The color of a pea is determined by the pairs of genes AA, Aa, aa. Pea plants with the gene pair aa are yellow, while others are green.
A total of 100 peas were planted. Out of the 100 peas, 45 had the gene pair AA and the rest had the gene pair aa. After one year, all the harvested peas were planted again. Each pea inherited one gene from its parent's gene pair, which could have been either AA or aa.
What fraction of the second-year peas will be yellow?
If the third-year peas are yellow, what is the probability that they inherited the genes from green-colored parents?
My solution:
Regarding the 1st question, my way of thinking goes like this:
First year peas can be either AA or aa. Probability of being AA is 0.45, Aa - 0, aa - 0.55.
Second year peas are AA if they inherit AA genes from both of the parents, so the probability is 0.45 * 0.45 = 0.2025. They are aa if they inherit aa genes from both of the parents, and so the probability is 0.55 * 0.55 = 0.3025. The probability of them being Aa is 1 - 0.2025 - 0.3025 = 0.495, then.
Since peas are yellow when their gene pair is aa, the fraction of second-year peas that are yellow is 0.3025.
However, when it comes to the second question I get quite lost since the peas can also be Aa and both either a or A genes can be inherited now.
Do you have any tips that would help me to advance in solving this problem? Any help appreciated.
Per definition of conditional probability:
$P(\text{parents are both Aa}|\text{pea is yellow})=\frac{P(\text{parents are Aa }and\text{ pea is yellow})}{P(\text{pea is yellow})}$
Now $P(\text{Parents are both Aa & pea is yellow}) = P(\text{First parent is Aa})\cdot P(\text{Second parent is Aa})\cdot P(\text{first parent gives a-gene})\cdot P(\text{second parent gives a-gene})=0.495\cdot0.495\cdot\frac12\cdot\frac12$
do you know how to continue from here?