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I am in Northumberland, UK. This is growing in my veggie plot and cannot find out what plant it is… looks like a triffid!! It is one stem growing out of the ground and has many stems growing on the side of the main one which lie along the soil with no roots appearing.
I placed a couple of pea pods beside the brick to show the size. Hope someone can help. Thank you.
Alan is right to call this a Cichorium. It is Cichorium endivia, Endive. It is a tropical plant that is widely cultivated for its edible leaves. However, when it flowers the leaves die, so you don't recognize the plant as Endive. The dying leaves can be seen on the picture below, from the Dutch wikipedia page.
What is this viny plant in the UK? - Biology
A vine (Latin vīnea "grapevine", "vineyard", from vīnum "wine") is any plant with a growth habit of trailing or scandent (that is, climbing) stems, lianas or runners. The word vine can also refer to such stems or runners themselves, for instance, when used in wicker work.  
In parts of the world, including the British Isles, the term "vine" usually applies exclusively to grapevines (Vitis),  while the term "climber" is used for all climbing plants. 
Our expertise is in plant primary metabolism, particularly seed biology and biochemical genetics. We are using molecular and chemical genetics, biochemistry and bioimaging to dissect the link between metabolism and growth in developing and in germinating seeds and seedlings.
We study three fundamental aspects of plant biology:
- how carbohydrate availability coordinates with growth during embryogenesis to regulate seed set, embryo morphogenesis and viability,
- how does abiotic stress affect reproductive growth, and
- what post-germination processes in the seedling control root growth and development for successful establishment of the new plant.
The lab is located at the School of Natural and Environmental Sciences - Devonshire Building. We are part of the Plant and Microbial Biology Group, within the Bioeconomy research focus area of the Biology cluster in SNES (https://www.ncl.ac.uk/nes/our-research/biology/plant-microbial-biology/).
The lab is always happy to host Stage 3 undergraduate, as well as taught post-graduate students for lab-based projects. We welcome informal enquiries about potential PhD projects related to our research from talented and enthusiastic researchers/students. Please contact Vasilios Andriotis with a CV and research interests to discuss opportunities to join the lab.
BIO1010 – Biology in Action (Stage 1)
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BIO1023 – Cells and Biomolecules (Stage 1, Module Leader)
MST1204 – Academic and Professional Skills for the Biosciences (Stage 1)
BIO2021 – Employability skills for Biologists (Stage 2)
BIO2035 – Plant Biology (Stage 2)
BIO3197 – Biological Literature Review (Stage 3)
BIO3198 – Biological Information Project (Stage 3)
BIO3199 – Biological Research Project (Stage 3)
BIO3048 – Current Research in Plant and Microbial Biology (Stage 3)
BIO3052 – Global challenges in plant science research (Stage 3)
A popular household fern may be the first known eusocial plant
A wild colony of staghorn ferns appears to demonstrate a division of labor, implying complex social organization. Image Credit: Ian Hutton.
Staghorn ferns are popular houseplants, sporting long, antler-like fronds that poke out from a brown, tissue-papery base. They may also be the first known example of a plant that exhibits a type of social organization—that is, the first plant thought to be eusocial.
In the dry forests of Lord Howe Island in in the Tasman Sea, wild staghorns grow in colonies clinging to the trunks of trees. A recent study in Ecology finds that nests of these ferns arguably share characteristics with those of ants or termites they’re composed of closely-related individuals that form morphologically distinct castes with different roles. All previous examples of eusociality are in animals. “Plants were completely off the radar,” says co-author K.C. Burns, an evolutionary biologist at Victoria University of Wellington, New Zealand. Finding a putative example of a eusocial fern shatters the notion that animals are somehow more socially complex, he says, and suggests that “the pathway to eusociality is open to both plants and animals.”
Burns noticed the fern colonies while hiking through the stunted tropical dry forests of Lord Howe Island, where trees are short and arboreal epiphytes grow at eye level. Each fern had two kinds of fronds. Strap fronds stand out, growing long and green from the center of each individual plant. Nest fronds, by contrast, are round and flat, spreading like green pancakes from the middle of each plant the fronds senesce to form the tissue-papery base.
Ferns growing at the top of a colony have upright strap fronds, shaped like rain gutters that channel water and nutrients into the core of the group. Their nest fronds are waxy, and appear to deflect water and leaf litter into the nest below. Ferns lower in the colony have strap fronds hanging slack toward the ground and nest fronds that are circular and spongy.
When Burns quantified the absorbency of nest fronds from 10 different colonies, he found that those growing at the bottom had thicker, more absorbent tissue than those at the top. “It’s plain as the nose on your face that they’re subdividing labor,” Burns says. “[Ferns] at top seem to be water and nutrient capturers ones below seem to store water.”
What’s more, when he randomly sampled five individual ferns across 10 colonies, Burns found that the number of reproductive fronds increased with the height of the colony, and 40% percent of ferns didn’t reproduce at all. Reproductive division of labor—typically a reproductive queen and nonreproductive workers and soldiers—is one classic criterion for eusociality. DNA analysis of the 10 fern colonies confirmed that the majority contained genetically identical individuals, though two nests held more than one genotype.
Eusociality typically requires two other conditions. One is overlapping generations, meaning one generation co-occurs with the next. The other is cooperative brood care, meaning collectively feeding and supporting offspring through division of labor. It’s unclear if those conditions were satisfied in the case of the ferns, says evolutionary biologist Sandra Rehan, at York University in Toronto Canada. There was no generational data in this study, she notes. And since the ferns spread asexually on shared roots, they don’t actually exhibit an active system of resource acquisition typical of brood care. But Uli Ernst, a behavioral ecologist at the the Apicultural State Institute at the University of Hohenheim, Germany, adds that since older and younger ferns (their clones) live together sharing water and nutrients, one could technically call these overlapping generations and brood care.
One key question is what defines an individual fern. If a colony can begin with a small plume of strap fronds sticking up from a few nest fronds and then spread asexually on the same roots, perhaps it is a single plant, Rehan says. Strawberries, too, spread on stolons, and a whole field can represent a summer’s growth of one plant.
The difference, Burns says, is that the whole strawberry patch looks the same. The ferns, by contrast, differ markedly in morphology and reproduction depending on their role in the colony. The next step, he adds, is to sample 40 more wild nests, quantifying their life history traits more thoroughly.
Drawing conclusions about staghorn social organization may ultimately hinge on the nuances of eusociality—some definitions frame the concept as more of a spectrum, notes evolutionary biologist Guy Cooper, at the University of Oxford in the UK, who was not involved in the fern study. The most extreme form, obligate eusociality, occurs when castes are permanent and irreversible, such that no individual could survive and reproduce alone. In these cases, natural selection shifts from acting on individuals, to driving adaptations at the colony level. At the other end of the spectrum is facultative eusociality, in which castes can be impermanent and individuals can survive and reproduce alone. Future work has yet to clarify on which end of the spectrum this staghorn falls. Either way, the fern is exciting, Cooper says. “It’s clearly eusocial in some sense of the word.”
Other recent papers recommended by Journal Club panelists:
Sergi Escribano/Getty Images
Pictured here is purple ice plant (Delosperma cooperi). Like Angelina sedum, this plant is a hardy perennial. Like Angelina sedum, it (Delosperma cooperi). Like Angelina sedum, this plant has succulent leaves and is suitable for xeriscaping.
Not only is this a drought-tolerant ground cover, but it positively does not like for its roots to be sitting in moisture. For that reason, ice plant can be challenging for someone experimenting with it for the first time. So unlike the other plants listed here, one would not classify this plant as easy to grow.
Still, its flowers are gorgeous, so why not give it a try? Its unusual foliage (the look of which gives it its common plant name) is moderately attractive, as well.
A-level Biology Paper 1 Specimen Set 2
How could you use the Michaelis constant to determine the type of inhibition occurring in an enzyme-catalysed reaction?
A group of scientists investigated the effect of grassy strips on the biodiversity of soil animals.
• They divided a field into plots measuring 25 m × 5 m, with a 5-metre-wide grassy strip of land between each plot.
• Each year, they planted wheat in each of the plots.
• In the fifth year, they removed samples of soil from each plot where wheat was growing and from the grassy strips around them.
• They sorted each soil sample by hand for 40 minutes to collect the soil animals within the sample
Graph is "Concentration of amylase in saliva" against "Number of copies of AMY1 gene in cells from each volunteer"
- Concentration of haemoglobin in blood = 150 g dm-3 .
- Volume of oxygen carried by fully saturated haemoglobin = 1.35 cm3 g-1 .
- Resting heart rate = 60 beats minute-1 .
- Volume of blood pumped out of left ventricle each beat = 60 cm3 .
One group of scientists tested a hypothesis that boron combines with sucrose to produce a sucrose-borate complex that is translocated more effectively than sucrose molecules.
They grew tomato plants in nutrient-poor sand. Prior to starting their experiment, they left the mature plants in a dark room for 48 hours.
For each plant, the scientists put one of its leaves into a solution of sucrose that was radioactively labelled. These leaves were left attached to the plants. They used two radioactively labelled sucrose solutions:
- solution A contained boron at a concentration of 10 parts per million.
- solution B contained no boron.
After a period of time, the scientists removed samples from parts of the plants, dried them in an oven and ground each into a powder. They then measured the radioactivity in each powdered sample. The scientists' results are shown in Table 4.
In a spin: Brexit spells trouble for UK vinyl industry
For vinyl lovers the act of placing the needle on a record and anticipating the crackle before the first bars ring out is a sensory and sensual thrill. A no-deal Brexit could nip that in the (ear)bud. Rob Mudge reports.
"I see you've sent my letters back,
And my LP records and they're all scratched."
So sang The Police on their track "Can't stand losing you" — or should that be EU? In the context of Brexit and a possible no deal, it may be more than a surface scratch for vinyl in terms of sales and exports from the UK to the EU and vice versa.
A decade ago, vinyl was essentially on its deathbed. The industry and consumers had embraced the CD as their format of choice. In 2007, a paltry 205,000 units were shifted, picked up mostly by analogue-craving aficionados who were in a, er, spin about music going digital.
Fast forward to today and vinyl sales in the UK have increased twentyfold, even outperforming downloads (4.2 million vinyl albums were sold in the UK last year) and are projected to keep on rising, thanks in part to acclaimed artists like the Arctic Monkeys or Jack White flying the flag for vinyl.
Press to play?
However that comeback could be short-lived in a no-deal Brexit scenario. Currently there are only a couple of plants still producing vinyl in the UK. The majority of records are pressed in mainland Europe. If the UK leaves the customs union, tariffs would be levied on all imports from Europe. That would have a damaging impact on sales and margin, especially for smaller, independent labels.
"Almost all the production of vinyl sold in the UK is done in EU territories whether that be the Czech Republic or Germany. All the pressing plants will remain in EU and the vinyl will have to be imported to the UK. Any delays and any additional costs go on to the cost of production and that obviously reduces profit," Ian Moss, director of public affairs at the British Phonographic Industry (BPI), told DW.
Some overseas pressing plants have said they would look into relocating parts of their operation to the UK, however given that many of the materials used in the manufacturing process also originate abroad, relocation may not be as easy as it sounds.
"Although the pressing happens externally, a lot of the movement happens from within warehouses in the UK. So it's effectively being shipped here and then shipped out around the world. So that I think is going to be one significant change as to whether self releases and small music businesses will look to handle warehousing outside of the UK," said Gee Davy, head of Legal & Business Affairs with the Association of Independent Music (AIM) in London.
Most of the vinyl pressing plants are in mainland Europe, like this one in the eastern German state of Saxony
For small, independent labels that additional burden and the costs could put them out of business. "The first thing is to make sure there is enough time for production to get everything across the Channel in a process which might be very difficult because of queues and delays at the border. Any increase in costs dents profits. And it's going to hit small producers, small bands. [Vinyl] is a really popular product, a premium product and it's a really important part of the mix of how labels get enough money to keep going and two thirds of our members are tiny businesses," said Moss.
No money, Mo problems
Assuming that Brexit raises production costs for record labels, consumers will bear the brunt, at least partially. The average LP is not cheap as it is, retailing at €20. Asking music fans to shell out even more could put them off vinyl.
Beyond that, Brexit in a no-deal scenario will undoubtedly put a strain on the wider UK economy, forcing consumers to think twice about what they spend their money on.
Currently if you order a record at, say, Roughtrade records in London, chances are it'll be with you the next day per express delivery. However, in a no-deal post-Brexit world that smooth supply chain would hit choppy waters. Custom tariffs, border controls and increased red tape could force many smaller, independent labels/stores and emerging artists to rethink their approach to vinyl once the costs outweigh the benefits.
Then there's the issue of freedom of movement. Were that to end it would prevent emerging artists, who rely increasingly on touring to make a living, from building an international audience in mainland Europe and being able to sell their merchandise, which increasingly includes vinyl, at the venues.
Even though it's not clear yet what type of Brexit the UK is heading for, the pound is already experiencing volatility. As the majority of records sold in the UK are imported from mainland Europe, the weakness of the pound means unfavorable exchange rates and increased costs for importers which result in higher vinyl prices.
So will vinyl's revival continue given all the uncertainties and unknowns surrounding Brexit? "Vinyl's a really popular product among fans. It's a small but significant part of the market. And it's a really good product. People like the artwork and inlay and the packaging. And all of that is unique and special to vinyl and that's why it's so popular," said Moss.
Specialist vinyl stores have enjoyed a revival — but they make have to rethink their business strategy in a no-deal Brexit scenario
That popularity, said AIM's Davy, has provided an economic spark for many small and specialist stores. "We've had reports from independent music stores within the UK that some of them have been able to expand their shops purely as a result of the vinyl revival."
Both Moss and Davy stress that the uncertainties surrounding Brexit coupled with a lack of guidance from the government make it almost impossible to predict the full impact on the industry.
However Davy firmly points the finger at the government for ignoring the importance of the arts and culture industry overall for too long.
"You know the soft power of music is phenomenal. I think a lot of Britain's trading strength has to do with its cultural strength and I think it's an absolute tragedy the way Brexit's unfolded for the music industry. And I think it's going to be hugely damaging for many years because the government hasn't focused enough on the cultural sector."
What is this viny plant in the UK? - Biology
Osmosis is the way in which plants take up water. This is how. Root hairs of the plant take in water from the soil by osmosis. The cell membrane of the root hair cell acts a partially permeable membrane (the cell wall is fully permeable) and because the cell sap inside the vacuole is a strong solution (low water concentration) water passes from the soil (high water concentration) into the root hair cell by osmosis. The concentration of the sap in the vacuole is now weaker as there is a high concentration of water. Water will now pass from this area of high concentration to the next cell which has a low water concentration by osmosis. In this way water continues to move along the cells of the root up the xylem to the leaf. all the time water is moving to areas of lower water concentration
As water enters plant cells it makes the cell swell up. The water moves into the plant cell vacuole and pushes against the cell wall. Eventually the cell contains as much water as it can hold. The strong cell wall stops the cell bursting. We say that cell is turgid. Turgid cells are useful implants as the give the plant support as they keep the stems of plants upright.
When plants are placed into a strong sugar or salts solution water will pass out of the cells by osmosis. As water passes out, the sap vacuole starts to shrink. These cells are no longer firm they are limp. We say that they are flaccid and the plant will wilt.
If a lot of water leaves the cells then the cytoplasm starts to peel away from the cell wall. We say that the cell has undergone plasmolysis.
Lonicera sempervirens ‘Blanche Sandman’
Lonicera sempervirens ‘Blanche Sandman’ was found and sold by Blanche in St Matthews in her tiny backyard nursery as a honeysuckle. Allen Bush said "When I had Holbrook Farm, I decided to put a name on it." It is known for its consistency of bloom and ability to thrive where planted. Like the species and its cultivars this Kentucky native is a long term bloomer that is available to pollinators and hummingbirds. The bloom is very showy with red, purple, and a bit more orange in the bloom depending on the season and environment compared to the native species which is mostly red flowers.The vining habit makes it a great plant for growing on structures and other plants. It can become particularly rangy and viny in shade but will still flower.
Lonicera sempervirens species and cultivars are readily propagated using "firm" wood treatby cuttings.
N-318 Ag. Science Center North
Lexington, KY 40546-0091
Top 26 Problems on Genetics
Q. 1. The round shape of pea seeds is dominant over the wrinkled shape of seeds. If two homozygous parents belonging to round and wrinkled character are crossed, then describe the F2 generation including test cross ratio.
In the test cross, F1 hybrid (Ww) is back crossed to one of the homozygous recessive parent like-
Thus test cross ratio shall be 1 round (Ww): 1 wrinkled (ww).
Q. 2. When a red fruited tomato plant is selfed, red and yellow fruited plants are produced in the ratio of 32:10. What is the genotype of the parent?
Note. In tomato red fruit colour is dominant over yellow.
Given ratio-Red: Yellow = 32: 10
Approximate phenotypic ratio = 3: 1
This 3:1 is the monohybrid ratio. The red fruited plant must be a hybrid one. The genotype is Rr. While crossing a pure bred red with pure yellow, this gives a hybrid but with red fruit.
Thus, the genotype of parent is Rr.
Q. 3. A normal wife carrying gene for colour-blindness marries a normal husband. Give the genotype and phenotype of their children. Further, if normal wife marries with colour blind husband what shall be their resulting generations ?
The wife (female) is represented by XX and husband (male) by XY chromosomes. The factors or genes for colour-blindness are present in X-chromosome.
In the first part of the question, wife (female) is carrier of colour blind gene and husband (male) is normal.
Therefore, in their generation by the marriage of normal wife (carrier) and normal husband, 50% of the sons will be normal and 50% will be colour-blind. Among daughters 50% will be pure normal and 50% normal (carrier).
In the 2nd part of question a pure wife marries with colorblind husband.
Therefore, as a result of marriage between pure wife and colorblind husband, their total sons will be normal and daughters will be carriers for colour-blindness.
Q. 4. Albinism, the inability to synthesise chlorophyll is a recessive character over normal green colour. If a normal green tobacco plant is crossed with albino plant, the green and albino seedlings are produced in the ratio of 22: 24. Determine the genotype of the parents.
Observed ratio = Green:albino 22:24
Assumed ratio = phenotype = 1:1
This is the ratio of monohybrid backcross.
(i) The albino plant must have both the recessive factors for green colour gg, otherwise its phenotype will be affected. Therefore, the genotype of albino plant is gg.
(ii) The green plant must be hybrid one i.e., Gg then we get the above ratio 1:1.
This experiment is a monohybrid back cross with recessive homozygous parent.
Q. 5. In fowls the gene for black feathers B is incompletely dominant over its allele for white feather b, the heterozygous being slate blue. Determine the F2 phenotypic and genotypic ratio of a cross between a black and a white bird. What will be the result when F1 bird produced in above cross is mated to a black bird and to a white bird ?
Q. 6. Bateson & Punnett demonstrated in sweet pea that when two white flowered varieties were crossed, F1 hybrids were all purple coloured but in F2 generation it segregated in to purple and white colour. Give the genotype of parents and grand parents.
Q. 7. In maize purple colour (P) is dominant over the green colour (p) and crinkled or cut character (c) is recessive to normal character (C) of leaves. What will be the phenotype and genotype of F1 and F2 generation if purple normal maize is crossed with green crinkled.
Thus, phenotypic ratio of F2 generation will be 9 Purple normal, 3 purple cut, 3 green normal and 1 green cut.
Q. 8. In fruit fly, Drosophila melanogaster, genes for normal developed wings are dominant over the genes for vestigial wings. Describe the genotypes of the parents and F2 genotypic and phenotypic ratio of the flies.
In fruit fly the normal fly will contain V and vestigial wings will be represented by v genotype.
After self fertilization Vv x Vv, the F2 will be:
Phenotypic ratio = 3 Normal: 1 Vestigial winged
Q. 9. In a cross between early and late rice, 3,570 early types and 1,200 late types appeared in segregating families. Find out the genotype of the parents and grand parents.
(i) The character pair early-late is governed by a single pair of Mendelian alleles.
(ii) The segregation is taking place in the ratio of 3:1
Phenotypes : 3 Early types : 1 Late type
Q. 10. Tall tomato plants are produced by the action of a dominant allele ‘D’ and dwarf plants by its recessive allele ‘d’. Hairy stems are produced by a dominant ‘H’ genes and hairless stems by its recessive genes ‘h’. A di-hybrid tall hairy plant is test crossed. The F2 progeny were observed to be 118 tall hairy, 112 tall hairless, 110 dwarf hairy and 120 dwarf hairless.
(b) What is the ratio of tall dwarf and hairy hairless.
(c) Are these 2 loci assorting independently of one another?
(a) This is a di-hybrid back cross ratio.
The phenotypes are as follows:
Assumed phenotypic ratio = 1:1:1:1
(b) The ratio of Tall : Dwarf = 1 : 1
The ratio of Hairy : Hairless = 1 : 1
(c) Yes. These two loci are assorting independently or freely with each other. Only because of this independent assortment four possible combinations are brought about in equal numbers.
Q. 11. A plant producing white, rotate shaped flowers is crossed with one producing cream, funnel shaped flower. Out of 76 offspring, 37 produce white, rotate shaped flowers and 39 produce cream, rotate shaped flowers. What are the genotypes of the parents?
1 (white rotate shaped) : 1 (cream rotate shaped)
Q. 12. Mating of a short tailed mouse with a normal long tailed mouse produce 1 : 1 ratio of short tailed and normal tailed mice. Crossing of two short tailed mice gives 2:1 ratio of short tail to normal tail using appropriate symbol diagram and explain these results. Is the gene for short tail a dominant or a recessive one? The gene for short tail is a dominant gene.
Let the genotype of the short tailed mouse be—Ss and normal long tailed mouse—ss
The given 1 : 1 ratio coincides with the derived ratio.
Phenotypic ratio : 2 short tailed : 1 normal long tailed.
The dominant gene for short tail condition when present in the homozygous form (SS) is lethal to the individual. So, the individual does not survive and we have 2:1 ratio of short tailed & normal long tailed mouse respectively.
Q. 13. In case of paddy an awned variety was crossed with an awnless one. In F1 all the plants were awned but in F2 20 plants were awnless and 310 awned. How can these results be explained genetically.
(i) Character awned-awnless is governed by 2 pair of alleles.
(ii) Assumed ratio is 15:1. This is the modified di-hybrid ratio, duplicate factors.
Q. 14. A breeder crossed two white flowered varieties of Pea. In F1 the colour of the flower appeared as pink. On raising F2 population the segregation took place as follows-
White flowered plants = 168
Pink flowered plants = 230
The assumed ratio is modified di-hybrid cross, complementary factors. Two pairs of genes are involved to govern the flower colour. When they remain alone, they produce white flowers. But when both the dominant factors come together, they interact to produce pink colour.
Q. 15. A variety of pepper having brown fruits was crossed with a variety having yellow fruits. All the F1 plants had red fruits. When the red fruited F1 plants were crossed together, the F2 consisted of 180 plants with red fruits, 60 with brown fruits, 58 with yellow fruits and 18 with green fruits. What appears to be the genetic basis of these fruit colours in pepper?
The fruit colour is governed by two different allelic pair of genes. Individually one dominant gene shows brown (B) and the other yellow (Y) colour. But when both the dominant genes meet together, after interaction produce red (BY) colour and the double recessive condition gives green colour (bbyy).
According to this hypothesis the segregation in F2 is in the ratio of 9:3:3:1 (interaction of genes).
Q. 16. In rice when purple pigmented plant was crossed with green pigmented plant, the F1 appeared green pigmented. In F2 the segregation of two types of pigmentation was as follows.
What may be the genetic basis for the segregation of pigmentation in F2.
Here is involved inhibitory factor. The segregation is in 13:3 ratio.
Let P = Purple pigmentation
I = Inhibits the effect of P
i = Does not inhibit the effect of P
Q. 17. In chicken the dominant gene R gives rose comb and the dominant gene P gives pea comb. When P and R are present together the comb form is walnut. The homozygous recessives of P and R produce single comb. Determine the comb form of the offspring of the following crosses:-
Q. 18. In summer squash the genes for white (W) colour is epistatic to that of an yellow (Y) which is dominant over green. Determine the fruit colour of the following crosses.
This is the case of modified di-hybrid ratio, epistatic factor. The gene ‘W’ acts as the epistatic factor and it will produce white coloured fruits irrespective of the presence of ‘Y’ (hypostatic factor) or ‘y’. If Y is present, the fruit colour will be Yellow and if ‘y’ is present, the fruit colour will be green.
Q. 19. A large number of crosses were made between yellow coated mice.
The progeny segregated as follows:
Explain the problem in genetical terms?
This is monohybrid modified lethal factor (2 : 1)
Q. 20. Assume that in man the difference in skin colour between negro and white is due to two pairs of factors. AABB is black and aabb is white and that any three of the colour producing factor produce dark skin and any one colour producing factor will produce light skin. What will be the skin colour of the offspring from a mating of two individuals of F1?
Phenotype : 1 Black : 4 Dark : 6 Intermediate : 4 Light : 1 White Progenies
Q. 21. In sweet peas C or P alone produces white flowers, the purple colour being due to the presence of both these factors. What will be the folower colour of the following crosses-
Q. 22. In flax blue flowered plants were crossed. The offspring segregated as 48 blue, 14 white and 21 green plants. What were the genotypes of the parents?
This is modified di-hybrid ratio supplementary factor. When both dominant genes are present it is blue. W-White dominant gene and G-Green dominant gene.
Q. 23. What would be the blood groups of the progeny of the following crosses:
50% OF THE PROGENY WILL BE ‘A’ GROUP
25% OF THE PROGENY WILL BE ‘AB’ GROUP
25% OF THE OFFSPRING WILL BE ‘B’ GROUP
Q. 24. In fowls gene for rose comb (R) and pea comb (P) if present together produce walnut comb. The recessive allele of both, when present together in homozygous condition produce single comb. What will be the comb character of the offspring of the following crosses in which the genotype of the parents are given?
What will be the genotype of the parent and offspring of the following crosses?
(i) Pea x Walnut → 3/8 walnut, 3/8 Pea, 1/8 rose, 1/8 single offspring.
(ii)Walnut x Single → 1/4 walnut, 1/4 rose, 1/4 pea, 1/4 single offspring.
(iii) Walnut X Rose 3/8 walnut, 3/8 rose, 1/8 pea, 1/8 single offspring.
(iv) Rose X Pea 1 walnut, 1 rose, 1 pea, 1 single offspring.
(v) Rose X Pea → 10 walnut, 15 pea offspring.
Q. 25. In snapdragons, red flower colour (R) is completely dominant over white (r) and broad leaves (B) are incompletely dominant over narrow (b). The heterozygous condition being intermediate leaf breadth. What would be the proportion of various phenotypes in the offspring of the following crosses
Q. 26. Determine the different gametes produced by the following: