CONCEPT CHECK 14.4 1. Beth and Tom each have a sibling with cystic fibrosis,but

Question

CONCEPT CHECK 14.4 1. Beth and Tom each have a sibling with cystic fibrosis,but neither Beth nor Tom nor any of their parents have the disease. Calculate the probability that if this couple has a child, the child will have cystic fibrosis. What would be the probability if a test revealed that Tom is a carrier but Beth is not? Explain your answers. 2. Explain how the change of a MAKE CONNECTIONS single amino acid in hemoglobin leads to the aggrega- tion of hemoglobin into long fibers. (Review Figures 5.14 5.18, and 5.19.) 3. Joan was born with six toes on each foot, a dominant trait called polydactyly. Two of her five siblings and her mother, but not her father, also have extra digits. What is Joan's genotype for the number-of-digits character? Ex- plain your answer. Use D and d to symbolize the alleles for this character 4. NS In Table 14.1, note the pheno- MAKE CONNECTIONS typic ratio of the dominant to recessive trait in the F2 gen- eration for the monohybrid cross involving flower color Then determine the phenotypic ratio for the offspring of the second-generation couple in Figure 14.15b. What ac- counts for the difference in the two ratios? For suggested answers, see Appendix A

Explanation / Answer

1). Assume that the gene coding for cystic fibrosis is “c” and a gene coding for the normal allele is, “C.” It is given that both Beth and Tom have siblings with cystic fibrosis, but their parents do not have the disease (means, they are carriers). So, both Beth and Tom have 50% probability of being carriers.

If both Beth and Tom are carriers, their children will have the following genotype and phenotype:

Cc* Cc à CC (1/4, normal), Cc (1/2, carriers), cc (1/4, affected)

The probability for each progeny to receive the recessive allele is, 50%. Thus, the probability for the progeny to be activated is = The probability for both the Beth and Tom to be carriers* the probability of receiving the recessive alleles from both the patents = ½*1/2*1/2*1/2 = 1/16

If only Tom is a carrier, but not Beth, no progeny will be affected by cystic fibrosis (0% probability)

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.