Dwarfism is recessive to normal size in cattle and many other species. In Herefo

Question

Dwarfism is recessive to normal size in cattle and many other species. In Hereford cattle a summary reveled that 4% of the cattle were dwarf and 96% of the cattle were normal size. Given the information, compute the expected gene frequency for the normal gene (N) and for the dwarf gene (n).

p(N) = ________________ (8, how is this?)

q(n)=__________________(.2, how is this?)

In the phenotypic distribution given above, what proportion of the calves would be expected to carry the dwarf gene, but be of normal size?

percent Nn =________________________ (32%, how is this?)

Explanation / Answer

First of all we have to start with the recessive homozygous genotype frequency.

So 4% of cattles are dwarf.

So recessive genotype frequency (q2nn) = .04

There fore allele frequency q(n) = root over(.04)

= 0.2

Therefore p(N) = 1- 0.2 As p +q =1 according to Hardy Wienberg Law

= 0.8

Now In the phenotypic distribution given above, what proportion of the calves would be expected to carry the dwarf gene, but be of normal size?

That means the genotype of the following calves would be - (Nn). Nn means normal size but which carries the dwarf gene that is -n. They are normal as "N" is dominant on -"n".

So we need to find out the genotype frequency of the heterozygotes. It will be-

2Pq = 2 X .8 X.2 ( As we know from Hardy Wienberg Law- P2NN +2Pq(Nn) +q2nn =1)

= .32

Therefore the percentage of Nn = 32%.

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