Imagine a recessive mutation in rabbits that causes wiry fur (whereas the dominant, normal allele causes fluffy fur). If two heterozygous rabbits are mated, what is the probability that the first 2 out of 4 offspring will have fluffly fur

the required probability is  0.03516

Step-by-step explanation:

Given the data in the question;

let F represent dominant allele that causes fluffy fur and

f represent corresponding recessive allele that causes wiry fur

here are the possible genotypes and corresponding phenotypes;

FF = Fluffy fur

Ff = Fluffy fur

ff = Wiry fur

genotype of parents is Ff and Ff

and there cross will be; Ff x Ff

now Offspring will be;

                                   F                      f

F                                FF                    Ff

f                                 Ff                     ff

so the probability that the offspring will have fluffy fur = 3/4

and the probability that the offspring will have wiry fur = 1/4

The probability that the first 2 out of 4 offspring will have fluffy fur will be;

⇒ 3/4 × 3/4 × 1/4 × 1/4 =  0.03516

 Therefore, the required probability is  0.03516

13.

Step-by-step explanation:

Here's the equation you can write:

s-7/3=2

s-7=6

s=13