Pigment of chicken feathers is regulated by two genes, the gene for feather pigment C, and a gene that inhibits pigment production, I. A mother chicken, who is homozygous dominant for gene C, and heterozygous for gene I, is crossed with a father chicken, who is heterozygous for the pigment gene, and heterozygous for gene I.
Inhibition is dominant to uninhibited, pigmented feathers are dominant to unpigmented feathers.
In a population of 150 chickens, produced from the same parents, how many chickens would you expect to have pigmented feathers?
For any given gene there are two alleles. It is given that the gene for inhibition I, is dominant over uninhibited which means that I is dominant over i. Gene for pigmented C, is dominant over unpigmented which means C is dominant over c.
Now, in order to solve the problem we first need to determine the parental genotypes of given chickens.
Mother: Mother is homozygous dominant for C (CC) and heterozygous for I (Ii). So, the genotype of mother is CCIi.
Father: Father is heterozygous for C (Cc) and heterozygous for I (Ii). So, the genotype of father is CcIi.
So, the cross will be CCIi x CcIi
The gametes produced by each parent will be:
Mother: CI, Ci
Father: CI, Ci, cI, ci
The punnett square showing the cross will be:
We can see that CCii and Ccii are the two genotypes out of eight genotypes formed in the punnett square where pigmented feathers are obtained while in other genotypes dominant allele of inhibition is present and hence they will not be pigmented. This means that probability of offspring with pigmented feathers is 2/8 or 25%....
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