Let's break down the problem step by step. ### Given Information - **Hypotrichosis** (h) is autosomal recessive. - **Normal** (H) is the dominant allele. - Two phenotypically normal parents have five children: two with hypotrichosis, three normal. --- ## **a. What are the genotypes of the parents? Explain.** ### **Key Points:** - Both parents are normal **phenotypically** (so they cannot be hh). - Two children have hypotrichosis (hh). The only way to get "hh" children is if both parents carry the "h" allele. - Therefore, both parents must be **heterozygous** (Hh). ### **Explanation:** - The cross is **Hh × Hh**. - Punnett square: | | H | h | |-----|---|---| | **H** | HH | Hh | | **h** | Hh | hh | - **Genotypic ratio:** 1 HH : 2 Hh : 1 hh - **Phenotypic ratio:** 3 normal : 1 hypotrichosis **Answer:** > The genotypes of both parents are **Hh** (heterozygous). This is because only Hh × Hh can produce hh (affected) children, and parents are phenotypically normal. --- ## **b. What is the chance that all the normal children are heterozygous?** ### **Let’s break this down:** - 5 children: 2 are hh (affected), 3 are normal (either HH or Hh). - For each normal child, what is the probability they are Hh? #### **Probability for a normal child to be Hh:** From the Punnett square above: - Probabilities among offspring: - HH: 1/4 - Hh: 2/4 = 1/2 - hh: 1/4 But **given they are normal** (i.e., not hh), their possible genotypes are **HH** and **Hh**. - Probability a normal child is **Hh**: - Probability of Hh among normals = P(Hh) / [P(HH) + P(Hh)] = (1/2) / (3/4) = 2/3 #### **Probability all 3 normal children are Hh:** - Each has a 2/3 chance of being Hh. - Probability all 3 are Hh = (2/3) × (2/3) × (2/3) = **8/27** --- ### **Summary of Answers:** **a.** The genotypes of the parents are both **Hh** (heterozygous), because only heterozygous normal parents can have children with the recessive condition. **b.** The chance that all the normal children are heterozygous (**Hh**) is **8/27**.