Problem 1

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 from this cross be?

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?

For help with this problem, review:

X-linked Color Blindness in Humans in Section 4.2.

Section 3.3 in Chapter 3.

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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 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, ¾ of the offspring will have the dominant trait and ¼ will have the recessive trait; thus, ¾ of the F₂ flies will have red eyes and ¼ 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 from their mother (X⁺) 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 from their mother (X⁺) and a Y chromosome from their father:

When these F₁ are intercrossed, ½ of the F₂ will be normal-bristle females, ¼ will be normal-bristle males, and ¼ will be forked-bristle males:

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

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Problem 2

The type of plumage found in mallard ducks is determined by three alleles at a single locus: M^R, which encodes restricted plumage; M, which encodes mallard plumage; and m^d, which encodes dusky plumage. The restricted phenotype is dominant over mallard and dusky; mallard is dominant over dusky (M^R > M > m^d). Give the expected phenotypes and proportions of offspring produced by the following crosses:

Solution Strategy

What information is required in your answer to the problem?

The phenotypes and proportions of progeny expected for each cross.

What information is provided to solve the problem?

For help with this problem, review:

Multiple Alleles in Section 4.3.

Solution Steps

We can determine the phenotypes and proportions of offspring by (1) determining the types of gametes produced by each parent and (2) combining the gametes of the two parents with the use of a Punnett square.

a.

b.

c.

d.

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Problem 3

In some sheep, horns are produced by an autosomal allele that is dominant in males and recessive in females. A horned female is crossed with a hornless male. One of the resulting F₁ females is crossed with a hornless male. What proportion of the male and female progeny from this cross will have horns?

Solution Strategy

What information is required in your answer to the problem?

Proportions of male and female progeny that have horns.

What information is provided to solve the problem?

For help with this problem, review:

Sex-Influenced and Sex-Limited Characteristics in Section 4.5.

Solution Steps

The presence of horns in these sheep is an example of a sex-influenced characteristic. Because the phenotypes associated with the genotypes differ between the two sexes, let’s begin this problem by writing out the genotypes and phenotypes for each sex. We will let H represent the allele that encodes horns and H⁺ represent the allele that encodes hornless. In males, the allele for horns is dominant over the allele for hornless, which means that males homozygous (HH) and heterozygous (H⁺H) for this gene are horned. Only males homozygous for the recessive hornless allele (H⁺H⁺) are hornless. In females, the allele for horns is recessive, which means that only females homozygous for this allele (HH) are horned; females heterozygous (H⁺H) and homozygous (H⁺H⁺) for the hornless allele are hornless. The following table summarizes the genotypes and their associated phenotypes:

In the problem, a horned female is crossed with a hornless male. From the preceding table, we see that a horned female must be homozygous for the allele for horns (HH), and that a hornless male must be homozygous for the allele for hornless (H⁺H⁺); so all the F₁ will be heterozygous; the F₁ males will be horned and the F₁ females will be hornless, as shown in the following diagram:

A heterozygous hornless F₁ female (H⁺H) is then crossed with a hornless male (H⁺H⁺):

Therefore, ½ of the male progeny will be horned, but none of the female progeny will be horned.