The chances of a second child to suffer from the same genetic disorder (cystic fibrosis) as the first child. Introduction: Cystic fibrosis is a genetic disorder which occurs due to the defect in the cystic fibrosis transmembrane conductance regulator (CFTR) gene. The most common defect observed in the patients with CFTR defective gene is delta-F508.

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Answer 1

Question

Chapter 2, Problem 1CS

Summary Introduction

To determine: The chances of a second child to suffer from the same genetic disorder (cystic fibrosis) as the first child.

Introduction: Cystic fibrosis is a genetic disorder which occurs due to the defect in the cystic fibrosis transmembrane conductance regulator (CFTR) gene. The most common defect observed in the patients with CFTR defective gene is delta-F508.

Expert Solution & Answer

Explanation of Solution

According to the given case, daughter M (a 6-year-old) is suffering from recurring bacterial lung infection since she was 18 months old. After a blood test, results showed that daughter M has a mutation in both copies of her cystic fibrosis gene. One copy carried ΔF508 mutation and the other copy that carried G551D. Testing of the parents showed that the mother M carries G551D defective allele and the father S carries ΔF508 defective allele.

Cystic fibrosis is an autosomal recessive disorder. It cannot be expressed in the presence of only one functional allele. According to the given case, both the mother and the father are the carriers of CFTR gene.

The chance of transfer of defective allele from mother to the child is 12 , and the chance of transfer of defective allele from father to child is 12. The total number of children is 2. By the product rule, the probability of second child to suffer from cystic fibrosis is as follows:

p(father)×p(mother)×p(Number of children)

=(12) ×(12) ×(2)

=(12) or 50%.

There is a 50 % chance that the second child will suffer from cystic fibrosis.

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Answer 2

Question

Chapter 2, Problem 1CS

Summary Introduction

To determine: The chances of a second child to suffer from the same genetic disorder (cystic fibrosis) as the first child.

Introduction: Cystic fibrosis is a genetic disorder which occurs due to the defect in the cystic fibrosis transmembrane conductance regulator (CFTR) gene. The most common defect observed in the patients with CFTR defective gene is delta-F508.

Expert Solution & Answer

Explanation of Solution

According to the given case, daughter M (a 6-year-old) is suffering from recurring bacterial lung infection since she was 18 months old. After a blood test, results showed that daughter M has a mutation in both copies of her cystic fibrosis gene. One copy carried ΔF508 mutation and the other copy that carried G551D. Testing of the parents showed that the mother M carries G551D defective allele and the father S carries ΔF508 defective allele.

Cystic fibrosis is an autosomal recessive disorder. It cannot be expressed in the presence of only one functional allele. According to the given case, both the mother and the father are the carriers of CFTR gene.

The chance of transfer of defective allele from mother to the child is 12 , and the chance of transfer of defective allele from father to child is 12. The total number of children is 2. By the product rule, the probability of second child to suffer from cystic fibrosis is as follows:

p(father)×p(mother)×p(Number of children)

=(12) ×(12) ×(2)

=(12) or 50%.

There is a 50 % chance that the second child will suffer from cystic fibrosis.

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Answer 3

Question

Chapter 2, Problem 1CS

Summary Introduction

To determine: The chances of a second child to suffer from the same genetic disorder (cystic fibrosis) as the first child.

Introduction: Cystic fibrosis is a genetic disorder which occurs due to the defect in the cystic fibrosis transmembrane conductance regulator (CFTR) gene. The most common defect observed in the patients with CFTR defective gene is delta-F508.

Expert Solution & Answer

Explanation of Solution

According to the given case, daughter M (a 6-year-old) is suffering from recurring bacterial lung infection since she was 18 months old. After a blood test, results showed that daughter M has a mutation in both copies of her cystic fibrosis gene. One copy carried ΔF508 mutation and the other copy that carried G551D. Testing of the parents showed that the mother M carries G551D defective allele and the father S carries ΔF508 defective allele.

Cystic fibrosis is an autosomal recessive disorder. It cannot be expressed in the presence of only one functional allele. According to the given case, both the mother and the father are the carriers of CFTR gene.

The chance of transfer of defective allele from mother to the child is 12 , and the chance of transfer of defective allele from father to child is 12. The total number of children is 2. By the product rule, the probability of second child to suffer from cystic fibrosis is as follows:

p(father)×p(mother)×p(Number of children)

=(12) ×(12) ×(2)

=(12) or 50%.

There is a 50 % chance that the second child will suffer from cystic fibrosis.

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Answer 4

Question

Chapter 2, Problem 1CS

Summary Introduction

To determine: The chances of a second child to suffer from the same genetic disorder (cystic fibrosis) as the first child.

Introduction: Cystic fibrosis is a genetic disorder which occurs due to the defect in the cystic fibrosis transmembrane conductance regulator (CFTR) gene. The most common defect observed in the patients with CFTR defective gene is delta-F508.

Expert Solution & Answer

Explanation of Solution

According to the given case, daughter M (a 6-year-old) is suffering from recurring bacterial lung infection since she was 18 months old. After a blood test, results showed that daughter M has a mutation in both copies of her cystic fibrosis gene. One copy carried ΔF508 mutation and the other copy that carried G551D. Testing of the parents showed that the mother M carries G551D defective allele and the father S carries ΔF508 defective allele.

Cystic fibrosis is an autosomal recessive disorder. It cannot be expressed in the presence of only one functional allele. According to the given case, both the mother and the father are the carriers of CFTR gene.

The chance of transfer of defective allele from mother to the child is 12 , and the chance of transfer of defective allele from father to child is 12. The total number of children is 2. By the product rule, the probability of second child to suffer from cystic fibrosis is as follows:

p(father)×p(mother)×p(Number of children)

=(12) ×(12) ×(2)

=(12) or 50%.

There is a 50 % chance that the second child will suffer from cystic fibrosis.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 2, Problem 1CS

Summary Introduction

To determine: The chances of a second child to suffer from the same genetic disorder (cystic fibrosis) as the first child.

Introduction: Cystic fibrosis is a genetic disorder which occurs due to the defect in the cystic fibrosis transmembrane conductance regulator (CFTR) gene. The most common defect observed in the patients with CFTR defective gene is delta-F508.

Expert Solution & Answer

Explanation of Solution

According to the given case, daughter M (a 6-year-old) is suffering from recurring bacterial lung infection since she was 18 months old. After a blood test, results showed that daughter M has a mutation in both copies of her cystic fibrosis gene. One copy carried ΔF508 mutation and the other copy that carried G551D. Testing of the parents showed that the mother M carries G551D defective allele and the father S carries ΔF508 defective allele.

Cystic fibrosis is an autosomal recessive disorder. It cannot be expressed in the presence of only one functional allele. According to the given case, both the mother and the father are the carriers of CFTR gene.

The chance of transfer of defective allele from mother to the child is 12 , and the chance of transfer of defective allele from father to child is 12. The total number of children is 2. By the product rule, the probability of second child to suffer from cystic fibrosis is as follows:

p(father)×p(mother)×p(Number of children)

=(12) ×(12) ×(2)

=(12) or 50%.

There is a 50 % chance that the second child will suffer from cystic fibrosis.

Answer 6

Chapter 2, Problem 1CS

Summary Introduction

To determine: The chances of a second child to suffer from the same genetic disorder (cystic fibrosis) as the first child.

Introduction: Cystic fibrosis is a genetic disorder which occurs due to the defect in the cystic fibrosis transmembrane conductance regulator (CFTR) gene. The most common defect observed in the patients with CFTR defective gene is delta-F508.

Expert Solution & Answer

Explanation of Solution

According to the given case, daughter M (a 6-year-old) is suffering from recurring bacterial lung infection since she was 18 months old. After a blood test, results showed that daughter M has a mutation in both copies of her cystic fibrosis gene. One copy carried ΔF508 mutation and the other copy that carried G551D. Testing of the parents showed that the mother M carries G551D defective allele and the father S carries ΔF508 defective allele.

Cystic fibrosis is an autosomal recessive disorder. It cannot be expressed in the presence of only one functional allele. According to the given case, both the mother and the father are the carriers of CFTR gene.

The chance of transfer of defective allele from mother to the child is 12 , and the chance of transfer of defective allele from father to child is 12. The total number of children is 2. By the product rule, the probability of second child to suffer from cystic fibrosis is as follows:

p(father)×p(mother)×p(Number of children)

=(12) ×(12) ×(2)

=(12) or 50%.

There is a 50 % chance that the second child will suffer from cystic fibrosis.

Answer 7

Chapter 2, Problem 1CS

Summary Introduction

To determine: The chances of a second child to suffer from the same genetic disorder (cystic fibrosis) as the first child.

Introduction: Cystic fibrosis is a genetic disorder which occurs due to the defect in the cystic fibrosis transmembrane conductance regulator (CFTR) gene. The most common defect observed in the patients with CFTR defective gene is delta-F508.

Answer 8

Expert Solution & Answer

Explanation of Solution

According to the given case, daughter M (a 6-year-old) is suffering from recurring bacterial lung infection since she was 18 months old. After a blood test, results showed that daughter M has a mutation in both copies of her cystic fibrosis gene. One copy carried ΔF508 mutation and the other copy that carried G551D. Testing of the parents showed that the mother M carries G551D defective allele and the father S carries ΔF508 defective allele.

Cystic fibrosis is an autosomal recessive disorder. It cannot be expressed in the presence of only one functional allele. According to the given case, both the mother and the father are the carriers of CFTR gene.

The chance of transfer of defective allele from mother to the child is 12 , and the chance of transfer of defective allele from father to child is 12. The total number of children is 2. By the product rule, the probability of second child to suffer from cystic fibrosis is as follows:

p(father)×p(mother)×p(Number of children)

=(12) ×(12) ×(2)