WORKED PROBLEMS

Problem 1

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

Solution Strategy

What information is required in your answer to the problem?

The sexual phenotype of a fly with sex chromosomes XXXYY.

What information is provided to solve the problem?

• The fly has sex chromosomes XXXYY.
• All autosomal chromosomes are normal.

For help with this problem, review:

Sex determination in Drosophila melanogaster in Section 4.1.

Solution Steps

Sex in fruit flies is predicted 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, so the X : A ratio in this case is , or 1.5. Thus, this fly is a metafemale.

Problem 2

In Drosophila melanogaster, forked bristles are caused by an allele (X^f) 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?

Solution Strategy

What information is required in your answer to the problem?

Phenotypes and proportions of the F₂ flies.

What information is provided to solve the problem?

• Forked bristles are X-linked recessive.
• Brown eyes are autosomal recessive.
• Phenotypes of the parents of the cross.
• The F₁ are intercrossed to produce the F₂.

For help with this problem, review:

X-linked Color Blindness in Humans in Section 4.2 Section 3.3 in Chapter 3.

Solution Steps

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-eye allele (bb). All of the offspring of this cross will be heterozygous (b⁺b) and will have red eyes:

The F₁ are then intercrossed to produce the F₂. Whenever two individual organisms heterozygous for an autosomal recessive characteristic are crossed, of the offspring will have the dominant trait and will have the recessive trait; thus, of the F₂ flies will have red eyes and will have brown eyes:

Next, we work out the results for the X-linked characteristic. A female that is homozygous for normal bristles (X⁺ X⁺) is crossed with a male that has forked bristles (X^fY). The female F₁ from this cross are heterozygous (X⁺X^f), receiving an X chromosome with a normal-bristle allele (X⁺) from their mother and an X chromosome with a forked-bristle allele (X^f) from their father. The male F₁ are hemizygous (X⁺ Y), receiving an X chromosome with a normal-bristle allele (X⁺) from their mother and a Y chromosome from their father:

When these F₁ are intercrossed, of the F₂ will be normal-bristle females, will be normal-bristle males, and will be forked-bristle males:

To obtain the phenotypic ratio in the F₂, we now combine these two crosses by using the multiplication rule of probability and the branch diagram: