Genetics: A Conceptual Approach
Genetics: A Conceptual Approach
6th Edition
ISBN: 9781319050962
Author: Benjamin A. Pierce
Publisher: W. H. Freeman
Question
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Chapter 2.3, Problem 42CQ

a.

Summary Introduction

To determine:

The proportion of genome between child and father that would be exactly the same, if no crossing over took place.

Introduction:

Crossing over refers to the exchange of genes between 2 chromosomes that results n chromatids that are non-identical and which comprises of the gametic (eggs or sperms) genetic material. Crossing over occurs at the time of Meiosis stage of prophase I and hence, the chromosomes are duplicated after swapping of genes.

a.

Expert Solution
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Explanation of Solution

In the given case, let P and Q be alleles of a single gene. Now,PfPf will denote the genotype of father and QmQm will denote the genotype of mother. If there is no crossing over then the cross between mother and father will produce the following offsprings:

Parents genotypePfPf×QmQm(Father)(Mother)OffspringsPfQm,PfQmPfQm,PfQm50% alleles of father and mother50% alleles of father and mother

From the given cross it can be said that there is only 50% transfer of alleles from the father and 50% transfer of alleles from the mother to the offsprings. Hence, exactly 50% alleles will be transferred to the child from father.

b.

Summary Introduction

To determine:

The proportion of genome between the child and mother that would be exactly the same, if no crossing over took place.

Introduction:

Crossing over refers to the exchange of genes between 2 chromosomes that results n chromatids that are non-identical and which comprises of the gametic (eggs or sperms) genetic material. Crossing over occurs at the time of Meiosis stage of prophase I and hence, the chromosomes are duplicated after swapping of genes.

b.

Expert Solution
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Explanation of Solution

In the given case, let P and Q be alleles of a single gene. Now,PfPf will denote the genotype of father and QmQm will denote the genotype of mother. If there is no crossing over then the cross between mother and father will produce the following offsprings:

Parents genotypePfPf×QmQm(Father)(Mother)OffspringsPfQm,PfQmPfQm,PfQm50% alleles of father and mother50% alleles of father and mother

From the given cross it can be said that there is only 50% transfer of alleles from the father and 50% transfer of alleles from the mother to the offspring’s. Hence, exactly 50% alleles will be transferred to the child from mother.

c.

Summary Introduction

To determine:

The proportion of genome of the two full siblings (having same two biological parents) that would be exactly the same if no crossing over took place.

Introduction:

Crossing over refers to the exchange of genes between 2 chromosomes that results n chromatids that are non-identical and which comprises of the gametic (eggs or sperms) genetic material. Crossing over occurs at the time of Meiosis stage of prophase I and hence, the chromosomes are duplicated after swapping of genes.

c.

Expert Solution
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Explanation of Solution

Here, suppose P and Q are the parents. Suppose A and B are the full siblings of these (P and Q) parents. Now, the gametes of P would be P and P and the gametes of Q would be Q and Q. Hence, the cross between the genes of mother and father would produce the following progeny:

Parents genotypesPP×QQ(Father)(Mother)OffspringsPQ,PQandPQ,PQ

Hence, there are 4 possible offsprings that can be produced from both the parents. Now, as there are two siblings, the sibling A’s probability of having the genotype PQ is one-fourth, PQ is one-fourth, PQ is one-fourth and PQ is also one-fourth. Similar is the case with the sibling B.

Case I: Now, if the genome of the siblings will be 100% identical, then the genotypes in both the sibling’s would be:

A=PQ and B=PQA=PQ and B=PQA=PQ and B=PQA=PQ and B=PQ

Hence, 4 possible cases can be possible in 100% identity.

Case II: Now, if the genome of the siblings will be 50% identical, then the genotypes in both the sibling’s would be:

A=PQ and B=PQ, A=PQ and B=PQA=PQ and B=PQ, A=PQ and B=PQA=PQ and B=PQ, A=PQ and B=PQA=PQ and B=PQ, A=PQ and B=PQ

Hence, there are 8 possible cases with 50% identity.

Case III: Now, if the genome of the siblings will be 0% identical, then the genotypes in both the sibling’s would be:

A=PQ and B=PQA=PQ and B=PQA=PQ and B=PQA=PQ and B=PQ

Hence, there are 4 possible cases with 0% identity.

Now, the total number of average genome that can be same between both the siblings should be calculated. The total progenies are 16 (all progenies of the 3 cases). Thus, the percentage of genome that is same between both the siblings can be calculated as follows:

% of same genome=(% of genome that is same between A and B in case I×4+case II×8+case III×4Total progenies)=1×4+12×8+0×416=816=12

Hence, the percentage of same genome in case of the two full sibling’s will be 12 or 50%.

d.

Summary Introduction

To determine:

The proportion of genome of the half siblings (having same two biological parents) that would be exactly the same if no crossing over took place.

Introduction:

Crossing over refers to the exchange of genes between 2 chromosomes that results n chromatids that are non-identical and which comprises of the gametic (eggs or sperms) genetic material. Crossing over occurs at the time of Meiosis stage of prophase I and hence, the chromosomes are duplicated after swapping of genes.

d.

Expert Solution
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Explanation of Solution

As P and Q are the parents (old) and let P and R are the new parents. Here, P is common between both the parents and thus, they would produce offsprings that would have only one parent in common biologically. The offsprings are A and S. The gametes of the parent P are P and P and parent Q are Q and Q and of the parent P are PP and parent R are RR. As the progenies of P and Q are:

Parents genotypesPP×QQ(Father)(Mother)OffspringsPQ,PQandPQ,PQ

Similarly, the progenies of P and R are:

Parents genotypesPP×RR(Father)(Mother)OffspringsPR,PRandPR,PR

Now, the offsprings A and S will be inheriting half of their genotypes from the parent P and half from the parent R.

Case I: Now, if the genome of the siblings will be 50% identical, then the genotypes in both the sibling’s would be:

A=PQ and S=PRA=PQ and S=PRA=PQ and S=PRA=PQ and S=PRA=PQ and S=PRA=PQ and S=PRA=PQ and S=PRA=PQ and S=PR

Hence, 8 possible cases can be possible in 50% identity.

Case II: Now, if the genome of the siblings will be 0% identical, then the genotypes in both the sibling’s would be:

A=PQ and S=PRA=PQ and S=PRA=PQ and S=PRA=PQ and S=PRA=PQ and S=PRA=PQ and S=PRA=PQ and S=PRA=PQ and S=PR

Hence, there are 8 possible cases with 0% identity.

Now, the total number of average genome that can be same between half of the siblings should be calculated. The total progenies are 16 (all progenies of both the cases). Thus, the percentage of genome that is same between half of the siblings can be calculated as follows:

% of same genome=(% of genome that is same between A and S in case I×8+case II×8Total progenies)=12×8+0×816=416=14

Hence, the percentage of same genome in case of half sibling’s will be 14 or 25%.

e.

Summary Introduction

To determine:

The proportion of genome between uncle and niece that would be exactly the same, if no crossing over took place.

Introduction:

Crossing over refers to the exchange of genes between 2 chromosomes that results n chromatids that are non-identical and which comprises of the gametic (eggs or sperms) genetic material. Crossing over occurs at the time of Meiosis stage of prophase I and hence, the chromosomes are duplicated after swapping of genes.

e.

Expert Solution
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Explanation of Solution

Pictorial representation: Fig.1: Relation between the uncle and offspring (niece).

Genetics: A Conceptual Approach, Chapter 2.3, Problem 42CQ , additional homework tip  1

Fig.1: Relation between the uncle and offspring (niece).

From the Fig.1 “Relation between the uncle and offspring (niece)” above, it can be concluded that the relation between individual’s A and E can be found with the help of the following data:

1. The genes that will be transferred to A and B will be of P and Q. Similarly, the genes from S and T will be transferred to C.

2. The relation between A and E can be determined after finding out the genes present in them.

The genes present in A and B will be of P and Q and so, the genotypes present in A and B will be PQ, PQ, PQ and PQ. Similarly, the genotypes present in C will be ST, ST, ST and ST. The gametes of the two individuals P, Q and S, T are P, P, Q, Q and S, S, T, T. The genotypes of the offspring E are as follows:

PS, PS, PT, PTPS, PS, PT, PTQS, QS, QT, QTQS, QS, QT, QT

Case I: Now, if the genome of the uncle (A) and niece (E) would be 50% identical, then the genotypes of A and E would be:

A=PQ and E=PS, PS, PT, PT or QS, QS, QT, QTA=PQ and E=PS, PS, PT, PT or QS, QS, QT, QTA=PQ and E=PS, PS, PT, PT or QS, QS, QT, QTA=PQ and E=PS, PS, PT, PT or QS, QS, QT, QT

Case II: Now, if the genome of the uncle (A) and niece (E) would be 0% identical, then the genotypes of A and E would be:

A=PQ and E=PS, PS, PT, PT or QS, QS, QT, QTA=PQ and E=PS, PS, PT, PT or QS, QS, QT, QTA=PQ and E= PS, PS, PT, PT or QS, QS, QT, QTA=PQ and E=PS, PS, PT, PT or QS, QS, QT, QT

Now, the total number of average genome that can be same between the uncle and niece should be calculated. The total progenies are 64 (all progenies of both the cases). Thus, the percentage of genome that is same between the uncle and niece can be calculated as follows:

% of same genome=(% of genome that is same between A and E in case I×32+case II×32Total progenies)=12×32+0×3264=1664=14

Hence, the percentage of same genome in case of uncle and niece will be 14 or 25%.

f.

Summary Introduction

To determine:

The proportion of genome between grandparent and grandchild that would be exactly the same, if no crossing over took place.

Introduction:

Crossing over refers to the exchange of genes between 2 chromosomes that results n chromatids that are non-identical and which comprises of the gametic (eggs or sperms) genetic material. Crossing over occurs at the time of Meiosis stage of prophase I and hence, the chromosomes are duplicated after swapping of genes.

f.

Expert Solution
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Explanation of Solution

Pictorial representation: Fig.2: Relation between the grandparent and grandchild.

Genetics: A Conceptual Approach, Chapter 2.3, Problem 42CQ , additional homework tip  2

Fig.2: Relation between the grandparent and grandchild.

From the Fig.2 “Relation between the grandparent and grandchild” above, it can be concluded that the relation between individual’s P and Q and E can be found on the basis of the following:

1. The genes that will be transferred to A and B will be of P and Q. Similarly, the genes from S and T will be transferred to C. The genes in E will be transferred from B and C. Hence, the relation between P and Q and E is of the grandparent and grandchild.

2. P and Q had inherited the genotypes as P and P and Q and Q from their parents. The father B will be having genotypes as PQ, PQ, PQ and PQ. The genotypes of the offspring E are as follows:

PS, PS, PT, PTPS, PS, PT, PTQS, QS, QT, QTQS, QS, QT, QT

Case I: Now, if the genome of the grandparents (P and Q) and grandchild (E) would be 50% identical, then the genotypes of P and Q and E would be:

P=PP and E=PS, PS, PT, PT or PS, PS, PT, PTQ=QQ and E=QS, QS, QT, QT or QS, QS, QT, QT

Case II: Now, if the genome of the grandparent (P and Q) and grandchild (E) would be 0% identical, then the genotypes of P and Q and E would be:

P=PP and E=QS, QS, QT, QT or QS, QS, QT, QTQ=QQ and E=PS, PS, PT, PT or PS, PS, PT, PT

Now, the total number of average genome that can be same between the grandparent and grandchild should be calculated. The total progenies are 32 (all progenies of both the cases). Thus, the percentage of genome that is same between the grandparent and grandchild can be calculated as follows:

% of same genome=(% of genome that is same between P and Q and E in case I×16+case II×16Total progenies)=12×16+0×1632=832=14

Hence, the percentage of same genome in case of grandparent and grandchild will be 14 or 25%.

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