Describe following on the basis of an autosomal recessive condition: The chance that the first child will be homozygous recessive for mutation, if both parents are heterozygous carriers of a mutant allele Parents who are heterozygous carriers for a recessive mutant allele have a normal child. Determine the chance that this child is a heterozygous carrier of the condition. If parents are heterozygous carriers of a mutant allele, and have the first child with homozygous recessive condition, calculate the probability that the second child of the couple will be homozygous recessive. Also calculate the probability that the second child will be a heterozygous carrier of the condition. Introduction: In an autosomal recessive condition, both copies of alleles are responsible for a trait and should be recessive (abnormal) to cause a disease. The diseased condition cannot be observed in an individual with a homozygous dominant or heterozygous dominant allele, but heterozygous dominant are the carrier for the trait. Phenylketonuria (PKU) is an autosomal recessive disorder and can be observed in individuals who have both the copies of alleles in a homozygous recessive condition.

Question

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Answer 1

Question

Chapter B, Problem 1P

Summary Introduction

To analyze:

Describe following on the basis of an autosomal recessive condition:

The chance that the first child will be homozygous recessive for mutation, if both parents are heterozygous carriers of a mutant allele

Parents who are heterozygous carriers for a recessive mutant allele have a normal child. Determine the chance that this child is a heterozygous carrier of the condition.

If parents are heterozygous carriers of a mutant allele, and have the first child with homozygous recessive condition, calculate the probability that the second child of the couple will be homozygous recessive. Also calculate the probability that the second child will be a heterozygous carrier of the condition.

Introduction:

In an autosomal recessive condition, both copies of alleles are responsible for a trait and should be recessive (abnormal) to cause a disease. The diseased condition cannot be observed in an individual with a homozygous dominant or heterozygous dominant allele, but heterozygous dominant are the carrier for the trait.

Phenylketonuria (PKU) is an autosomal recessive disorder and can be observed in individuals who have both the copies of alleles in a homozygous recessive condition.

Expert Solution & Answer

Explanation of Solution

Pedigree chart of cross between both heterozygous parents (Aa × Aa) is as follows:

♂

♀

A

a

A

AA

Aa

a

Aa

aa

The phenotypic ratio is 3:1(3 normal and 1 diseased), and genotypic ratio is 1:2:1.

Homozygous dominant (AA) is 14,

Heterozygous carrier (Aa) is 24,

Homozygous recessive (aa) is 14.

The chance that the first child of the couple will be homozygous recessive (aa) for the mutation is 14.

In this part, they have asked to estimate the probability of a normal child being a carrier for the mutation.

The chance of having normal (AA / Aa) child is 34 and out of this, the chance of the child being a heterozygous carrier (Aa) for the condition is 23. So, the chance of a normal child being a carrier for the condition is 23.

Since the birth of the second child is independent of the birth of first child so, the chance of the second child of the couple will be homozygous recessive (aa) for the mutation is 14.

The chance of the second child of the couple will be heterozygous carrier (Aa) for the mutation is 23.

Conclusion

The chance that the first child of the couple will be homozygous recessive (aa) for the mutation is 14.

The chance of the normal child being a carrier for the condition is 23.

The chance that the second child of the couple will be homozygous recessive (aa) for the mutation is 14.

The chance that the second child of the couple will be a heterozygous carrier (Aa) for the mutation is 23.

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Students have asked these similar questions

A sample of blood was taken from the above individual and prepared for haemoglobin analysis. However, when water was added the cells did not lyse and looked normal in size and shape. The technician suspected that they had may have made an error in the protocol – what is the most likely explanation? The cell membranes are more resistant than normal. An isotonic solution had been added instead of water. A solution of 0.1 M NaCl had been added instead of water. Not enough water had been added to the red blood cell pellet. The man had sickle-cell anaemia.

With reference to their absorption spectra of the oxy haemoglobin intact line) and deoxyhemoglobin (broken line) shown in Figure 2 below, how would you best explain the reason why there are differences in the major peaks of the spectra? Figure 2. SPECTRA OF OXYGENATED AND DEOXYGENATED HAEMOGLOBIN OBTAINED WITH THE RECORDING SPECTROPHOTOMETER 1.4 Abs < 0.8 06 0.4 400 420 440 460 480 500 520 540 560 580 600 nm 1. The difference in the spectra is due to a pH change in the deoxy-haemoglobin due to uptake of CO2- 2. There is more oxygen-carrying plasma in the oxy-haemoglobin sample. 3. The change in Mr due to oxygen binding causes the oxy haemoglobin to have a higher absorbance peak. 4. Oxy-haemoglobin is contaminated by carbaminohemoglobin, and therefore has a higher absorbance peak 5. Oxy-haemoglobin absorbs more light of blue wavelengths and less of red wavelengths than deoxy-haemoglobin

Answer 2

Question

Chapter B, Problem 1P

Summary Introduction

To analyze:

Describe following on the basis of an autosomal recessive condition:

The chance that the first child will be homozygous recessive for mutation, if both parents are heterozygous carriers of a mutant allele

Parents who are heterozygous carriers for a recessive mutant allele have a normal child. Determine the chance that this child is a heterozygous carrier of the condition.

If parents are heterozygous carriers of a mutant allele, and have the first child with homozygous recessive condition, calculate the probability that the second child of the couple will be homozygous recessive. Also calculate the probability that the second child will be a heterozygous carrier of the condition.

Introduction:

In an autosomal recessive condition, both copies of alleles are responsible for a trait and should be recessive (abnormal) to cause a disease. The diseased condition cannot be observed in an individual with a homozygous dominant or heterozygous dominant allele, but heterozygous dominant are the carrier for the trait.

Phenylketonuria (PKU) is an autosomal recessive disorder and can be observed in individuals who have both the copies of alleles in a homozygous recessive condition.

Expert Solution & Answer

Explanation of Solution

Pedigree chart of cross between both heterozygous parents (Aa × Aa) is as follows:

♂

♀

A

a

A

AA

Aa

a

Aa

aa

The phenotypic ratio is 3:1(3 normal and 1 diseased), and genotypic ratio is 1:2:1.

Homozygous dominant (AA) is 14,

Heterozygous carrier (Aa) is 24,

Homozygous recessive (aa) is 14.

The chance that the first child of the couple will be homozygous recessive (aa) for the mutation is 14.

In this part, they have asked to estimate the probability of a normal child being a carrier for the mutation.

The chance of having normal (AA / Aa) child is 34 and out of this, the chance of the child being a heterozygous carrier (Aa) for the condition is 23. So, the chance of a normal child being a carrier for the condition is 23.

Since the birth of the second child is independent of the birth of first child so, the chance of the second child of the couple will be homozygous recessive (aa) for the mutation is 14.

The chance of the second child of the couple will be heterozygous carrier (Aa) for the mutation is 23.

Conclusion

The chance that the first child of the couple will be homozygous recessive (aa) for the mutation is 14.

The chance of the normal child being a carrier for the condition is 23.

The chance that the second child of the couple will be homozygous recessive (aa) for the mutation is 14.

The chance that the second child of the couple will be a heterozygous carrier (Aa) for the mutation is 23.

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Answer 3

Question

Chapter B, Problem 1P

Summary Introduction

To analyze:

Describe following on the basis of an autosomal recessive condition:

The chance that the first child will be homozygous recessive for mutation, if both parents are heterozygous carriers of a mutant allele

Parents who are heterozygous carriers for a recessive mutant allele have a normal child. Determine the chance that this child is a heterozygous carrier of the condition.

If parents are heterozygous carriers of a mutant allele, and have the first child with homozygous recessive condition, calculate the probability that the second child of the couple will be homozygous recessive. Also calculate the probability that the second child will be a heterozygous carrier of the condition.

Introduction:

In an autosomal recessive condition, both copies of alleles are responsible for a trait and should be recessive (abnormal) to cause a disease. The diseased condition cannot be observed in an individual with a homozygous dominant or heterozygous dominant allele, but heterozygous dominant are the carrier for the trait.

Phenylketonuria (PKU) is an autosomal recessive disorder and can be observed in individuals who have both the copies of alleles in a homozygous recessive condition.

Expert Solution & Answer

Explanation of Solution

Pedigree chart of cross between both heterozygous parents (Aa × Aa) is as follows:

♂

♀

A

a

A

AA

Aa

a

Aa

aa

The phenotypic ratio is 3:1(3 normal and 1 diseased), and genotypic ratio is 1:2:1.

Homozygous dominant (AA) is 14,

Heterozygous carrier (Aa) is 24,

Homozygous recessive (aa) is 14.

The chance that the first child of the couple will be homozygous recessive (aa) for the mutation is 14.

In this part, they have asked to estimate the probability of a normal child being a carrier for the mutation.

The chance of having normal (AA / Aa) child is 34 and out of this, the chance of the child being a heterozygous carrier (Aa) for the condition is 23. So, the chance of a normal child being a carrier for the condition is 23.

Since the birth of the second child is independent of the birth of first child so, the chance of the second child of the couple will be homozygous recessive (aa) for the mutation is 14.

The chance of the second child of the couple will be heterozygous carrier (Aa) for the mutation is 23.

Conclusion

The chance that the first child of the couple will be homozygous recessive (aa) for the mutation is 14.

The chance of the normal child being a carrier for the condition is 23.

The chance that the second child of the couple will be homozygous recessive (aa) for the mutation is 14.

The chance that the second child of the couple will be a heterozygous carrier (Aa) for the mutation is 23.

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Answer 4

Question

Chapter B, Problem 1P

Summary Introduction

To analyze:

Describe following on the basis of an autosomal recessive condition:

The chance that the first child will be homozygous recessive for mutation, if both parents are heterozygous carriers of a mutant allele

Parents who are heterozygous carriers for a recessive mutant allele have a normal child. Determine the chance that this child is a heterozygous carrier of the condition.

If parents are heterozygous carriers of a mutant allele, and have the first child with homozygous recessive condition, calculate the probability that the second child of the couple will be homozygous recessive. Also calculate the probability that the second child will be a heterozygous carrier of the condition.

Introduction:

In an autosomal recessive condition, both copies of alleles are responsible for a trait and should be recessive (abnormal) to cause a disease. The diseased condition cannot be observed in an individual with a homozygous dominant or heterozygous dominant allele, but heterozygous dominant are the carrier for the trait.

Phenylketonuria (PKU) is an autosomal recessive disorder and can be observed in individuals who have both the copies of alleles in a homozygous recessive condition.

Expert Solution & Answer

Explanation of Solution

Pedigree chart of cross between both heterozygous parents (Aa × Aa) is as follows:

♂

♀

A

a

A

AA

Aa

a

Aa

aa

The phenotypic ratio is 3:1(3 normal and 1 diseased), and genotypic ratio is 1:2:1.

Homozygous dominant (AA) is 14,

Heterozygous carrier (Aa) is 24,

Homozygous recessive (aa) is 14.

The chance that the first child of the couple will be homozygous recessive (aa) for the mutation is 14.

In this part, they have asked to estimate the probability of a normal child being a carrier for the mutation.

The chance of having normal (AA / Aa) child is 34 and out of this, the chance of the child being a heterozygous carrier (Aa) for the condition is 23. So, the chance of a normal child being a carrier for the condition is 23.

Since the birth of the second child is independent of the birth of first child so, the chance of the second child of the couple will be homozygous recessive (aa) for the mutation is 14.

The chance of the second child of the couple will be heterozygous carrier (Aa) for the mutation is 23.

Conclusion

The chance that the first child of the couple will be homozygous recessive (aa) for the mutation is 14.

The chance of the normal child being a carrier for the condition is 23.

The chance that the second child of the couple will be homozygous recessive (aa) for the mutation is 14.

The chance that the second child of the couple will be a heterozygous carrier (Aa) for the mutation is 23.

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Answer 5

Chapter B, Problem 1P

Summary Introduction

To analyze:

Describe following on the basis of an autosomal recessive condition:

The chance that the first child will be homozygous recessive for mutation, if both parents are heterozygous carriers of a mutant allele

Parents who are heterozygous carriers for a recessive mutant allele have a normal child. Determine the chance that this child is a heterozygous carrier of the condition.

If parents are heterozygous carriers of a mutant allele, and have the first child with homozygous recessive condition, calculate the probability that the second child of the couple will be homozygous recessive. Also calculate the probability that the second child will be a heterozygous carrier of the condition.

Introduction:

In an autosomal recessive condition, both copies of alleles are responsible for a trait and should be recessive (abnormal) to cause a disease. The diseased condition cannot be observed in an individual with a homozygous dominant or heterozygous dominant allele, but heterozygous dominant are the carrier for the trait.

Phenylketonuria (PKU) is an autosomal recessive disorder and can be observed in individuals who have both the copies of alleles in a homozygous recessive condition.

Expert Solution & Answer

Explanation of Solution

Pedigree chart of cross between both heterozygous parents (Aa × Aa) is as follows:

♂

♀

A

a

A

AA

Aa

a

Aa

aa

The phenotypic ratio is 3:1(3 normal and 1 diseased), and genotypic ratio is 1:2:1.

Homozygous dominant (AA) is 14,

Heterozygous carrier (Aa) is 24,

Homozygous recessive (aa) is 14.

The chance that the first child of the couple will be homozygous recessive (aa) for the mutation is 14.

In this part, they have asked to estimate the probability of a normal child being a carrier for the mutation.

The chance of having normal (AA / Aa) child is 34 and out of this, the chance of the child being a heterozygous carrier (Aa) for the condition is 23. So, the chance of a normal child being a carrier for the condition is 23.

Since the birth of the second child is independent of the birth of first child so, the chance of the second child of the couple will be homozygous recessive (aa) for the mutation is 14.

The chance of the second child of the couple will be heterozygous carrier (Aa) for the mutation is 23.

Conclusion

The chance that the first child of the couple will be homozygous recessive (aa) for the mutation is 14.

The chance of the normal child being a carrier for the condition is 23.

The chance that the second child of the couple will be homozygous recessive (aa) for the mutation is 14.

The chance that the second child of the couple will be a heterozygous carrier (Aa) for the mutation is 23.

Answer 6

Chapter B, Problem 1P

Summary Introduction

To analyze:

Describe following on the basis of an autosomal recessive condition:

The chance that the first child will be homozygous recessive for mutation, if both parents are heterozygous carriers of a mutant allele

Parents who are heterozygous carriers for a recessive mutant allele have a normal child. Determine the chance that this child is a heterozygous carrier of the condition.

If parents are heterozygous carriers of a mutant allele, and have the first child with homozygous recessive condition, calculate the probability that the second child of the couple will be homozygous recessive. Also calculate the probability that the second child will be a heterozygous carrier of the condition.

Introduction:

In an autosomal recessive condition, both copies of alleles are responsible for a trait and should be recessive (abnormal) to cause a disease. The diseased condition cannot be observed in an individual with a homozygous dominant or heterozygous dominant allele, but heterozygous dominant are the carrier for the trait.

Phenylketonuria (PKU) is an autosomal recessive disorder and can be observed in individuals who have both the copies of alleles in a homozygous recessive condition.

Expert Solution & Answer

Explanation of Solution

Pedigree chart of cross between both heterozygous parents (Aa × Aa) is as follows:

♂

♀

A

a

A

AA

Aa

a

Aa

aa

The phenotypic ratio is 3:1(3 normal and 1 diseased), and genotypic ratio is 1:2:1.

Homozygous dominant (AA) is 14,

Heterozygous carrier (Aa) is 24,

Homozygous recessive (aa) is 14.

The chance that the first child of the couple will be homozygous recessive (aa) for the mutation is 14.

In this part, they have asked to estimate the probability of a normal child being a carrier for the mutation.

The chance of having normal (AA / Aa) child is 34 and out of this, the chance of the child being a heterozygous carrier (Aa) for the condition is 23. So, the chance of a normal child being a carrier for the condition is 23.

Since the birth of the second child is independent of the birth of first child so, the chance of the second child of the couple will be homozygous recessive (aa) for the mutation is 14.

The chance of the second child of the couple will be heterozygous carrier (Aa) for the mutation is 23.

Conclusion

The chance that the first child of the couple will be homozygous recessive (aa) for the mutation is 14.

The chance of the normal child being a carrier for the condition is 23.

The chance that the second child of the couple will be homozygous recessive (aa) for the mutation is 14.

The chance that the second child of the couple will be a heterozygous carrier (Aa) for the mutation is 23.

Answer 7

Chapter B, Problem 1P

Summary Introduction

To analyze:

Describe following on the basis of an autosomal recessive condition:

The chance that the first child will be homozygous recessive for mutation, if both parents are heterozygous carriers of a mutant allele

Parents who are heterozygous carriers for a recessive mutant allele have a normal child. Determine the chance that this child is a heterozygous carrier of the condition.

If parents are heterozygous carriers of a mutant allele, and have the first child with homozygous recessive condition, calculate the probability that the second child of the couple will be homozygous recessive. Also calculate the probability that the second child will be a heterozygous carrier of the condition.

Introduction:

In an autosomal recessive condition, both copies of alleles are responsible for a trait and should be recessive (abnormal) to cause a disease. The diseased condition cannot be observed in an individual with a homozygous dominant or heterozygous dominant allele, but heterozygous dominant are the carrier for the trait.

Phenylketonuria (PKU) is an autosomal recessive disorder and can be observed in individuals who have both the copies of alleles in a homozygous recessive condition.

Answer 8

Expert Solution & Answer

Explanation of Solution

Pedigree chart of cross between both heterozygous parents (Aa × Aa) is as follows:

♂

♀

A

a

A

AA

Aa

a

Aa

aa

The phenotypic ratio is 3:1(3 normal and 1 diseased), and genotypic ratio is 1:2:1.

Homozygous dominant (AA) is 14,

Heterozygous carrier (Aa) is 24,

Homozygous recessive (aa) is 14.

The chance that the first child of the couple will be homozygous recessive (aa) for the mutation is 14.

In this part, they have asked to estimate the probability of a normal child being a carrier for the mutation.

The chance of having normal (AA / Aa) child is 34 and out of this, the chance of the child being a heterozygous carrier (Aa) for the condition is 23. So, the chance of a normal child being a carrier for the condition is 23.

Since the birth of the second child is independent of the birth of first child so, the chance of the second child of the couple will be homozygous recessive (aa) for the mutation is 14.

The chance of the second child of the couple will be heterozygous carrier (Aa) for the mutation is 23.