Let's understand the inheritance pattern of pea plants for determining probabilities. We know round peas (R) are dominant over wrinkled peas (r).
### Second to Third Generation Cross
When second-generation pea plants, each with the genotype Rr, are crossed, they form the following Punnett square:
[tex]\[
\begin{array}{c|c|c}
& R & r \\
\hline
R & RR & Rr \\
\hline
r & Rr & rr \\
\end{array}
\][/tex]
This table gives us the probabilities of the various genotypes in the third generation:
- [tex]\( RR \)[/tex] (homozygous dominant - round)
- [tex]\( Rr \)[/tex] (heterozygous - round)
- [tex]\( rr \)[/tex] (homozygous recessive - wrinkled)
Probability of [tex]\(RR\)[/tex]:
[tex]\[
\text{Number of } RR \text{ genotypes} = 1 \text{ (out of 4 possibilities)} \implies P(RR) = \frac{1}{4} = 0.25
\][/tex]
Probability of [tex]\(Rr\)[/tex]:
[tex]\[
\text{Number of } Rr \text{ genotypes} = 2 \text{ (out of 4 possibilities)} \implies P(Rr) = \frac{2}{4} = 0.50
\][/tex]
Probability of round peas (either [tex]\(RR\)[/tex] or [tex]\(Rr\)[/tex]):
[tex]\[
P(\text{Round}) = P(RR) + P(Rr) = 0.25 + 0.50 = 0.75
\][/tex]
Probability of [tex]\(rr\)[/tex]:
[tex]\[
\text{Number of } rr \text{ genotypes (wrinkled)} = 1 \text{ (out of 4 possibilities)} \implies P(rr) = \frac{1}{4} = 0.25
\][/tex]
### Summary:
(a) The probability that a third-generation offspring has round peas is [tex]\(0.75\)[/tex].
(b) The probability that a third-generation offspring has wrinkled peas is [tex]\(0.25\)[/tex].
