The questions given are below. a is pretty straight forward. The probabilities end up the same as those of the parent generation. I am having a hard time wrapping my head around how to do part b. I tried defining the possible parent allele combinations as events $A-D$ and conditioning on $U_i$ $A_i$ but got stuck on how to move forward from there. I don't think Bayes law will help. Some help would be much appreciated!

In humans (and many other organisms), genes come in pairs. A certain gene comes in two types (alleles): type a and type A. The genotype of a person for that gene is the types of the two genes in the pair: AA, Aa, or aa (aA is equivalent to Aa). Assume that the Hardy-Weinberg law applies here, which means that the frequencies of AA, Aa, aa in the population are $p^2, 2p(1 − p), (1 − p)^2$ respectively, for some $p$ with $0 < p < 1$.

When a woman and a man have a child, the child’s gene pair consists of one gene contributed by each parent. Suppose that the mother is equally likely to contribute either of the two genes in her gene pair, and likewise for the father, independently. Also suppose that the genotypes of the parents are independent of each other (with probabilities given by the Hardy-Weinberg law).

(a) Find the probabilities of each possible genotype (AA, Aa, aa) for a child of two random parents. Explain what this says about stability of the Hardy-Weinberg law from one generation to the next. Hint: Condition on the genotypes of the parents.

(b) A person of type AA or aa is called homozygous (for the gene under consideration), and a person of type Aa is called heterozygous (for that gene). Find the probability that a child is homozygous, given that both parents are homozygous. Also, find the probability that a child is heterozygous, given that both parents are heterozygous.

My attempt:

$A_1 =$ Event that Dad is AA, Mom AA

$A_2 =$ Event that Dad is AA, Mom aa

$A_3 =$ Dad aa, Mom AA

$A_4 =$ Dad aa, Mom aa

H = Event that child is homozygous

$$ H = P(\text{Child}~ AA \cup \text{Child}~ aa \mid A_1 \cup A_2 \cup A_3 \cup A_4)$$ $$ A_1 \cup A_2 \cup A_3 \cup A_4 = \cup_i A_i $$

Since Child AA and Child aa are disjoint I think that I can separate them and add them together.

$$ H = P(\text{Child}~ AA \mid \cup_i A_i) + P(\text{Child}~ aa \mid \cup_i A_i)$$

I decided I should start by trying the definition of conditional probability on $$P(\text{Child}~AA \mid \cup_i A_i)$$

$$P(\text{Child}~AA \mid \cup_i A_i) = \frac{P(\text{Child}~AA\cap \cup_i A_i)}{P(\cup_i A_i)} $$

I am not really sure what to do with this intersection. I think the denominator should be easy to compute since you are just multiplying the parents probabilities for each event then adding everything together. I'm not really sure where to go from here.

  • $\begingroup$ The question itself is not that hard. As what you mentioned in the first paragraph, please show more details of your work so that we can know which part exactly you stuck at. $\endgroup$ – BGM Oct 2 '18 at 11:13
  • $\begingroup$ @BGM thanks for the comment. I edited the question with my work so far. $\endgroup$ – Ammar J Oct 2 '18 at 12:15
  • $\begingroup$ Those events $A_i$ are also disjoint. By the distributive law, en.wikipedia.org/wiki/… we have $P(C \cap (\cup_i A_i)) = P(\cup_i (C \cap A_i))$ and thus you can break this probability of disjoint union into sum of individual probabilities, and you know how to evaluate each term. $\endgroup$ – BGM Oct 2 '18 at 17:00

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