Answer:
Let us assume:
Spotted coat color = ss (homozygous recessive)
Solid coat color = SS (homozygous dominant)
and, the heterozygous = Ss
,
and the gametes formed would be S and s for each parent, so the next generation would be :
Male/Female S s
S <u><em>SS(Solid coat color pups) </em></u> <em><u> Ss(Solid coat color pups)
</u></em>
s <u><em> Ss(Solid coat color pups)</em></u> <u><em> ss(Spotted coat color pups)
</em></u>
<em>
B</em>y this punnet square the ratio of phenotype is : 3 : 1 of solid : spotted
On this basis
i) 4 with solid fur and 1 with spotted fur out of first litter of five:
Using binomial expansion equation, where n = 5, x = 4, p = 0.75, q = 0.25. The answer is 0.396 = 39.6% of the time.
ii) A first litter of six pups, four with solid fur and two with spotted fur, and then a second litter of five pups, all with solid fur. For the first litter, n = 6, x = 4, p = 0.75, q = 0.25; for the second litter, n = 5, x = 5, p = 0.75, q = 0.25.
Due to the litters are in a specified order, one need to use the product rule and multiply the probability of the first litter times and the second litter. The answer is 0.070, or 7.0%.
iii) The product rule has been applied and multiply the probability of the first pup (0.75) times the probability of the remaining four.
By using binomial expansion equation the probability of the remaining four, where n = 4, x = 3, p = 0.75, q = 0.25. The probability of the first litter is 0.316.
Similarly the probability of the remaining five by binomial expansion equation, where n = 5, x = 4, p = 0.75, q = 0.25. The probability of the second litter is 0.025.
The probability of the first litter (0.316) times the probability of second litter (0.025). The answer is 0.008, or 0.8%.
