Problem

Solutions For An Introduction to Genetic Analysis Chapter 17 Problem 73P

Step-by-Step Solution

Solution 1

In the fruit fly Drosophila, a cross was made between two mutant flies. One fly was homozygous for the recessive mutation of bent wings [b] while the other fly was homozygous for the recessive mutation of eyeless [e]. These two alleles are of different genes which are closely linked. They are present on the autosomal chromosome 4.

The progeny of this cross had the wild type phenotype. One female was then crossed with a male which is homozygous recessive for both the alleles. In the progeny of the second cross, one female had the wild type phenotype.

1. Homozygous:

It is a chromosomal condition in which the organism has two identical alleles.

Mutation:

Any change in the sequence of the DNA which results in alteration from wild type.

Allele:

One variant form of a gene is called an allele.

Closely linked:

This indicates that two genes are close to each other on the same chromosome and are transmitted together during the process of meiosis.

Recessive:

This is a characteristic of an allele which is not expressed in the presence of the dominant allele.

Wild type:

The common characteristics of a particular species found naturally are collectively known as wild type.

Crossing over:

The exchange of parts of chromosomal material between homologous chromosomes during the process of meiosis is known as crossing over.

Nondisjunction:

This is a process in which homologous chromosomes fail to separate and go into one daughter cell. It can also occur when sister chromatids fail to separate and move into one daughter cell. This process leads to extra chromosomes in gametes or no chromosomes in some gametes.

Testcross:

When an experimental organism is crossed with a homozygous recessive organism, it is known as a testcross. It helps us to determine the genotype of the experimental organism.

Phenotype:

A phenotype is the outward characteristics of an organism.

Genotype:

A genotype is the collective term for the genes in the organism.

2. No, the problem is not connected with sex linkage. Since the genes are on an autosome, sex chromosomes do not come into the picture.

3. The normal species of Drosophila melanogaster has 8 chromosomes.

4. The pedigree summarizing the details of the three crosses presented in the question can be shown in the following way:

Parents: be+/be+ x b+e/b+e cross 1

F 1 : be+/b+e x be/be cross 2

Progeny:

• Parental - be+/be

• Parental - b+e/be

• Unexpected recombinant - b+e+/be

• Unexpected recombinant - be/be

Rare wild type x be/be cross 3

Progeny:

• 1/6 wild type

• 1/6 bent, eyeless

• 1/3 bent

• 1/3 eyeless

5. The gametes produced by both parents in cross 1 are:

be+ and

b+e

6. The chromosome 4 constitution of the progeny of cross 1 would be:

be+/ b+e

7. It is not surprising that the progeny of cross-1 have wild type phenotype. This indicates that both the genes are recessive and are present on different genes. The two genes complement each other.

8. The chromosome 4 constitution of the male tester used in cross 2 would be:

be/ be

The gametes he can produce would be be.

9. With respect to chromosome 4, the gametes of the female palrent in cross 2 in the absence of nondisjunction would be:

The two common gametes would be:

be+ and

b+e

The two rare gametes would be:

b+e+ and

be

10. The first and second divisions in meiosis is shown below:

Normal

Picture 1

First division nondisjunction: all gametes are aneuploid

Picture 5

Second division nondisjunction: half the gametes are aneuploid

Picture 4

11. Yes, in the nondisjunction of first division of meiosis, all gametes are aneuploid. In the second division nondisjunction, half the gametes are aneuploid and half are normal.

12. Aneuploid gametes might give rise to viable progeny. Since the chromosome 4 is quite small, nondisjunction may not be lethal. The progeny would be monosomic and trisomic.

13. The gametes from part 9 and 10 are listed below. When crossed with a homozygous recessive male, the phenotype of the offspring is also given:

Female gamete

Male gamete

phenotype

b+e

be

eyeless

be+

be

bent

b+e+

be

Wild type

be

be

Bent, eyeless

b+e/ be+

be

Wild type

-

be

Bent, eyeless

be+/ be+

be

bent

b+e/ b+e

be

eyeless

14. The phenotypic ratio in the progeny of cross 3 is shown as thirds and sixths. This ratio points to the process of meiosis in a trisomic.

15. Yes, there is significance to the fact that the crosses concern genes on a very small chromosome. Research with artificial chromosomes has shown that very small chromosomes segregate improperly at higher rates than longer chromosomes. Scientists think that intertwining of homologous chromosomes is needed till anaphase begins. Short chromosomes cannot intertwine like longer chromosomes. Hence, they have a higher rate of nondisjunction. In this problem, since small chromosomes would have very less genetic material, it may not make much difference to the individual.

16. The progeny expected from cross 3 under the two hypotheses can be drawn in the following way:

Rare wild type x eb/eb cross 3

If the rare wild type has resulted from recombination, then the cross becomes:

b + e + /be x be/be

Progeny:

Parental [wild type]- b+e+/be

Parental[bent, eyeless] - be/be

Rare recombinant [eyeless]- b+e/be

Rare recombinant [bent]- be+/be

If the rare wild type is from nondisjunction, then the cross becomes:

be + /b + e/be x be/be/be

Progeny:

Wild type- be+/b+e/be

Bent - be+/be/be

Eyeless- b+e/be/be

Bent- be+/be

Eyeless - b+e/be

Bent, eyeless - be/be

a) Cross 1:

Parents: be+/be+ x b+e/b+e

F 1 : be+/b+e

Cross 2:

Parents: be+/b+e; X/X x be/be; XY

F 1 : be+/be; XX

b + e/be; XY

The ratio expected is 1:1 along with one rare observed b+e+ X/X.

The common progeny expected from cross 2 are b+e/be and be+/be.

b) The rare female could have occurred by the process of crossing over which would give gametes of b+e+ genotype. The rare female could have also occurred by the process of nondisjunction that gave a gamete of the genotype be+/b+e. Such a gamete might give rise to progeny which was viable.

c) If the female had been wild type b+e+/be, as a result of crossing over, the progeny would be:

Parental: b+e+/be – wild type

be/be - bent, eyeless

Recombinant: be+/be – bent

b+e/be – eyeless

These results do not match the results given in the problem. So, the rare female was not the result of recombination/crossing over.

If the female had been the product of nondisjunction, be+ b+e/be, the progeny when crossed with be/be would be:

• 1/6 - b+e/be – eyeless

• 1/6 - be+/be/be – bent

• 1/6- b+e/be/be – eyeless

• 1/6 - be+/be – bent

• 1/6 - be/be - bent, eyeless

• 1/6 - be+ b+e/be – wild type

The ratio is 2 bent: 2 eyeless: 1 bent eyeless: 1 wild type

These results match the observed results indicating that the female was a result of nondisjunction.

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