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n maize coloured endosperm (C) is dominant over colourless (c) and full endosperm (R) is dominant over shrunken (r). When a dihybrid of F1-generation was test crossed it produced four phenotypes in the following percentage
Coloured and Full = 45%
Coloured – Shrunken = 5%
Colourless – Full = 4%
Colourless  – Shrunken = 46%
From these data what would be distance between the two non allelic genes :-
  • a)
    48 unit
  • b)
    9 unit
  • c)
    4 unit
  • d)
    12 unit
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
n maize coloured endosperm (C) is dominant over colourless (c) and ful...
Since non allelic genes are mentioned that means recombination will take place and distance is equal to recombinant frequency=5+4/100*100 = 9 %=9unit.
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n maize coloured endosperm (C) is dominant over colourless (c) and ful...
To determine the distance between the two non-allelic genes, we can use the phenotypic ratios from the test cross of the dihybrid F1 generation. The four phenotypes observed are:

1. Coloured and Full: 45%
2. Coloured and Shrunken: 5%
3. Colourless and Full: 4%
4. Colourless and Shrunken: 46%

Let's analyze these phenotypic ratios and determine the distance between the two genes.

**Step 1: Determine the parental genotypes**
We know that maize coloured endosperm (C) is dominant over colourless (c) and full endosperm (R) is dominant over shrunken (r). Based on the phenotypic ratios, we can deduce the parental genotypes as follows:

1. Coloured and Full: This phenotype can be produced by the genotype CR/CR or CR/Cr.
2. Coloured and Shrunken: This phenotype can be produced by the genotype CR/rr.
3. Colourless and Full: This phenotype can be produced by the genotype Cc/CR or Cc/Cr.
4. Colourless and Shrunken: This phenotype can be produced by the genotype Cc/rr or cc/rr.

**Step 2: Determine the recombinant genotypes**
To calculate the distance between the two non-allelic genes, we need to determine the recombinant genotypes. In this case, the recombinant genotypes are the ones that deviate from the parental genotypes. From the phenotypic ratios, we can identify two recombinant phenotypes:

1. Coloured and Shrunken: This phenotype is a recombinant as it deviates from the parental genotypes CR/CR or CR/Cr.
2. Colourless and Full: This phenotype is a recombinant as it deviates from the parental genotypes Cc/CR or Cc/Cr.

**Step 3: Calculate the recombination frequency**
The recombination frequency is the percentage of recombinant phenotypes among the total phenotypes observed. In this case, the recombinant phenotypes are:

1. Coloured and Shrunken: 5%
2. Colourless and Full: 4%

The total percentage of these recombinant phenotypes is 5% + 4% = 9%.

**Step 4: Determine the distance between the genes**
The recombination frequency is a measure of the distance between genes on a chromosome. It is expressed as a percentage. In this case, the recombination frequency is 9%.

Since the recombination frequency represents the percentage of recombinant phenotypes, it also represents the percentage of crossing over events that occurred between the two genes during meiosis.

The distance between genes can be calculated using the formula:

Distance = (Recombination frequency / 2) * 100

In this case, the distance = (9 / 2) * 100 = 4.5 units.

Therefore, the correct answer is option B) 9 units.
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n maize coloured endosperm (C) is dominant over colourless (c) and full endosperm (R) is dominant over shrunken (r). When a dihybrid of F1-generation was test crossed it produced four phenotypes in the following percentageColoured and Full = 45%Coloured – Shrunken = 5%Colourless – Full = 4%Colourless – Shrunken = 46%From these data what would be distance between the two non allelic genes :-a)48 unitb)9 unitc)4unitd)12 unitCorrect answer is option 'B'. Can you explain this answer?
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n maize coloured endosperm (C) is dominant over colourless (c) and full endosperm (R) is dominant over shrunken (r). When a dihybrid of F1-generation was test crossed it produced four phenotypes in the following percentageColoured and Full = 45%Coloured – Shrunken = 5%Colourless – Full = 4%Colourless – Shrunken = 46%From these data what would be distance between the two non allelic genes :-a)48 unitb)9 unitc)4unitd)12 unitCorrect answer is option 'B'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about n maize coloured endosperm (C) is dominant over colourless (c) and full endosperm (R) is dominant over shrunken (r). When a dihybrid of F1-generation was test crossed it produced four phenotypes in the following percentageColoured and Full = 45%Coloured – Shrunken = 5%Colourless – Full = 4%Colourless – Shrunken = 46%From these data what would be distance between the two non allelic genes :-a)48 unitb)9 unitc)4unitd)12 unitCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for n maize coloured endosperm (C) is dominant over colourless (c) and full endosperm (R) is dominant over shrunken (r). When a dihybrid of F1-generation was test crossed it produced four phenotypes in the following percentageColoured and Full = 45%Coloured – Shrunken = 5%Colourless – Full = 4%Colourless – Shrunken = 46%From these data what would be distance between the two non allelic genes :-a)48 unitb)9 unitc)4unitd)12 unitCorrect answer is option 'B'. Can you explain this answer?.
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