Answer:
look at the pictures
Explanation:
1. If we cross the parental generation which is RR x rr, the probability of the F1 generation to get rough coat is 100% and they will be a carrier of the recessive trait smooth coat.
Since the phenotypic ratio is 100% heterozygous Rr, in crossing the F1 to get the F2, we will use the genotype Rr. To get the F2, use the cross Rr x Rr.
The phenotypic ration for F2 is 3:1. There is 75% to get rough coat and 25% smooth. The answer is based on the result on the Punnett square. On the other hand, the genotypic ratio is 1:2:1. There is 25% probability to get RR genotype, 50% Rr, and 25% rr.
2. Since the two parental mice got 6 albino offspring and 5 brown mice offspring, it is approximately 50%. it takes a parental mice who is Brown that is carrying an albino trait crossed with an albino to get offspring with almost the same number. Therefore, the genotype of the brown mice is Aa.
