**Question 5:**What are the genotypes of individuals in the following autosomal recessive trait, attached earlobe?**Pedigree Diagram Explanation:**- The pedigree chart consists of three generations labeled A, B, and C. - "A1" and "A2" at the top. A1 has genotype "Ww" and A2 has genotype "ww".- Generation B includes individuals B1 to B5. - B1: Shaded square representing a male with an attached earlobe (homozygous recessive, "ww").- B2: Shaded circle representing a female with an attached earlobe (homozygous recessive, "ww").- B3: Unshaded square representing a male without an attached earlobe.- B4 and B5 are unshaded squares without attached earlobes, indicating heterozygous or homozygous dominant.- Generation C includes individuals C1 to C5.- C1 and C4: Shaded circles with attached earlobes (homozygous recessive, "ww").- C2, C3, and C5: Unshaded circles and squares without attached earlobes.**Questions:**A. What is the genotype of individual C3?B. Individuals B4 & B5 are going to have another child. What is the probability that their next child will have an attached earlobe?C. How many children does A1 have?

Name: **Question 5:**What are the genotypes of individuals in the following
Uploaded: 2024-03-20T06:43:04.000Z
Channel: Bartleby
Description: **Question 5:**What are the genotypes of individuals in the following autosomal recessive trait, attached earlobe?**Pedigree Diagram Explanation:**- The pedigree chart consists of three generations labeled A, B, and C. - "A1" and "A2" at the top. A1

Since the trait is an autosomal recessive trait the affected individuals will have the genotype (ww) i.e., two copies of the recessive allele are required to express the trait. Individuals having atleast one copy of the dominant gene (W) will not show the trait. Hence, people without the trait will have genotype either WW or Ww. A. Individual C3 is unaffected. Therefore, the possible genotype of C3 is either WW or Ww. Since, one of the parents of C3 have the trait i.e., B4 has genotype ‘ww’. Hence, the only possible genotype of C3 is Ww. (Because he will inherit a ‘w’ from B4). B. The genotype of B4 - ww. Since, B5 do not have the trait, it can have genotype WW or Ww. From the pedigree, B4 and B5 have an affected son (C4). It means that C4 receive two recessive (w) alleles from each parent. Hence, the genotype of B5 is Ww. Cross between B4 and B5: Parents: Ww x ww Offspring: W w w Ww (no trait) ww (attached earlobe) w Ww (no trait) ww (attached earlobe) Therefore, the probability that the next child will have attached earlobe is 2/4 = 1/2. C. A1 has five male children and one female children. So, in total A1 has six (6) children. Answers: 1. Genotype of individual C3 is Ww. 2. The required probability is 1/2. 3. A1 has 6 children.