Question

The genetic map for three genes A, B and C is as follows: A-B = 10 map units, B-C = 5 map units and A-C = 15 map units. In an individual of genotype AbC/aBc, the percentage of gametes expected to be of ABC?

Answer 1

Explanation:

Genes that are sufficiently close together on a chromosome will tend to "stick together," and the versions (alleles) of those genes that are together on a chromosome will tend to be inherited as a pair more often than not. This phenomenon is called genetic linkage. When genes are linked, genetic crosses involving those genes will lead to ratios of gametes (egg and sperm) and offspring types that are not what we'd predict from Mendel's law of independent assortment. That is the reason why the frequency of offsprings with the parental phenotype is different.

The distance between a and b = no. of recombinants/total no. of offsprings * 100

= 350/1000 *100

= 35 map units.

So, the correct answer is '35 map units'.

Answer 2

In an individual of genotype AbC/aBc, 16.67% of gametes expected to be of ABC.

Given,

The genetic map for three genes A, B and C is as follows: A-B = 10 map units, B-C = 5 map units and A-C = 15 map units.

To Find,

In an individual of genotype AbC/aBc, the percentage of gametes expected to be of ABC?

Solution,

We know that,

Genetype = AbC/aBc

Formula used = $2^{A}n$,

              where n = number of heterozygous alleles

Here, AbC and aBc are 3 heterozygous alleles

So, the gametes formed are of 6 types:

      ABc

      abC

      ABC

      abc

      AbC

      aBc

Expected Percentage = $\frac{1}{6}$ x 100 = $\frac{100}{6}$  = 16.67 %

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