Genetics: From Genes to Genomes, 5th edition
Genetics: From Genes to Genomes, 5th edition
5th Edition
ISBN: 9780073525310
Author: Leland H. Hartwell, Michael L. Goldberg, Janice A. Fischer, Leroy Hood, Charles F. Aquadro
Publisher: McGraw-Hill Education
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Chapter 5, Problem 8P

CCDD and ccdd individuals were crossed to each other, and the F1 generation was backcrossed to the ccdd parent. 997 CcDd, 999 ccdd, 1 Ccdd, and 3 ccDd offspring resulted.

a. How far apart are the c and d loci?
b. What progeny and in what frequencies would you expect to result from testcrossing the F1 generation from a CCdd × cc DD cross to ccdd?
c. In a typical meiosis, how many crossovers occur between genes C and D?
d. Assume that the C and D loci are on the same chromosome, but the offspring from the testcross described above were 498 CcDd, 502 ccdd, 504 Ccdd, and 496 ccDd. How would your answer to part (c) change?
Expert Solution
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Summary Introduction

a.

To determine:

The distance between the c and d loci in the given question.

Introduction:

Recombination frequency (RF) is the rate of occurrence of recombination between a pair of linked genes. It helps in determining the distance between two genes and in generating linkage map.

Explanation of Solution

It is given that CCDD and ccdd individuals were crossed to each other, and the F1 generation was backcrossed to the ccdd parent. The offsprings of the F2 generation are as follows:

CcDd- 997

ccdd-999

Cc dd-1

ccDd-3

From the given data, it can be concluded that individuals with genotypes CcDd and ccdd are parental types. Individuals with genotypes Ccdd, and ccDd are recombinant types. The formula to calculate the recombination frequency or gene distance is as follows:

Recombination frequency(RF)=(Number of recombinant progenyTotal number of progeny)×100

Here, the total number of recombinant progeny is 4, and the total number of progeny is 2000. Put these values in the above formula:

RF=(42000)=0.2%or 0.2 m.u.

Thus, the map distance between loci c and d is about 0.2 m.u. (map units).

Expert Solution
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Summary Introduction

b.

To determine:

The progeny and its frequency that comes from testcrossing the F1 generation from a CCdd × cc DD cross to ccdd.

Introduction:

The testcross involves the crossing of an individual with a phenotypically recessive individual. This cross is performed to determine the type of zygosity of the former individual by observing the proportions of the phenotypes of the offspring.

Explanation of Solution

The following table explains the cross from the given data:

Genotype Number Type of progeny
CcDd 997 Parental
ccdd 999 Parental
Ccdd 1 Recombinant
ccDd 3 Recombinant
Total 2000

The frequency of parental progeny is as follows:

(997+9992000)×100=99.8%

The frequency of the recombinant progeny is as follows:

(1+32000)×100=0.2%

Thus, the expected genotypes are CcDd, Ccdd, ccDd, and ccdd. The parental proportions are about 99.8%, and recombinant proportion is 0.2%.

Expert Solution
Check Mark
Summary Introduction

c.

To determine:

The number of crossovers that occurs between genes C and D in typical meiosis.

Introduction:

The exchange of genes between two chromosomes is termed as crossing over. This results in non-identical chromatids that have the genetic material of gametes.

Explanation of Solution

In typical meiosis, there is a very less possibility of crossover occurring between two genes C and D, which are very close to each other. So, there is always a chance for the two genes to inherit together during meiosis. The crossing over results, as shown in part (b), is 0.2% recombination frequency. Thus, in further generations, the offspring will also have the traits as that of the parental type.

Expert Solution
Check Mark
Summary Introduction

d.

To determine:

The number of crossing over that occurs between genes C and D when C and D loci are on the same chromosome, but the offspring from the testcross described in part (b) were 498 CcDd, 502 ccdd, 504 Ccdd, and 496 ccDd.

Introduction:

The presence of two different genes on the same chromosome is termed as linkage. The genes which are close to each other and present on the same chromosome are said to be linked. The linked genes tend to inherit together during meiosis.

Explanation of Solution

It is given that the C and D loci are on the same chromosome. The offspring from testcrossing the F1 generation are as follows:

CcDd-498

ccdd-502

Ccdd-504

ccDd-496

The following table explains the cross from the given data:

Genotype Number Type of progeny
CcDd 498 Parental
ccdd 502 Parental
Ccdd 504 Recombinant
ccDd 496 Recombinant
Total 2000

The frequency of parental progeny is as follows:

(498+5022000)×100=50%

The frequency of the recombinant progeny is as follows:

(504+4962000)×100=50%

The frequency of parental progeny and the recombinant progeny is equal; that is, 50%. Thus, the two genes assort independently or located far apart on the same chromosome. There is a possibility that these genes undergo crossing over during meiosis and produce recombinant offspring.

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Chapter 5 Solutions

Genetics: From Genes to Genomes, 5th edition

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