Worked Problems

1. A fruit fly has XXXYY sex chromosomes; all the autosomal chromosomes are normal. What sexual phenotype will this fly have?

Sex in fruit flies is determined by the X:A ratio — the ratio of the number of X chromosomes to the number of haploid autosomal sets. An X:A ratio of 1.0 produces a female fly; an X:A ratio of 0.5 produces a male. If the X:A ratio is greater than 1.0, the fly is a metafemale; if it is less than 0.5, the fly is a metamale; if the X:A ratio is between 1.0 and 0.5, the fly is an intersex.

This fly has three X chromosomes and normal autosomes. Normal diploid flies have two autosomal sets of chromosomes; so the X:A ratio in this case is 3/2 or 1.5. Thus, this fly is a metafemale.

2. Color blindness in humans is most commonly due to an X-linked recessive allele. Betty has normal vision, but her mother is color blind. Bill is color blind. If Bill and Betty marry and have a child together, what is the probability that the child will be color blind?

Because color blindness is an X-linked recessive characteristic, Betty's color-blind mother must be homozygous for the colorblind allele (XCXC). Females inherit one X chromosome from each of their parents; so Betty must have inherited a color-blind allele from her mother. Because Betty has normal color vision, she must have inherited an allele for normal vision (X+) from her father; thus Betty is heterozygous (X+Xc). Bill is color blind. Because males are hemizygous for X-linked alleles, he must be (XcY). A mating between Betty and Bill is represented as:

Betty X Bill X+Xc XcY

Thus, V4 of the children are expected to be female with normal color vision, % female with color blindness, V4 male with normal color vision, and % male with color blindness.













3. Chickens, like all birds, have ZZ-ZW sex determination. The bar-feathered phenotype in chickens results from a Z-linked allele that is dominant over the allele for nonbar feathers. A barred female is crossed with a nonbarred male. The F1 from this cross are intercrossed to produce the F2. What will the phenotypes and their proportions be in the F1 and F2 progeny?

With the ZZ-ZW system of sex determination, females are the heterogametic sex, possessing a Z chromosome and a W chromosome; males are the homogametic sex, with two Z chromosomes. In this problem, the barred female is hemizygous for the bar phenotype (Z^W). Because bar is dominant over nonbar, the nonbarred male must be homozygous for nonbar (ZbZb). Crossing these two chickens, we obtain:

barred female X nonbarred male ZB W ZbZb v v

4. In Drosophila melanogaster, forked bristles are caused by an allele (XO that is X linked and recessive to an allele for normal bristles (X+). Brown eyes are caused by an allele (b) that is autosomal and recessive to an allele for red eyes (b+). A female fly that is homozygous for normal bristles and red eyes mates with a male fly that has forked bristles and brown eyes. The F1 are intercrossed to produce the F2. What will the phenotypes and proportions of the F2 flies be from this cross?

This problem is best worked by breaking the cross down into two separate crosses, one for the X-linked genes that determine the type of bristles and one for the autosomal genes that determine eye color.

Let's begin with the autosomal characteristics. A female fly that is homozygous for red eyes (b+b+) is crossed with a male with brown eyes. Because brown eyes are recessive, the male fly must be homozygous for the brown-eyed allele (bb). All of the offspring of this cross will be heterozygous (b+b) and will have brown eyes:

b+b+ X bb P red eyes brown eyes

Thus, all the males in the F1 will be barred (ZBZb), and all the females will be nonbarred (ZbW).

The F1 are now crossed to produce the F2:

nonbarred female X barred male ZbW ZBZb

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