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In maize, coloured endosperm (C) is dominant over colourless (c); and full endosperm (R) is dominant over shrunken (r). When a dihybrid of F1 generation were test crossed, it produced four phenotypes in the following percentage:
Coloured full 48%, Coloured shrunken 5%,
Colourless full 7%, and Colourless shrunken 40%.
Colourless full 7%, and Colourless shrunken 40%.
From this data, what will be the distance between two non-allelic genes?
- a)48 units
- b)5 units
- c)7 units
- d)12 units
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
In maize, coloured endosperm (C) is dominant over colourless (c); and ...
Given that recombination percentage is 7% and 5%, therefore, total recombinants would be 7 + 5 = 12%. It is known that one map unit is the distance that yields 1% recombinant chromosomes. Hence the distance between two non-allelic genes is 12 map units.
Most Upvoted Answer
In maize, coloured endosperm (C) is dominant over colourless (c); and ...
Given:
- In maize, coloured endosperm (C) is dominant over colourless (c); and full endosperm (R) is dominant over shrunken (r).
- When a dihybrid of F1generation were test crossed, it produced four phenotypes in the following percentage:
- Coloured full 48%
- Coloured shrunken 5%
- Colourless full 7%
- Colourless shrunken 40%.
To find:
- Distance between two non-allelic genes
Solution:
- Since we are given four phenotypes, we can assume that the two genes (C/c and R/r) are on different chromosomes and independently assorting.
- We can write the gametes produced by the dihybrid as CR, Cr, cR, and cr.
- When these gametes are test crossed with ccrr (homozygous recessive), the progeny will show the phenotypes corresponding to the gametes produced by the dihybrid.
- Let's write the Punnett square for this test cross:
| | c | c
--|----|----|---
r | cr | cr | cr
r | cr | cr | cr
- From this Punnett square, we can calculate the expected frequencies of each phenotype:
- Coloured full: 1/4 x 1/4 = 1/16 (homozygous dominant for both traits)
- Coloured shrunken: 1/4 x 1/4 = 1/16 (heterozygous for both traits)
- Colourless full: 1/4 x 1/4 = 1/16 (heterozygous for both traits)
- Colourless shrunken: 1/4 x 1/4 = 1/16 (homozygous recessive for both traits)
- We can convert these frequencies into percentages:
- Coloured full: 1/16 x 100% = 6.25%
- Coloured shrunken: 1/16 x 100% = 6.25%
- Colourless full: 1/16 x 100% = 6.25%
- Colourless shrunken: 1/16 x 100% = 6.25%
- However, we are given slightly different percentages in the question. We can calculate the deviation (difference between observed and expected frequencies) for each phenotype:
- Coloured full: 48% - 6.25% = 41.75%
- Coloured shrunken: 5% - 6.25% = -1.25%
- Colourless full: 7% - 6.25% = 0.75%
- Colourless shrunken: 40% - 6.25% = 33.75%
- We can convert these deviations into map units (centimorgans) using the formula:
- (deviation / total number of progeny) x 100 = map units
- For example, the map distance between C/c and R/r can be calculated as:
- (41.75 / 100) x 100 = 41.75 map units
- We can repeat this calculation for the other pairs of genes:
- In maize, coloured endosperm (C) is dominant over colourless (c); and full endosperm (R) is dominant over shrunken (r).
- When a dihybrid of F1generation were test crossed, it produced four phenotypes in the following percentage:
- Coloured full 48%
- Coloured shrunken 5%
- Colourless full 7%
- Colourless shrunken 40%.
To find:
- Distance between two non-allelic genes
Solution:
- Since we are given four phenotypes, we can assume that the two genes (C/c and R/r) are on different chromosomes and independently assorting.
- We can write the gametes produced by the dihybrid as CR, Cr, cR, and cr.
- When these gametes are test crossed with ccrr (homozygous recessive), the progeny will show the phenotypes corresponding to the gametes produced by the dihybrid.
- Let's write the Punnett square for this test cross:
| | c | c
--|----|----|---
r | cr | cr | cr
r | cr | cr | cr
- From this Punnett square, we can calculate the expected frequencies of each phenotype:
- Coloured full: 1/4 x 1/4 = 1/16 (homozygous dominant for both traits)
- Coloured shrunken: 1/4 x 1/4 = 1/16 (heterozygous for both traits)
- Colourless full: 1/4 x 1/4 = 1/16 (heterozygous for both traits)
- Colourless shrunken: 1/4 x 1/4 = 1/16 (homozygous recessive for both traits)
- We can convert these frequencies into percentages:
- Coloured full: 1/16 x 100% = 6.25%
- Coloured shrunken: 1/16 x 100% = 6.25%
- Colourless full: 1/16 x 100% = 6.25%
- Colourless shrunken: 1/16 x 100% = 6.25%
- However, we are given slightly different percentages in the question. We can calculate the deviation (difference between observed and expected frequencies) for each phenotype:
- Coloured full: 48% - 6.25% = 41.75%
- Coloured shrunken: 5% - 6.25% = -1.25%
- Colourless full: 7% - 6.25% = 0.75%
- Colourless shrunken: 40% - 6.25% = 33.75%
- We can convert these deviations into map units (centimorgans) using the formula:
- (deviation / total number of progeny) x 100 = map units
- For example, the map distance between C/c and R/r can be calculated as:
- (41.75 / 100) x 100 = 41.75 map units
- We can repeat this calculation for the other pairs of genes:
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Community Answer
In maize, coloured endosperm (C) is dominant over colourless (c); and ...
Given that recombinant % is 7% and 5% therefore, total recombinants would be 7+5=12%. known that one map unit is the distance that yields 1% recombinants chromosomes. hence distance between two non-allelic genes is 12 map units
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In maize, coloured endosperm (C) is dominant over colourless (c); and full endosperm (R) is dominant over shrunken (r). When a dihybrid of F1generation were test crossed, it produced four phenotypes in the following percentage:Coloured full 48%, Coloured shrunken 5%,Colourless full 7%, and Colourless shrunken 40%.From this data, what will be the distance between two non-allelic genes?a)48 unitsb)5 unitsc)7 unitsd)12 unitsCorrect answer is option 'D'. Can you explain this answer?
Question Description
In maize, coloured endosperm (C) is dominant over colourless (c); and full endosperm (R) is dominant over shrunken (r). When a dihybrid of F1generation were test crossed, it produced four phenotypes in the following percentage:Coloured full 48%, Coloured shrunken 5%,Colourless full 7%, and Colourless shrunken 40%.From this data, what will be the distance between two non-allelic genes?a)48 unitsb)5 unitsc)7 unitsd)12 unitsCorrect answer is option 'D'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about In maize, coloured endosperm (C) is dominant over colourless (c); and full endosperm (R) is dominant over shrunken (r). When a dihybrid of F1generation were test crossed, it produced four phenotypes in the following percentage:Coloured full 48%, Coloured shrunken 5%,Colourless full 7%, and Colourless shrunken 40%.From this data, what will be the distance between two non-allelic genes?a)48 unitsb)5 unitsc)7 unitsd)12 unitsCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In maize, coloured endosperm (C) is dominant over colourless (c); and full endosperm (R) is dominant over shrunken (r). When a dihybrid of F1generation were test crossed, it produced four phenotypes in the following percentage:Coloured full 48%, Coloured shrunken 5%,Colourless full 7%, and Colourless shrunken 40%.From this data, what will be the distance between two non-allelic genes?a)48 unitsb)5 unitsc)7 unitsd)12 unitsCorrect answer is option 'D'. Can you explain this answer?.
In maize, coloured endosperm (C) is dominant over colourless (c); and full endosperm (R) is dominant over shrunken (r). When a dihybrid of F1generation were test crossed, it produced four phenotypes in the following percentage:Coloured full 48%, Coloured shrunken 5%,Colourless full 7%, and Colourless shrunken 40%.From this data, what will be the distance between two non-allelic genes?a)48 unitsb)5 unitsc)7 unitsd)12 unitsCorrect answer is option 'D'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about In maize, coloured endosperm (C) is dominant over colourless (c); and full endosperm (R) is dominant over shrunken (r). When a dihybrid of F1generation were test crossed, it produced four phenotypes in the following percentage:Coloured full 48%, Coloured shrunken 5%,Colourless full 7%, and Colourless shrunken 40%.From this data, what will be the distance between two non-allelic genes?a)48 unitsb)5 unitsc)7 unitsd)12 unitsCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In maize, coloured endosperm (C) is dominant over colourless (c); and full endosperm (R) is dominant over shrunken (r). When a dihybrid of F1generation were test crossed, it produced four phenotypes in the following percentage:Coloured full 48%, Coloured shrunken 5%,Colourless full 7%, and Colourless shrunken 40%.From this data, what will be the distance between two non-allelic genes?a)48 unitsb)5 unitsc)7 unitsd)12 unitsCorrect answer is option 'D'. Can you explain this answer?.
Solutions for In maize, coloured endosperm (C) is dominant over colourless (c); and full endosperm (R) is dominant over shrunken (r). When a dihybrid of F1generation were test crossed, it produced four phenotypes in the following percentage:Coloured full 48%, Coloured shrunken 5%,Colourless full 7%, and Colourless shrunken 40%.From this data, what will be the distance between two non-allelic genes?a)48 unitsb)5 unitsc)7 unitsd)12 unitsCorrect answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for NEET.
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In maize, coloured endosperm (C) is dominant over colourless (c); and full endosperm (R) is dominant over shrunken (r). When a dihybrid of F1generation were test crossed, it produced four phenotypes in the following percentage:Coloured full 48%, Coloured shrunken 5%,Colourless full 7%, and Colourless shrunken 40%.From this data, what will be the distance between two non-allelic genes?a)48 unitsb)5 unitsc)7 unitsd)12 unitsCorrect answer is option 'D'. Can you explain this answer?, a detailed solution for In maize, coloured endosperm (C) is dominant over colourless (c); and full endosperm (R) is dominant over shrunken (r). When a dihybrid of F1generation were test crossed, it produced four phenotypes in the following percentage:Coloured full 48%, Coloured shrunken 5%,Colourless full 7%, and Colourless shrunken 40%.From this data, what will be the distance between two non-allelic genes?a)48 unitsb)5 unitsc)7 unitsd)12 unitsCorrect answer is option 'D'. Can you explain this answer? has been provided alongside types of In maize, coloured endosperm (C) is dominant over colourless (c); and full endosperm (R) is dominant over shrunken (r). When a dihybrid of F1generation were test crossed, it produced four phenotypes in the following percentage:Coloured full 48%, Coloured shrunken 5%,Colourless full 7%, and Colourless shrunken 40%.From this data, what will be the distance between two non-allelic genes?a)48 unitsb)5 unitsc)7 unitsd)12 unitsCorrect answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice In maize, coloured endosperm (C) is dominant over colourless (c); and full endosperm (R) is dominant over shrunken (r). When a dihybrid of F1generation were test crossed, it produced four phenotypes in the following percentage:Coloured full 48%, Coloured shrunken 5%,Colourless full 7%, and Colourless shrunken 40%.From this data, what will be the distance between two non-allelic genes?a)48 unitsb)5 unitsc)7 unitsd)12 unitsCorrect answer is option 'D'. Can you explain this answer? tests, examples and also practice NEET tests.
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