BIOL 354

Homework Chapter 5

Solutions

1. Solve problem 19, p. 101 in your textbook.

Remember that sex determination in birds is according to the ZZ-ZW system, where males are ZZ and females are ZW. Because of the non-reciprocity of the crosses it appears that this trait is sex-linked with black-eyed being dominant to pink-eyed. Using the symbols Z^B for the black-eyed allele and Z^b for the pink-eyed allele the parents and offspring in each cross should have the following genotypes:

Cross	Parents	Progeny
a.	ZbW X ZbZb	½ ZbW and ½ ZbZb
b.	ZbW X ZBZB	½ ZBW and ½ ZBZb
c.	ZBW X ZbZb	½ ZbW and ½ ZBZb

2. Solve problem 21, p. 101 in your textbook.

To solve this problem it is best to draw a pedigree to visualize the data as follows:

1. The best explanation for an inheritance pattern like this is sex-linked dominant. A father with the trait married to a normal woman passes it along to all his daughters and none of his sons and a mother with the trait (A in this pedigree) married to a normal father passes it on to approximately half of her children without regard to sex. Sex-linked dominant traits act just like autosomal dominant traits when the mother is the affected parent. It is also possible, but not likely, that this trait is an autosomal dominant trait. If it were autosomal dominant we wouldn't expect only daughters to be affected, although it could happen. Lastly, it could be autosomal recessive, but that is even less likely because we would have to assume that B was a carrier and for a rare trait like this the chance of a carrier from the general population marrying an affected individual is very unlikely.
2. To determine the probability we must assign genotypes to A and B. A must be heterozygous for brown teeth and B must be hemizygous for white teeth. The mother, therefore, will be the sole determiner of whether a child will have brown teeth or white teeth. Because she is heterozygous she has a 1/2 chance of passing on the allele for brown teeth, therefore their next child has a 1/2 chance of having brown teeth.

3. Sex determination in the dioecious plant Melandrium album is by the XY method. A sex-linked gene governs leaf size, the dominant B allele producing broad leaves, and the recessive b allele producing narrow leaves. Pollen grains bearing the recessive allele are inviable. What phenotypic results are expected from the following crosses (give phenotype ratios)?

	Seed Parent	Pollen Parent	Results
a.	Homozygous broad-leaf	Narrow-leaf	all male broad-leaf
b.	Heterozygous broad-leaf	Narrow-leaf	all male, ½ broad-leaf and ½ narrow-leaf
c.	Heterozygous broad-leaf	Broad-leaf	½ female broad-leaf, ¼ male broad-leaf, ¼ male narrow-leaf.

4. The following typical pedigrees, based on studies by Wirth, B. et al. (Genomics 2: 263, 1988) show the pattern of transmission of retinitis pigmentosum (RP), a condition that results in gradual loss of peripheral and night vision:

1. Account genetically for the inheritance of RP in each of the three families.

In all three families the inheritance of this trait appears to most likely be sex-linked recessive. This is because it shows up in males almost exclusively and only shows up in females when their father is affected and their mother in normal (although a carrier). It is almost impossible to consider that the trait is autosomal recessive because we would then have to assume that several individuals marrying into the family were heterozygous (e.g. II-5 and IV-6 in family A, II-10 and III-8 in family B, and II-14 in family C), which is highly unlikely.

A mating between IV-4 and IV-9 produces six children with the phenotypes as shown:

2. If your explanation in (a) is correct why are all of the daughters of this mating unaffected?

It should never be assumed that a given phenotype is always due to the same gene or mutation. In this case it appears that the gene or mutation that causes RP is different for family A than it is for families B and C. Therefore, only sons in this mating would be expected to have RP, whereas daughters are now carriers for both genes that cause RP. It might be tempting to invoke an explanation based on lack of penetrance, but this would be inconsistent with the data from the pedigrees before. If lack of penetrance were a problem for this trait then it would have showed up somewhere in the first pedigrees and it doesn't.

5. The pedigree shows the pattern of transmission of two rare human traits, cataracts and pituitary dwarfism.

1. What is the mode of inheritance for each of these traits? Explain.

Cataracts is an autosomal dominant gene and Pituitary Dwarfism is an autosomal recessive trait.

2. IV-1 marries IV-6 and they have five children, three dwarfs with no cataracts and two dwarfs with Cataracts. Does this verify the hypothesis formulated in (a)? Give the genotypes of the parents of this marriage.

Yes, it does agree. The genotypes of the parents must be ccdd for IV-1 and Ccdd for IV-6 where the symbols are as follows: C, cataracts; c, normal; D, normal; d, pituitary dwarfism.

3. What phenotype would you expect among the progeny from the following marriages: i.) III-5 X IV-1 and ii) III-4 X IV-5?

The genotypes for III-5 and IV-1 are CcDd and ccdd, respectively. This is a classic dihybrid testcross, so the progeny should have an equal chance of the following four phenotypes: Cataracts only, Dwarf only, both Cataracts and Dwarf, or Normal.

The genotypes for III-4 and IV-5 are not entirely certain. III-4 could be either ccDD or ccDd and IV-5 could be CcDD, CCDD, CcDd, or CCDd. The following table shows the possible outcomes:

Parents	Progeny
ccDD X CcDD or CcDd	½ Normal:½ Cataracts
ccDD X CCDD or CCDd	all Cataracts
ccDd X CcDD	½ Normal:½ Cataracts
ccDd X CcDd	3 Cataracts & Dwarf: 1 Cataracts: 3 Dwarf; 1 Normal
ccDd X CCDD	all Cataracts
ccDd X CCDd	3 Cataracts: 1 Cataracts & Dwarf

4. What is the probability of: I) II-6 being heterozygous for the alleles for both traits and ii) II-2 being homozygous for the allele for dwarfism?

Probability that II-6 is heterozygous for both traits is 1. Based on the pedigree II-6 must be heterozygous for both.

Probability that II-2 is homozygous for dwarfism is 0. If she were homozygous for dwarfism she would be affected.

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