In: Biology
Feathered (F-) shank in chickens is dominant to clean (ff) shanks and white feathers (I-) is dominant to black (ii) feathers. Comb structure is controlled by 2 epistatic genes R and P with alternate alleles r and p. R-P- gives the phenotype walnut, R-pp pea, rrP- rose and rrpp single. All of these genes are independently assorting. A white, feathered shank, walnut comb chicken was mated with a white, feathered, pea comb rooster and produced a black, clean, single comb chick.
Using this information, predict the proportion of additional progeny that would have the following phenotype: White, clean, walnut.
a.18/256
b.36/256
c.4/256
d.81/256
e.12/256
Lets first write the genotypes of parents:
parent1 (rooster): F_I_R_pp (F_ for feathered, I_ for white, R_pp for pea comb)
parent2 (hen): F_I_R_P_ (F_ for feathered, I_ for white, R_P_ for walnut)
the genotype of one of the offsprings: ffiirrpp. (ff for clean, ii for black, rrpp for single comb)
The offspring is totally recessive. Look closely that both the parents have F allele, but whether f allele was also present was unknown. But the chick shows recessive phenotype, meaning that recessive alleles were in fact present in the parents. Otherwise, the chick could not have this particular phenotype.
So, the genotypes of the parents now stand:
parent1 (rooster) : FfIiRrpp
parent2 (hen) : FfIiRrPp
Desired phenotype: white, clean, walnut. Its corresponding genotype would be: ffI_R_P_
To calculate the probability of arriving this genotype from the parents' genotype we just deduced. We need to individually determine the probabilities of each character and then apply the multiplication rule of probability. Essentially this becomes the probability of ffI_R_P_ or p( ffI_R_P_)= p(ff)*p(I_)*p(R_)*p(P_) = p(ff)*p(II or Ii)*p(RR or Rr)*p(PP or Pp) = 1/4*3/4*3/4*2/4=18/256. hence option (a) would be correct.
Individual calculations are shown in the image below: