The possible genotypes of children born if a individual who is heterozygous for a recessive mutation in enzyme 1 and enzyme 2 marries an individual having the same genotype. Introduction: Metabolic pathways are catalyzed using enzymes. These enzymes are synthesized based on information provided by the genes.. Any type of change in these genes can hinder the synthesis of these enzymes or lead to synthesis of faulty enzymes. Heterozygosity is when same copies of alleles are present for a gene and homozygosity is when different copies of alleles are present for a gene.

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Answer 1

Question

Chapter 10, Problem 6QP

Summary Introduction

To determine: The possible genotypes of children born if a individual who is heterozygous for a recessive mutation in enzyme 1 and enzyme 2 marries an individual having the same genotype.

Introduction: Metabolic pathways are catalyzed using enzymes. These enzymes are synthesized based on information provided by the genes.. Any type of change in these genes can hinder the synthesis of these enzymes or lead to synthesis of faulty enzymes. Heterozygosity is when same copies of alleles are present for a gene and homozygosity is when different copies of alleles are present for a gene.

Expert Solution

Explanation of Solution

It is given that both the individuals are heterozygous for a recessive mutation of enzymes 1 and 2. To find the possible genotypes of their children, genotypes of parents can be assumed as:

R represents normal enzyme 1 and r shows mutated enzyme 1. D shows normal enzyme 2 and d shows mutated enzyme 2.

Both the parents are heterozygous for the recessive mutation in both the enzymes. So the gametes produced by these parents will be RD, Rd, rD, and rd.

Genotype of the children from these gametes can be determined as follows:

	RD	Rd	rD	rd
RD	RRDD	RRDd	RrDD	RrDd
Rd	RRDd	RRdd	RrDd	Rrdd
rD	RrDD	RrDd	rrDD	rrDd
rd	RrDd	Rrdd	rrDd	rrdd

GenotypesRRDD, RRDd, RrDD, RrDd, RRDd, RrDd, RrDD, RrDd, and RrDd will not show any mutation in any of the enzymes.

Genotypes rrDD, rrDd, and rrDd will show mutation in enzyme 1 only.

Genotypes RRdd, Rrdd, and Rrdd will show mutation in enzyme 2 only.

Genotype rrdd will show mutation in both the enzymes 1 and 2.

Summary Introduction

To determine: The activity ofenzyme 1 and enzyme 2 for all the genotypes that are produced in children if an individual who is heterozygous for a recessive mutation in both, enzyme 1 and enzyme 2, marries an individual having the same genotype, assuming that there is 0% activity for mutant alleles and 50% activity in normal alleles.

Introduction: Metabolic pathways are catalyzed using enzymes. These enzymes are synthesized based on information provided by the genes.. Any type of change in these genes can hinder the synthesis of these enzymes or lead to synthesis of faulty enzymes. Heterozygosity is when same copies of alleles are present for a gene and homozygosity is when different copies of alleles are present for a gene.

Expert Solution

Explanation of Solution

The possible genotypes of children born from parents who are heterozygous for a recessive mutation in both the enzymes are determined as follows:

R represents normal enzyme 1 and r shows mutated enzyme 1. D shows normal enzyme 2 and d shows mutated enzyme 2.

Both the parents are heterozygous for the recessive mutation in both the enzymes. So the gametes produced by these parents will be RD, Rd, rD, and rd.

Genotype of the children from these gametes can be determined as follows:

	RD	Rd	rD	rd
RD	RRDD	RRDd	RrDD	RrDd
Rd	RRDd	RRdd	RrDd	Rrdd
rD	RrDD	RrDd	rrDD	rrDd
rd	RrDd	Rrdd	rrDd	rrdd

Genotypes RRDD, RRDd, RrDD, RrDd, RRDd, RrDd, RrDD, RrDd, and RrDd will not show any mutation in any of the enzymes. Hence, both the enzymes will show 50% activity in these genotypes

Genotypes rrDD, rrDd, and rrDd will show mutation in enzyme 1 only. Hence, in children with these genotypes, enzyme 1 will show 0% activity and enzyme 2 will show 50 % activity.

Genotypes RRdd, Rrdd, and Rrdd will show mutation in enzyme 2 only. Hence, in children with these genotypes, enzyme 1 will show 50% activity and enzyme 2 will show 0% activity.

Genotype rrdd will show mutation in both the enzymes 1 and 2. Hence, in children with these genotypes, both, enzyme 1 and enzyme 2 will show 0% activity.

Summary Introduction

To determine: Whether compound C will be made or not in each genotype of children born, if an individual who is heterozygous for a recessive mutation in both, enzyme 1 and enzyme 2, marries an individual having the same genotype, and if compound C not produced, the compounds that will be in excess.

Introduction: Metabolic pathways are catalyzed using enzymes. These enzymes are synthesized based on information provided by the genes.. Any type of change in these genes can hinder the synthesis of these enzymes or lead to synthesis of faulty enzymes. Heterozygosity is when same copies of alleles are present for a gene and homozygosity is when different copies of alleles are present for a gene.

Expert Solution

Explanation of Solution

Genotype RRDD will not show any mutation in any of the enzymes. So, all the compounds will be present in their adequate amounts.

Genotype RRDd will not show any mutation in any of the enzymes. So, all the compounds will be present in their adequate amounts.

Genotype RrDD will not show any mutation in any of the enzymes. So, all the compounds will be present in their adequate amounts

Genotype RrDd will not show any mutation in any of the enzymes. So, all the compounds will be present in their adequate amounts

Genotype RRDd will not show any mutation in any of the enzymes. So, all the compounds will be present in their adequate amounts

Genotype RRdd will show mutation in enzyme 2 only. Hence, compound C will not be produced while there will be an excess production of compound B. Compound A will be present in normal amounts.

Genotype RrDd will not show any mutation in any of the enzymes. So, all the compounds will be present in their adequate amounts

Genotype Rrdd will show mutation in enzyme 2 only. Hence, compound C will not be produced while there will be an excess production of compound B. Compound A will be present in normal amounts.

Genotype RrDD will not show any mutation in any of the enzymes. So, all the compounds will be present in their adequate amounts

Genotype RrDd will not show any mutation in any of the enzymes. So, all the compounds will be present in their adequate amounts

Genotype rrDD will show mutation in enzyme 1 only. So there will be no production of compound B as mutated enzyme 1 will not be able to act on compound A. Absence of compound B will cause no production of compound C. Here, compound A will be present in excess.

Genotype rrDd will show mutation in enzyme 1 only. So there will be no production of compound B as mutated enzyme 1 will not be able to act on compound A. Absence of compound B will cause no production of compound C. Here, compound A will be present in excess.

Genotype RrDd will not show any mutation in any of the enzymes. So, all the compounds will be present in their adequate amounts

Genotype Rrdd will show mutation in enzyme 2 only. Hence, compound C will not be produced while there will be an excess of compound B. Compound A will be present in normal amounts.

Genotype rrDd will show mutation in enzyme 1 only. So there will be no production of compound B as mutated enzyme 1 will not be able to act on compound A. Absence of compound B will cause no production of compound C. Here, compound A will be present in excess.

Genotype rrdd will show mutation in both, enzyme 1 and enzyme 2. Hence no compound will be metabolized and compound C will not be produced.

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Answer 2

Question

Chapter 10, Problem 6QP

Summary Introduction

To determine: The possible genotypes of children born if a individual who is heterozygous for a recessive mutation in enzyme 1 and enzyme 2 marries an individual having the same genotype.

Introduction: Metabolic pathways are catalyzed using enzymes. These enzymes are synthesized based on information provided by the genes.. Any type of change in these genes can hinder the synthesis of these enzymes or lead to synthesis of faulty enzymes. Heterozygosity is when same copies of alleles are present for a gene and homozygosity is when different copies of alleles are present for a gene.

Expert Solution

Explanation of Solution

It is given that both the individuals are heterozygous for a recessive mutation of enzymes 1 and 2. To find the possible genotypes of their children, genotypes of parents can be assumed as:

R represents normal enzyme 1 and r shows mutated enzyme 1. D shows normal enzyme 2 and d shows mutated enzyme 2.

Both the parents are heterozygous for the recessive mutation in both the enzymes. So the gametes produced by these parents will be RD, Rd, rD, and rd.

Genotype of the children from these gametes can be determined as follows:

	RD	Rd	rD	rd
RD	RRDD	RRDd	RrDD	RrDd
Rd	RRDd	RRdd	RrDd	Rrdd
rD	RrDD	RrDd	rrDD	rrDd
rd	RrDd	Rrdd	rrDd	rrdd

GenotypesRRDD, RRDd, RrDD, RrDd, RRDd, RrDd, RrDD, RrDd, and RrDd will not show any mutation in any of the enzymes.

Genotypes rrDD, rrDd, and rrDd will show mutation in enzyme 1 only.

Genotypes RRdd, Rrdd, and Rrdd will show mutation in enzyme 2 only.

Genotype rrdd will show mutation in both the enzymes 1 and 2.

Summary Introduction

To determine: The activity ofenzyme 1 and enzyme 2 for all the genotypes that are produced in children if an individual who is heterozygous for a recessive mutation in both, enzyme 1 and enzyme 2, marries an individual having the same genotype, assuming that there is 0% activity for mutant alleles and 50% activity in normal alleles.

Introduction: Metabolic pathways are catalyzed using enzymes. These enzymes are synthesized based on information provided by the genes.. Any type of change in these genes can hinder the synthesis of these enzymes or lead to synthesis of faulty enzymes. Heterozygosity is when same copies of alleles are present for a gene and homozygosity is when different copies of alleles are present for a gene.

Expert Solution

Explanation of Solution

The possible genotypes of children born from parents who are heterozygous for a recessive mutation in both the enzymes are determined as follows:

R represents normal enzyme 1 and r shows mutated enzyme 1. D shows normal enzyme 2 and d shows mutated enzyme 2.

Both the parents are heterozygous for the recessive mutation in both the enzymes. So the gametes produced by these parents will be RD, Rd, rD, and rd.

Genotype of the children from these gametes can be determined as follows:

	RD	Rd	rD	rd
RD	RRDD	RRDd	RrDD	RrDd
Rd	RRDd	RRdd	RrDd	Rrdd
rD	RrDD	RrDd	rrDD	rrDd
rd	RrDd	Rrdd	rrDd	rrdd