Chapter 12, Problem 35P

The Drosophila chromosome 4 is extremely small; virtually no recombination occurs between genes on this chromosome. You have available three differently marked chromosome 4’s: one has a recessive allele of the gene eyeless (ey), causing very small eyes; one has a recessive allele of the cubitus interruptus (ci) gene, which causes disruptions in the veins on the wings; and the third carries recessive alleles of both genes. Drosophila adults can survive with two or three, but not with one or four, copies of chromosome 4.

a.	How could you use these three chromosomes to find Drosophila mutants with defective meioses causing an elevated rate of nondisjunction?
b.	Would your technique allow you to discriminate nondisjunction occurring during the first meiotic division from nondisjunction occurring during the second meiotic division?
c.	What progeny types would you expect if a fly recognizably formed from a gamete produced by nondisjunction were testcrossed to a fly homozygous for a chromosome 4 carrying both ey and ci?
d.	Geneticists have isolated so-called compound 4th chromosomes in which two entire chromosome 4s are attached to the same centromere. How can such chromosomes be used to identify mutations causing increased meiotic nondisjunction? Are there any advantages relative to the method you described in part (a)?

Summary Introduction

a.

To determine:

The way to use the three chromosomes to identify the mutant strains of Drosophila with defective meiotic cell divisions that resulted in elevated nondisjunction rate.

Introduction:

The three marked fourth chromosomes can be depicted as ci⁺ ey, ci ey⁺, and ci ey. Drosophila can survive with two or three copies of chromosome 4, but not with single or four copies.

Explanation of Solution

Mate the potential meiotic mutants having genotype ci⁺ ey/ci ey⁺ with homozygotes having genotype ey ci/ey ci. The normal segregants should be ci ey⁺/ey ci and ci⁺ ey/ey ci. In meiosis I, nondisjunction will be seen as the progeny having genotype ci⁺ ey/ci ey⁺/ey ci. The null-4 gametes that do not have any copy of chromosome number 4 would form zygotes with only a single copy of chromosome 4 that would not survive.

Summary Introduction

b.

To determine:

Whether the technique discussed in part (a) would allow discriminating nondisjunction occurring during meiosis I from nondisjunction during meiosis II.

Introduction:

The genotype of potential meiotic mutants will be ci⁺ ey/ci ey⁺ and homozygotes will be ey ci/ey ci.

Explanation of Solution

The mating between the potential meiotic mutants having genotype ci⁺ ey/ci ey⁺ and homozygotes having genotype ey ci/ey ci will detect nondisjunction. However, this method will not differentiate between nondisjunction occurring during meiosis I and nondisjunction during meiosis II.

Summary Introduction

c.

To determine:

The progeny types formed by the crossing between a fly that developed from a gamete produced by nondisjunction and homozygote fly.

Introduction:

The testcross can be depicted as $c{i}^{+}ey/cie{y}^{+}/eyci\times eyci/eyci$. In a trisomic fly, two of the three copies of chromosome pair normally at metaphase I and the third copy randomly assort to one pole or the other.

Explanation of Solution

In a trisomic fly there are three different ways to pair the chromosome 4. The first option involves $\frac{1}{3}$ $c{i}^{+}ey$ segregating from $cie{y}^{+}$ with $eyci$ independently assorting. This condition generates $\frac{1}{6}$ probability of $\left(\frac{1}{2}c{i}^{+}ey/eyci:\frac{1}{2}cie{y}^{+}\right)$ and $\frac{1}{6}$ probability of $\left(\frac{1}{2}c{i}^{+}ey:\frac{1}{2}cie{y}^{+}/eyci\right)$. The second option is $\frac{1}{3}$ of $c{i}^{+}ey$ segregating from $eyci$ with $cie{y}^{+}$ independently assorting. This condition generates $\frac{1}{6}$ probability of $\left(\frac{1}{2}c{i}^{+}ey/cie{y}^{+}:\frac{1}{2}eyci\right)$ and $\frac{1}{6}$ probability of $\left(\frac{1}{2}c{i}^{+}ey:\frac{1}{2}cie{y}^{+}/eyci\right)$. The third option is $\frac{1}{3}$ of $cie{y}^{+}$ segregating from $eyci$ with $c{i}^{+}ey$ independently assorting. This condition generates $\frac{1}{6}$ probability of $\left(\frac{1}{2}cie{y}^{+}/c{i}^{+}ey:\frac{1}{2}eyci\right)$ and $\frac{1}{6}$ probability of $\left(\frac{1}{2}cie{y}^{+}:\frac{1}{2}eyci/c{i}^{+}ey\right)$. The first option provides $\frac{2}{12}e{y}^{+}:\frac{2}{12}c{i}^{+}$, the second option gives $\frac{1}{12}wild\text{ type}:\frac{1}{12}ciey:\frac{1}{12}c{i}^{+}:\frac{1}{12}e{y}^{+}$, and the third also gives $\frac{1}{12}wild\text{ type}:\frac{1}{12}ciey:\frac{1}{12}c{i}^{+}:\frac{1}{12}e{y}^{+}$. The final result can be summarized as $\frac{1}{3}c{i}^{+}:\frac{1}{3}e{y}^{+}:\frac{1}{6}wild\text{ type}:\frac{1}{6}ciey$

Summary Introduction

d.

To determine:

The way by which compound 4^th chromosomes can be used to identify mutations and its advantages relative to the method described in part (a).

Introduction:

The genotype of a fly with attached fourth chromosomes that are not marked can be depicted as $c{i}^{+}e{y}^{+}c{i}^{+}e{y}^{+}$.

Explanation of Solution

The compound 4^th chromosomes can be used in crosses to assay potential mutants. For example, in cross between $c{i}^{+}ey/cie{y}^{+}\times c{i}^{+}e{y}^{+}c{i}^{+}e{y}^{+}$, all the normal progeny would have three copies of chromosome 4 and would be wild type. Nondisjunction in meiosis II would be seen as unusual $e{y}^{+}$ or $c{i}^{+}$ progeny. The compound 4^th chromosomes can be used to differentiate between nondisjunction occurring during meiosis I and meiosis II.