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The Genetics of Sex Practice Problems

By — McGraw-Hill Professional
Updated on Aug 21, 2011

Review the following concepts if needed:

The Genetics of Sex Practice Problems

Practice 1

An autosomal recessive gene tra, when homozygous, transforms a Drosophila female (X=X) into a phenotypic male. All such "transformed" males are sterile. The gene is without effect in males (X=Y). A cross is made between a female heterozygous at the tra locus and a male homozygous recessive at the same locus (tra/tra). What is the expected sex ratio in the F1 and F2?

Solution 1

We will use a slash mark (/) to separate alleles or homologous chromosomes, and a comma (,) to separate one gene locus from another.

Solved Problems

This is the same as the original parental mating and hence we expect 3/4 males : 1/4 females. Correcting these proportions for the frequency of this mating, we have (1/2)(3/4) = 3/8 males : (1/2)(1/4) = 1/8 females. Summary of the F2 from both matings: males = (5/16) + 3/8 = 11/16; females = 3/16 + 1/8 = 5/16.

Practice 2

There is a dominant sex-linked gene B that places white bars on an adult black chicken as in the Barred Plymouth Rock breed. Newly hatched chicks, which will become barred later in life, exhibit a white spot on the top of the head. (a) Diagram the cross through the F2 between a homozygous barred male and a nonbarred female. (b) Diagram the reciprocal cross through the F2 between a homozygous nonbarred male and a barred female. (c) Will both of the above crosses be useful in sexing F1 chicks at hatching?

Solution 2

Solved Problems

Practice 3

A recessive sex-linked gene h prolongs the blood-clotting time, causing hemophilia. From the information in the pedigree, answer the following questions.     (a)  If II2 marries a normal man, what is the chance of her first child being a hemophilic boy?     (b)  Suppose her first child is actually hemophilic. What is the probability that her second child will be a hemophilic boy?     (c)  If II3 marries a hemophilic man, what is the probability that her first child will be normal?     (d)  If the mother of I1 was phenotypically normal, what phenotype was her father?     (e)  If the mother of I1 was hemophilic, what phenotype was her father?

Solved Problems

Solution 3

(a)   Since II4 is a hemophilic male (hY), the hemophilic allele is on an X chromosome that he received from his mother (I1). But I1 is phenotypically normal and therefore must be heterozygous or a carrier of hemophilia of genotype Hh. Both I2 and II1 are normal males (HY). Therefore, the chance of II2 being a carrier female (Hh) is 1/2. When a carrier woman marries a normal man (HY), 25% of their children are expected to be hemophilic boys (hY). The combined probability that she is a carrier and will produce a hemophilic boy is (1/2)(1/4) = 1/8.

(b)   Because her first child was hemophilic, she must be a carrier. One-quarter of the children from carrier mothers (Hh) × normal fathers (HY) are expected to be hemophilic boys (hY).

(c) II3 (like II2) has a 50% chance of being a carrier of hemophilia (Hh). If she marries a hemophilic man (hY), 1/2 of their children (both boys and girls) are expected to be hemophilic. The combined chance of II3 being a carrier and producing a hemophilic child is (1/2)(1/2) = 1/4. Therefore, the probability that her first child is normal is the complementary fraction, 3/4.

(d) It is impossible to deduce the phenotype of the father of I1 from the information given because the father could be either normal or hemophilic and still produce a daughter (I1) who is heterozygous normal (Hh), depending upon the genotype of the normal mother:

    Solved Problems

(e) In order for a hemophilic mother (hh) to produce a normal daughter (Hh), her father must possess the dominant normal allele (HY) and therefore would have normal blood-clotting time.

Practice 4

The recessive incompletely sex-linked gene called "bobbed" (bb) causes the bristles of Drosophila to be shorter and of smaller diameter than the normal bristles produced by its dominant wild-type allele (bb+). Determine the phenotypic expectations of the F1 and F2 when bobbed females are crossed with each of the two possible heterozygous males.

Solution 4

Recall that an incompletely sex-linked gene has an allele on the homologous portion of the Y chromosome in a male. The wild-type allele in heterozygous males may be on either the X or the Y, thus making possible two types of crosses.

Solved Problems

Practice 5

Let us consider two sex-influenced traits simultaneously, pattern baldness and short index finger, both of which are dominant in men and recessive in women. A heterozygous bald man with long index finger marries a heterozygous long-fingered, bald woman. Determine the phenotypic expectations for their children.

Solution 5

Let us first select appropriate symbols and define the phenotypic expression of the three genotypes in each sex.

Solved Problems

Practice 6

Cock-feathering in chickens is a trait limited to expression only in males and is determined by the autosomal recessive genotype hh. The dominant allele H produces hen-feathered males. All females are hen-feathered regardless of genotype. A cock-feathered male is mated to three females, each of which produces a dozen chicks. Among the 36 progeny are 15 hen-feathered males, 18 hen-feathered females, and 3 cock-feathered males. What are the most probable genotypes of the three parental females?

Solution 6

In order for both hen-feathered (H-) and cock-feathered (hh) males to be produced, at least one of the females had to be heterozygous (Hh) or recessive (hh). The following female genotype possibilities must be explored:

(a)  2 HH, 1 Hh     (b)  1 HH, 2 Hh    (c)  1 HH, 1 Hh, 1 hh     (d)  3 Hh    (e)  2 Hh, 1 hh     (f)  1 Hh, 2 hh     (g)  2HH, 1 hh     (h)  2 hh, 1 HH

Obviously, the more hh or Hh hen genotypes, proportionately the more cock-feathered males that are expected in the progeny. The ratio of 15 hen-feathered males : 3 cock-feathered males is much greater than the 1 : 1 ratio expected when all three females are heterozygous (Hh).

Solved Problems

Possibility (d) is therefore excluded. Possibilities (e) and (f), which both contain one or more hh genotypes in addition to one or more Hh genotypes, must also be eliminated because these matings would produce even more cock-feathered males than possibility (d). In possibility (g), the 2 HH : 1 hh hens are expected to produce an equivalent ratio of 2 hen-feathered (Hh) : 1 cock-feathered (hh) males. This 2 : 1 ratio should be expressed in the 18 male offspring as 12 hen-feathered : 6 cock-feathered. These numbers compare fairly well with the observed 15 : 3, but possibility (h) would be even less favorable because even more cock-feathered males would be produced. Let us see if one of the remaining three possibilities will give us expected values closer to our observations.

Solved Problems

Set the observation of 3 cock-feathered males equal to the 1/6, then 5 × 3 = 15 hen-feathered males should represent the 5/6. These expectations agree perfectly with the observations and therefore it is most probable that two of the females were HH and one was Hh.

Practice 7

Suppose that a hen's ovaries are destroyed by disease, allowing its rudimentary testes to develop. Further suppose that this hen was carrying the dominant sex-linked gene B for barred feathers, and upon sex reversal was then crossed to a nonbarred female. What phenotypic proportions are expected in the F1 and F2?

Solution 7

Remember that sex determination in chickens is by the ZW method and that sex reversal does not change this chromosomal constitution. Furthermore, at least one sex chromosome (Z) is essential for life.

Solved Problems

Practice 8

In monoecious corn, a recessive gene called "tassel-seed" (ts), when homozygous, produces only seedswhere the staminate inflorescence (tassel) normally appears. No pollen is produced. Thus, individuals of genotype ts / ts are functionally reduced to a single sex, that of the female. On another chromosome, the recessive gene called "silkless" (sk), when homozygous, produces ears with no pistils (silks). Without silks, none of these ears can produce seed and individuals of genotype sk/sk are reduced to performing only male functions (production of pollen in the tassel). The recessive gene for tassel-seed is epistatic to the silkless locus.    (a)  What sex ratio is expected in the F1 and F2 from the cross ts/ts; sk+/sk+ (female) × ts+/ts+ ; sk/sk(male)?     (b)  How could the genes for tassel-seed and silkless be used to establish male and female plants (dioecious) that would continue, generation after generation, to produce progeny in the ratio of 1 male : 1 female?

Solution 8

Solved Problems

Practice 9

Pollen tubes containing the same self-incompatibility allele as that found in the diploid tissue of the style grow so slowly that fertilization cannot occur before the flower withers. Pollen produced by a plant of genotype S1S3 would be of two types, S1 and S3. If this pollen were to land on the stigma of the same plant (S1S3), none of the pollen tubes would grow. If these pollen grains (S1 and S3) were to alight on a stigma of genotype S1S2, then only the tubes containing the S3 allele would be compatible with the alleles in the tissue of the style. If these pollen grains were to alight on a stigma of genotype S2S4, all of the pollen tubes would be functional. Four plant varieties (A, B, C, and D) are crossed, with the results listed in the table below.

Solved Problems

Notice that two additional varieties (E and F) appear in the progeny. Determine the genotypes for all six varieties in terms of four self-sterility alleles (S1, S2, S3, and S4).

Solution 9

None of the genotypes are expected to be homozygous for the self-incompatibility alleles because pollen containing the same allele present in the maternal tissue is not functional and therefore homozygosity is prevented. Thus, six genotypes are possible with four self-incompatibility alleles: S1S2, S1S3, S1S4, S2S3, S2S4, S3S4. Crosses between genotypes with both alleles in common produce no progeny (A × A, B × B, etc.). Crosses between genotypes with only one allele in common produce offspring in the ratio of 1 : 1 (e.g., S1S2 female × S1S3 male = 1/2 S1S3 : 1/2 S2S3). Crosses between genotypes with none of their self-incompatibility alleles in common produce progeny in the ratio 1 : 1 : 1 : 1 (e.g., S1S2 × S3S4 = 1/4 S1S3 : 1/4 S1S4 : 1/4 S2S3 : 1/4 S2S4. Turning now to the table of results, we find the cross B female × A male produces offspring in the ratio 1 : 1 : 1 : 1 and therefore neither B nor A have any alleles in common. If we assume that variety B has the genotype S1S4, then variety A must have the genotype S2S3 (the student's solution to this problem may differ from the one presented here in the alleles arbitrarily assigned as a starting point). The cross C male × A female produces offspring in the ratio 1 : 1, indicating one pair of alleles in common. Since we have already designated variety A to be of genotype S2S3, let us arbitrarily assign the genotype S1S2 to variety C. The cross D female × Amale also indicates that one allele is held in common by these two varieties. Let us assign the genotype S1S3 to variety D. The genotype for variety E may now be determined from the cross C female × B male.

Solved Problems

Likewise, the genotype for variety F may now be determined from the cross D female × B male.

Solved Problems

Summary of genotypes for all six varieties:

    A = S2S3 B = S1S4 C = S1S2 D = S1S3 E = S2S4 F = S3S4

The student should confirm that the other results shown in the table are compatible with the genotypic assumptions shown above.

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