instead of looking to molecules, he—like his uncle His––favoured the idea that the physical movement of the sperm cells or an activation of the oocyte, which he likened to the stimulation of a muscle by neuronal impulses, was responsible for the process of fertilization: “Like the muscle during the activation of its nerve, the oocyte will, when it receives appropriate impulses, become a chemically and physically very different entity” ( Miescher, 1874 ). For nuclein itself, Miescher considered that it might be a source material for other molecules, such as lecithin––one of the few other molecules with a high phosphorous content known at the time ( Miescher, 1870, 1871a, 1874 ). Miescher clearly preferred the idea of nuclein as a repository for material for the cell—mainly phosphorous—rather than as a molecule with a role in encoding information to synthesize such materials. This idea of large molecules being source material for smaller ones was common at the time and was also contemplated for proteins ( Miescher, 1870 ).

The entire section of Miescher's 1874 paper in which he discusses the physiological role of nuclein reads as though he was deliberately trying to assemble evidence against nuclein being the key molecule in fertilization and heredity. This disparaging approach towards the molecule that he himself had discovered might also be explained, at least to some extent, by his pronounced tendency to view his results so critically tellingly, he published only about 15 papers and lectures in a career spanning nearly three decades.

The modern understanding that fertilization is achieved by the fusion of two germ cells only became established in the final quarter of the nineteenth century. Before that time, the almost ubiquitous view was that the sperm cell, through mere contact with the egg, in some way stimulated the oocyte to develop—the physicalist viewpoint. His was a key advocate of this view and firmly rejected the idea that a specific substance might mediate heredity. We can only speculate as to how Miescher would have interpreted his results had he worked in a different intellectual environment at the time, or had he been more independent in the interpretation of his results.

We can only speculate as to how Miescher would have interpreted his results had he worked in a different intellectual environment at the time…

Miescher's refusal to accept nuclein as the key to fertilization and heredity is particularly tragic in view of several studies that appeared in the mid-1870s, focusing the attention of scientists on the nuclei. Leopold Auerbach (1828–1897) demonstrated that fertilized eggs contain two nuclei that move towards each other and fuse before the subsequent development of the embryo ( Auerbach, 1874 ). This observation strongly suggested an important role for the nuclei in fertilization. In a subsequent study, Oskar Hertwig (1849–1922) confirmed that the two nuclei—one from the sperm cell and one from the oocyte—fuse before embryogenesis begins. Furthermore, he observed that all nuclei in the embryo derive from this initial nucleus in the zygote ( Hertwig, 1876 ). With this he had established that a single sperm fertilizes the oocyte and that there is a continuous lineage of nuclei from the zygote throughout development. In doing so, he delivered the death blow to the physicalist view of fertilization.

By the mid-1880s, Hertwig and Kölliker had already postulated that the crucial component of the nucleus that mediated inheritance was nuclein—an idea that was subsequently accepted by several scientists. Sadly, Miescher remained doubtful until his death in 1895 and thus failed to appreciate the true importance of his discovery. This might have been an overreaction to the claims by others that sperm heads are formed from a homogeneous substance Miescher had clearly shown that they also contain other molecules, such as proteins. Moreover, Miescher's erroneous assumption that nuclein occurred only in the outer shell of the sperm head resulted in his failure to realize that stains for chromatin, which stain the centres of the heads, actually label the region where there is nuclein although he later realized that virtually the entire sperm head is composed of nuclein and associated protein ( Miescher, 1892a Schmiedeberg & Miescher, 1896 ).

Unfortunately, not only Miescher, but the entire scientific community would soon lose faith in DNA as the molecule mediating heredity. Miescher's work had established DNA as a crucial component of all cells and inspired others to begin exploring its role in heredity, but with the emergence of the tetranucleotide hypothesis at the beginning of the twentieth century, DNA fell from favour and was replaced by proteins as the prime candidates for this function. The tetranucleotide hypothesis—which assumed that DNA was composed of identical subunits, each containing all four bases—prevailed until the late 1940s when Edwin Chargaff (1905–2002) discovered that the different bases in DNA were not present in equimolar amounts ( Chargaff et al, 1949, 1951 ).

Unfortunately, not only Miescher, but the entire scientific community would soon lose faith in DNA as the molecule mediating heredity

Just a few years before, in 1944, experiments by Avery and colleagues had demonstrated that DNA was sufficient to transform bacteria ( Avery et al, 1944 ). Then in 1952, Al Hershey (1908–1997) and Martha Chase (1927–2003) confirmed these findings by observing that viral DNA—but no protein—enters the bacteria during infection with the T2 bacteriophage and, that this DNA is also present in new viruses produced by infected bacteria ( Hershey & Chase, 1952 ). Finally, in 1953, X-ray images of DNA allowed Watson and Crick to deduce its structure ( Watson & Crick, 1953 ) and thus enable us to understand how DNA works. Importantly, these experiments were made possible by advances in bacteriology and virology, as well as the development of new techniques, such as the radioactive labelling of proteins and nucleic acids, and X-ray crystallography—resources that were beyond the reach of Miescher and his contemporaries.

In later years (Fig 5), Miescher's attention shifted progressively from the role of nuclein in fertilization and heredity to physiological questions, such as those concerning the metabolic changes in the bodies of salmon as they produce massive amounts of germ cells at the expense of muscle tissue. Although he made important and seminal contributions to different areas of physiology, he increasingly neglected to explore his most promising line of research, the function of DNA. Only towards the end of his life did he return to this question and begin to reconsider the issue in a new light, but he achieved no further breakthroughs.

One area, however, where he did propose intriguing hypotheses—although without experimental data to support them—was the molecular underpinnings of heredity. Inspired by Darwin's work on fertilization in plants, Miescher postulated, for instance, how information might be encoded in biological molecules. He stated that, “the key to sexuality for me lies in stereochemistry,” and expounded his belief that the gemmules of Darwin's theory of pangenesis were likely to be “numerous asymmetric carbon atoms [present in] organic substances” ( Miescher, 1892b ), and that sexual reproduction might function to correct mistakes in their “stereometric architecture”. As such, Miescher proposed that hereditary information might be encoded in macromolecules and how mistakes could be corrected, which sounds uncannily as though he had predicted what is now known as the complementation of haploid deficiencies by wild-type alleles. It is particularly tempting to assume that Miescher might have thought this was the case, as Mendel had published his laws of inheritance of recessive characteristics more than 25 years earlier. However, there is no reference to Mendel's work in the papers, talks or letters that Miescher has left to us.

Miescher proposed that hereditary information might be encoded in macromolecules and how mistakes could be corrected…

What we do know is that Miescher set out his view of how hereditary information might be stored in macromolecules: “In the enormous protein molecules […] the many asymmetric carbon atoms allow such a colossal number of stereoisomers that all the abundance and diversity of the transmission of hereditary [traits] may find its expression in them, as the words and terms of all languages do in the 24–30 letters of the alphabet. It is therefore completely superfluous to see the sperm cell or oocyte as a repository of countless chemical substances, each of which should be the carrier of a special hereditary trait (de Vries Pangenesis). The protoplasm and the nucleus, that my studies have shown, do not consist of countless chemical substances, but of very few chemical individuals, which, however, perhaps have a very complex chemical structure” ( Miescher, 1892b ).

This is a remarkable passage in Miescher's writings. The second half of the nineteenth century saw intense speculation about how heritable characteristics are transmitted between the generations. The consensus view assumed the involvement of tiny particles, which were thought to both shape embryonic development and mediate inheritance ( Mayr, 1982 ). Miescher contradicted this view. Instead of a multitude of individual particles, each of which might be responsible for a specific trait (or traits), his results had shown that, for instance, the heads of sperm cells are composed of only very few compounds, chiefly DNA and associated proteins.

He elaborated further on his theory of how hereditary information might be stored in large molecules: “Continuity does not only lie in the form, it also lies deeper than the chemical molecule. It lies in the constituent groups of atoms. In this sense I am an adherent of a chemical model of inheritance à outrance [to the utmost]” ( Miescher, 1893b ). With this statement Miescher firmly rejects any idea of preformation or some morphological continuity transmitted through the germ cells. Instead, he clearly seems to foresee what would only become known much later: the basis of heredity was to be found in the chemical composition of molecules.

To explain how this could be achieved, he proposed a model of how information could be encoded in a macromolecule: “If, as is easily possible, a protein molecule comprises 40 asymmetric carbon atoms, there will be 2 40 , that is, approximately a trillion isomerisms [sic]. And this is only one of the possible types of isomerism [not considering other atoms, such as nitrogen]. To achieve the incalculable diversity demanded by the theory of heredity, my theory is better suited than any other. All manner of transitions are conceivable, from the imperceptible to the largest differences” ( Miescher, 1893b ).

Miescher's ideas about how heritable characteristics might be transmitted and encoded encapsulate several important concepts that have since been proven to be correct. First, he believed that sexual reproduction served to correct mistakes, or mutations as we call them today. Second, he postulated that the transmission of heritable traits occurs through one or a few macromolecules with complex chemical compositions that encode the information, rather than by numerous individual molecules each encoding single traits, as was thought at the time. Third, he foresaw that information is encoded in these molecules through a simple code that results in a staggeringly large number of possible heritable traits and thus explain the diversity of species and individuals observed.

Miescher's ideas about how heritable characteristics might be transmitted and encoded encapsulate several important concepts that have since been proven to be correct

It is a step too far to suggest that Miescher understood what DNA or other macromolecules do, or how hereditary information is stored. He simply could not have done, given the context of his time. His findings and hypotheses that today fit nicely together and often seem to anticipate our modern understanding probably appeared rather disjointed to Miescher and his contemporaries. In his day, too many facts were still in doubt and too many links tenuous. There is always a danger of over-interpreting speculations and hypotheses made a long time ago in today's light. However, although Miescher himself misinterpreted some of his findings, large parts of his conclusions came astonishingly close to what we now know to be true. Moreover, his work influenced others to pursue their own investigations into DNA and its function ( Dahm, 2008 ). Although DNA research fell out of fashion for several decades after the end of the nineteenth century, the information gleaned by Miescher and his contemporaries formed the foundation for the decisive experiments carried out in the middle of the twentieth century, which unambiguously established the function of DNA.

As such, perhaps the most tragic aspect of Miescher's career was that for most of his life he firmly believed in the physicalist theories of fertilization, as propounded by His and Ludwig among others, and his reluctance to combine the results from his rigorous chemical analyses with the 'softer' data generated by cytologists and histologists. Had he made the link between nuclein and chromosomes and accepted its key role in fertilization and heredity, he might have realized that the molecule he had discovered was the key to some of the greatest mysteries of life. As it was, he died with a feeling of a promising career unfulfilled ( His, 1897 ), when, in actual fact, his contributions were to outshine those of most of his contemporaries.

…he died with a feeling of a promising career unfulfilled […] when, in actual fact, his contributions were to outshine those of most of his contemporaries

It is tantalizing to speculate the path that Miescher's investigations—and biology as a whole—might have taken under slightly different circumstances. What would have happened had he followed up on his preliminary results about the role of DNA in different physiological conditions, such as cell proliferation? How would his theories about fertilization and heredity have changed had he not been misled by the mistaken identification of what appeared to him to be a multitude of small nuclei in the oocyte? Or how would he have interpreted his findings concerning nuclein had he not been influenced by the views of his uncle, but also those of the wider scientific establishment?

There is a more general lesson in the life and work of Friedrich Miescher that goes beyond his immediate successes and failures. His story is that of a brilliant researcher who developed innovative experimental approaches, chose the right biological systems to address his questions and made ground-breaking discoveries, and who was nonetheless constrained by his intellectual environment and thus prevented from interpreting his findings objectively. It therefore fell to others, who saw his work from a new point of view, to make the crucial inferences and thus establish the function of DNA.


THE MICROBIAL WORLD: A HISTORICAL PERSPECTIVE

As you may imagine, this story has to be much condensed because the number of people intimately involved in solving the molecular nature and mode of action of DNA is very large indeed. But it makes a great story none-the-less.

Friedrich Miescher, was 25 in 1869 when he discovered DNA and he was

totally unknown. He discovered DNA in the nucleus of the cell. He suggested that the nucleus might have a unique chemical composition and have important function. And this was at a time when most investigators believed that there was nothing unique about the nucleus and that it was a relatively unimportant cell structure.

In 1866 Ernst Haeckel, a well respected anatomist of the time, made some rather profound statements relative to cell structure which were taken seriously immediately. He said the nucleus provides for the transmission of hereditary characters, the external plasma on the other hand is for accommodation or adaptation to the external world. Attention thereafter remained focused on the nucleus, the nature of its chemistry and the nature of it‘s function..

After many years of hard chemistry and much argument, from about 1909 to somewhere along the early 1940s it was an almost undisputed dogma that the four nucleotide bases, ATGC, were present in equal proportions in the nucleic acids. Remember that the technological advancement in the machinery of modern science was still a long way off. This idea of equal amounts of the bases in nucleic acids led to the formulation of what became known as the tetranucleotide hypothesis for the structure of nucleic acids. This came into question when other more refined studies using more sophisticated techniques showed that the molecular weight of DNA was a million or more, whereas the molecular weight of a tetranucleotide is less than 10,000. But no well understood or generally accepted function was evident for DNA.

To continue the story, we’ll start with Gregor Mendel, a monk who, in the mid 1800s, enjoyed playing in the garden mating peas. To make a long and complicated story short, the take home message of these sexual dalliances was that there exists a mechanism by which plants, and all creatures in general, pass on their physical characteristics to their offspring. We call that mechanism heredity, and it is carried out by the faithful transfer, and I underline faithful, meaning precise transfer, of genes coding for all your characteristics, seen and unseen, to your offspring. Your genes say who you are. Of course, Nurture plays a great role in our final outcome, but we’ll disregard that in our story. And I take great pride in saying that microbiology had a great voice in the story of DNA and heredity.

Our next protagonist in the story is Charles Darwin, a biologist who proposed the theory of evolution. Basically what Darwin stated was that the natural force driving evolution was the struggle for existence. Thomas Malthus had said that natural populations produced more individuals than they could feed. The competition for food then resulted in the death of those least capable of competing . From this Darwin concluded that those organisms most capable of adapting to current external conditions were most likely to survive and reproduce. Chance variations that gave advantages for survival were maintained through reproduction those that were disadvantageous disappeared as a result of the death of the organisms. Over long periods of time, this interminable process gave rise to new varieties and new species.

Using this theory we can explain how it is we are so alike at the molecular level, yet as a group we can have such a great array of physical characteristics, from monkeys, to chickens to ants and worms. We are alike at the molecular level because we share a great percentage of our genes with other animals and plants. We are so similar to the higher apes that we share 97% or more DNA homology, meaning that our genes are extremely similar, and in many instances, identical. This is readily understood when we realize that we arose from one primordial cell which evolved, through the eons of time, and eventually gave rise to the multitude of biological types that exist on earth today. If time permits we can return to this interesting topic later in the course.

Now, if all living things arose from one primordial cell, it is understandable why we should be so alike at the molecular level. We inherit in the process of evolution all the genes of our parents. These genes are undergoing changes, mutations we call them, and new genes appear as we were going along through the evolutionary process. By evolution I mean nothing more than the selection of the fittest form that adapts to any change in the environment. It makes sense that it is only the fittest forms which will survive and out-live all other forms. As an example consider our current problem of global warming. Only those forms capable of tolerating, and multiplying, at higher temperatures will survive. All others will perish.

The person next on my list is an Englishman, Fred Griffith, a medical officer who worked in the pathology laboratory of the British Ministry of Health. He was a microbiologist working in the 1920s on the transfer of virulence in the organism that causes pneumonia. He was the first to describe Smooth and Rough forms of the pneumococcus, and to associate these with virulence. By virulence I mean the ability of an organism to cause disease in man. One of the characteristics of the pneumococcus, the causative agent of bacterial pneumonia, is that the organisms found in the sputum of patients ill with pneumonia have a capsule covering the cell wall. The presence of a capsule is easy to detect on Petri plates because the cells form very smooth colonies. They are called buttery in appearance. A capsule is also readily visible on cells in the microscope, especially after staining. This capsule is variable in chemical composition among the different strains of pneumococci and these different capsular types as they are called can be differentiated by simple serological means, and are designated by Roman numerals.

Griffith’s discovery of the S and R forms of the pneumococci was made in 1923. He made the fundamental observation that, when large numbers of avirulent R cells are injected into mice, it is often possible to recover S cells which are fully virulent and have the same capsule type as the one from which the R cells were derived. He concluded that the virulence character is retained, but not expressed in R cells, and that the animal body acts as a selective medium in which only S forms can multiply. After much experimentation, he concluded that the S & R variation is a reversible mutational change.

Griffith was interested in the role of the capsule types in virulence. He observed that when a smooth or S strain originating from a capsule type III strain of pneumococcus is serially subcultured, that is, transferred from a growth plate to a fresh one, a rough colony, called R, will occasionally appear on the plate. The cells making up this colony lack the capsule, and that makes the colony rough. And R colonies bread true, that is, they always gave rise to R colonies. But most interesting of all, these R cells are avirulent, they do not kill mice, whereas the S smooth cells from which they originated do.

In 1928 Fred Griffith made a startling observation. When a very large number of R cells derived from a Smooth type I was inoculated under the skin of a mouse together with a heat-killed S type II, the mouse died within a few days. Mice receiving heat-killed S type II alone survived. Mice receiving only the R cells also survived. Blood from the dead mice yielded only S type II cells. But remember that the type II smooth cells inoculated into the animal had been heat killed, so this was not the source of the new type II S cells. They concluded, after many control experiments, that the dead type II cells had secreted something which conferred on the R cells the ability to make a new type of capsule. They discovered that the transferred property was heritable, thus they had generated a new species. They called this transfer of genetic material from one microbial cell to another cell, Transformation. Now keep in mind this was the first time in history that transfer of genetic material had been seen in bacteria. No one had even considered this possibility. It was amazing.

Others got in the picture and found they could get the same capsule type transformation in vitro, in the test tube, simply by mixing heat killed type II S cells with R type I S and plating the mixture. Still others showed they could do the transformation experiment by mixing cell free extracts of heat killed S cells with R recipient cells. Clearly there was something in this cell free extract that was acting like a gene, I.e., transferring heritable characters. This something was given the name Transforming Principle.

In 1944 Ostwald T. Avery, Colin MacLeod and Maclin McCarty, three microbiologists working at the Rockefeller Institute in N.Y. succeeded in purifying the transforming principle form the pneumococcus and found it to be composed solely of DNA. This was somewhat heretical because until that time it was generally believed that the only thing complex enough in a cell to behave like a gene was protein. You see, the nucleic acids have only four component subunits, A ,T, G, C, thus limiting the possibilities of variation within tetra nucleotides whereas proteins can be made up from over 20 different amino acids, giving rise to the possibility for an infinite number of possible permutations in structure, and thus a limitless number of genes. In any case, the story goes that if these guys had not been working at the Rockefeller Institute in N.Y., they might not have been able to succeed in characterizing the Transforming Principle. You see, the Rockefeller is a very in-house and friendly place and everyone knows what everyone is doing. Pepsin, an enzyme that degrades proteins specifically had recently been purified there. Some of this pure pepsin was given to Avery and he was told, “Here this will degrade your transforming principle.” So Avery tried to degrade their semi purified transforming principle by incubating it with pepsin and then trying it out on a transformation experiment. Well, the enzyme did not degrade the transforming principle. So now they started looking for other enzymes that would degrade their transforming principle. Ribonuclease, an enzyme that degrades ribonucleic acid specifically, did not have an effect on the transforming principle. McCartyi purified DNAase, the DNA-degrading enzyme, and transformation was eliminated. He found that DNAase would degrade their transforming principle. They succeeded in purifying the transforming principle and found it to be composed of DNA. Ite’s biological activity was so grreat that it was capable of bringing about transfromation even when used in a final dilution of less that 1 in 100,000,000 in the reaction system.

The paper was not received with great acclaim. For one it was published in a medical journal, not a microbiological or biochemical one. Most important of all, people were not convinced that a molecule as simple as DNA could have sufficient flexibility to encode the thousands of proteins in a cell. Only proteins contained this coding potential. So the Avery paper did not make a big stink.

Yet, some people read it an accepted it. Edwin Chargaff changed his entire research project to study nucleic acids after reading Avery’s paper. Chargaff made some very fundamental discoveries regarding DNA structure. He used a simple technique to measure the relative amounts of the four nucleotides in DNA samples from a variety of vertebrates and many bacteria. While some species had DNA in which adenine and thymine predominated, others had DNA with more guanine and cytosine. The possibility thus presented itself that no two DNA molecules had the same composition. Further he showed that invariably the concentration of adenine always equaled that of thymine and the same for guanine and cytosine that were always the same in concentration. This came to be known as Chargaff’s rule and was of tremendous significance in generating the model of DNA.

It took an experiment with a biological system completely different from that used by Avery to win universal acceptance for the genetic role of DNA. Alfred Hershey and Martha Chase used a virus that infects bacteria, called bacteriophage, or phage for short, to prove DNA was the carrier of genetic information. Very briefly, phages are bacteria-specific viruses composed solely of protein and either RNA or DNA, nothing else. Now, proteins contain the element S whereas nucleic acids do not contain S, and nucleic acids contain phosphorous whereas proteins do not. Hershey and Chase labeled phage T2, which infects the bacterium E. coli, with either radioactive S (S35) which is incorporated into the proteins of T2 or with radioactive phosphorous (P32) which enters the nucleic acid of T2. They then incubated the individually labeled T2 with E. coli, allowed a few minutes for the phages to attach and do their thing, and then blended the mixture and centrifuged the released phage bodies. They found that only P32 labeled DNA entered the cells while the S35 labeled protein remained outside the cells and was removed by blending. Since T2 phage progeny were produced by the cells, this proves that DNA is the genetic material.

A few years later, Bob Romig, a Microbiology professor at UCLA at the time, did the clincher experiment. He did the very laborious experiment of isolating the DNA for a B. subtilis phage intact, I.e, as one large circular molecule, and used it to transform a culture of B. subtilis, which gave rise to a burst of phage particles.

And then in 1953 came the paper by James Watson and Francis Crick with their proposal for the structure of DNA. This paper is so important because it readily, and very precisely explains how genes work, how they replicate and how they can change, that is mutate. It is probably they most important piece of scientific work in the 20th century. In the entire history of biological science, only Darwin’s recognition of the existence and fundamental mechanism of evolution has equaled the discovery in importance. And this was all said in one page in a paper in Nature.

Basically DNA is made up of two enormously long chains of specifically ordered ATGC molecules linked together forming a double helix. DNA serves as a dictionary for writing the genetic information necessary to make a cell and have it divide faithfully. It uses a triplet code for each amino acid in a protein, I.e., three specific nucleotides encode for one amino acid in the final protein product of the gene. It was a young biochemist at the National Institutes of Health named Marshall Nirenberg, who first discovered that the triplet UUU in RNA codes for the amino acid phenylalanine. This was the very first letter of the DNA alphabet, and the remaining triplets were identified mainly by Severo Ochoa and Gobind Khorana and others who devised methods for synthesizing polymers of defined structure of ATG or C in all possible triplet permutations. This work was of such importance that Nirenberg and Khorana shared the Nobel Prize in 1968.

AS An aside let me say that all was not always rosy for DNA having any role in heredity. Especially when Wendell Stanley crystallized tobacco mosaic virus in 1935 and found it to be composed primarily of protein with lesser amount of RNA. Further, changes in the protein structure of the virus were discovered, whereas the amounts of phosphorous in the nucleic acids were virtually identical, suggesting that it stayed constant while the protein changed. It was thus thought that the diversity of proteins meant that they comprised the genes of the virus.

Replication of DNA occurs by the separation of the Watson and crick strands of the double helix, followed by the synthesis of the complementary copies of the two separated strands yielding two identical double helixes. When the entire molecule is copied, one helix is passed to the daughter cell and one stays behind with the mother.

Let me tell you about what has been called the most beautiful experiment in biology. It was performed by two very young men, one a post doc and the other a beginning assistant professor at Oregon state University. The experiment is called the meselson, stahl experiment in honor of the inventors. What these guys wanted to do was to test the idea proposed by Watson and Crick that dna replication entailed the separation of the two strands of the double helix followed by copying these two strands. Not everyone was convinced that strand separation could be so easy. Max Delbruck thought that unzipping the dna could give rise to knots.

Meselson and stahl used a centrifuge technique that allowed them to separate molecules according to slight deifferences in molecular weight. following a centrifugal spin, heavier molecules end up nearer the bottom of the test tube than lighter ones which remain closer to the top. Because Nitrogen atoms are a component of dna, and because nitrogen exists in two forms, a heavy atom n15 and the natural n14, Meselson and stahl could label their dna with n15 by simply growing the bacteria in n15 ammonium sulphate. All their DNA would be labeled with n15 and would form a heavy band of DNA migrating further down in the centrifuge tube. From this culture they took a sample and replaced the medium to now contain n14 ammonium sulphate ensuring that the next round of dna replication would make use of light n14. If, as Watson and Crick predicted, unzipping followed by copying, then the two nascent dna strands in the experiment should be hyb rids composed of one heavy n 15 parental strand and one light nascent n14 strand and the dna band should migrate higher up in the centrifuge tube after the spin. Further growth of the culture results in the continual unzipping and copying the double stranded dna molecule. So by the third generation, separating the hybrid dna bands and copying them will result in one pure n14 double helix and one hybrid helix. This n14 band will migrate even higher in the tube than the hiybrid because of its lighter weight. I think the illustratoin will make this clear.

Mutations occur by the incorporation of an occasional error in DNA during the replication process, I.e., the insertion of a g instead of an a in one strand as it is being copied will give rise to a change in the amino acid in the protein product, thus changing the protein and possibly affecting its function. Not all mutations need be harmful. Only those mutational changes that alter the physical structure of the protein will affect its function.

So you see that the Watson-Crick model of DNA really explains the functioning of genes and the process of heredity as a whole. It certainly was worth the nobel Prize.

This same exact mechanism of heredity is seen in bacteria, baboons, parrots and elephants and ants. We are really very similar in our basics. Accept evolution and accept darwin and accept dna. It’s all here to stay.

The presence of bacteria 3 billion years ago may indicate the presence of a functional genetic code at that time. The remarkable similaritiy in codon base sequences recognized by bacterial, amphibian and mammalian protein synthesizing systems, suggest that most , if not all, forms of life on this planet use almost the same genetic language, and that the language has been used, possibly with few major changes, for at least 500 Billion years.


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