An answer adequate for full credit has been provided in bold-face type for each question. Please note that alternative answers may be possible for some questions. Additional comments about the question or answers that were not acceptable have been added in regular type in some cases.

1. (25 points, 5 for each part) Please write a brief description of each of the following genetic phenomena:

a. Dosage compensation

Dosage compensation is a mechanism that keeps the functional level of expression of genes on the homogametic sex chromosome (X or Z) in a constant relationship to the level of expression of autosomal genes in both sexes.

The mechanisms that are involved are very different in mammals , where one of the two X-chromosomes is invactivated in each somatic cell of the female body, and in Drosophila, where the rate of transcription in both X-chromosomes is reduced in the female, relative to the rate of transcription from the single X-chromosome in the male.

b. Maternal effect gene

Maternal effect genes are those for which the phenotype of the offspring is determined by the genotype of the mother, and not by the genotype of the offspring.

A maternal effect gene typically exerts its effects by introducing specific substances into the egg cell that influence the pattern of development of the embryos that develop from the eggs. Some answers confused maternal effect with uniparental inheritance, for example of mitochondrial genomes. Some also confused it with the parent-specific aspects of imprinting.

c. Imprinting

Imprinting is the selective inactivation at a specific genetic locus of the allele received from one of the parents, such that the progeny is functionally hemizygous at that locus. However, there is no permanent change in the allele, and when it is passed on to progenyby an individual of the sex that does not imprint at that locus, it is once again fully active. Depending on the genetic locus, imprinting may occur either in the female or in the male.

In the example we studied, the IGF-2 locus was imprinted in the female, such that only the allele received from the father was active in the progeny. However, when male progeny pass the imprinted allele to their progeny, it is once again fully active and functional.

d. Position effect

The term position effect refers to modification of the expression of a particular genetic locus determined by its position within a chromosome.

Translocation of a genetic locus to a position close to the centromere often results in sharply reducing the level of expression at that locus. A number of answers confused position effect with interference to crossing over, and some others confused it with map distance and rate of crossing over.

e. Polygenic inheritance

Polygenic inheritance refers to additive effects to a particular phenotype by alleles from more than one genetic locus. In extreme cases, this may result in what appears to be continuous variation, rather than discrete stepwise patterns of inheritance.

The example that we examined in greatest detail was inheritance of seed color in wheat, which appears to involve additive effects of alleles at no less than three separate genetic loci (which is consistent with the hexaploid nature of modern wheat). Some answers confused polygenic inheritance with polyploidy.

2. (25 points, 5 for each part) A female Drosophila from a true-breeding strain exhibiting recessive mutant phenotype a is mated to a male from a true-breeding strain exhibiting recessive mutant phenotype b. What phenotypic ratio would you epect in each of the following situations? Identify the phenotypes that correspond to the numbers in each of your ratios.

a. The male F2 progeny when genetic loci a and b are on separate autosomes.

9/16 wild type, 3/16 ab+, 3/16 a+b, 1/16 ab

The F2 ratio is the same in males and females. The 9:3:3:1 ratio is one that you should have memorized. However, if you do not remember it, you can get there easily with the forked line approach using a 3/4 dominant to 1/4 recessive phenotypic distribution for each of the loci.

b. The male F2 progeny when a is autosomal and b is sex-linked.

3/8 wild type, 1/8 ab+, 3/8 a+b, 1/8 ab

The F2 males each have only one X chromosome , obtained from their mothers, who are heterozygous b+/b. They are therefore 1/2 b+ and 1/2 b. Within each of those groups, 3/4 are wild type at autosomal locus a and 1/4 exhibit the mutant phenotype.

c. The female progeny of a test cross of an F1 male when a is autosomal and b is sex-linked.

1/2 wild type, 1/2 ab+

The F1 males all received from their mother an X chromosome carrying a wild type allele at locus b. Thus, all F2 females will have at least one wild type allele at locus b. The F1 males are heterozygous at locus a. The test cross mating is with a female that is fully recessive at both loci. Thus, the progeny of the test cross are 1/2 a+b+ and 1/2 ab+.

d. The male progeny of a test cross of an F1 female when a and b are on the same autosome and separated by 20 centimorgans.

10% wild type, 40% a+b, 40% ab+, 10% ab

Since each true-breeding parent exhibited only one mutant phenotype, the two mutations must be in a trans arrangement in the F1 female, who is an obligate heterozygote at both loci. A separation of 20 centimorgans corresponds to a total frequency of recombination of 20%, which means that there will be 10% each of the two classes of recombinant gametes (a+b+ and ab). The remaining 80% of parental gametes will also be equally divided between the two possible types. In a test cross, the phenotypes of the progeny exactly reflect the genotypes of the gametes produced by the fly that is being test crossed.

e. The female progeny of a test cross of an F1 male when a and b are on the same autosome and separated by 20 centimorgans.

50% a+b, 50% ab+

There is no recombination in a male Drosophila. Thus, only the parental combinations will be present.

3 (25 points, 5 for each part). Briefly explain the value of each of the following for the study of genetics. Include in your answer a sufficient description so that it is clear you know what each item is.

a. Chromosomal banding

A staining technique is utilized that gives each chromosome characteristic banded pattern. This makes it possible to identify any one of the human chromosomes in the absence the others, for example in a somatic cell hybrid that contains only one or a few human chromosomes in a background of chromosomes from another species.

Banding also makes it easier to identify deleted, duplicated, or translocated regions of individual chromosomes, and to identify specific monosomies and trisomies with greater reliability.

b. Merozygote

A merozygote is a bacterial cell that has been made partially diploid through the incorporation of a plasmid that contains a second copy of a small segment of the bacterial genome. The use of merozygotes permits the study of regulatory interactions among bacterial genes.

Merozygotes are particularly useful for determining whether regulatory sequences have to be on the same physical chromosome as the genes they are regulating, or whether they can achieve their effects in a trans configuration.

c. Ordered tetrad

An ordered tetrad is a spore capsule in which the haploid spores generated by meiosis are in a linear order that allows determination of which subsets were separated at eacj meiotic division. Use of ordered tetrads allows precise analysis of all four products of meiosis, including verification of the reciprocal nature of recombination, and the identification of which strands of the meiotic tetrad were involved in double crossover events.

d. Temperature-sensitive mutation

A temperature sensitive (ts) mutation is one that results in a gene product that is non-functional at an elevated temperature, but still capable of funciton at a reduced temperature. The use of ts mutations allows organisms with otherwise lethal mutatations to be propagated at low termperatures and the effects of loss of the gene products to be studied at higher temperatures.

e. lod score.

A lod score is a mathematical formulation used to evaluate data suggesting that two genetic loci may be linked. It consists of the log base 10 of the ordered probability of obtaining an observed set of births based on the assumption of a specified degree of linkage, divided by the ordered probability of obtaining the same set of births based on the assumption that there is no linkage. The use of lod scores makes it possible to combine data from more than one pedigree in order to obtain statistically adequate data. This is particularly useful in studies of genetic linkage in humans, where family sizes are small and selective breeding cannot be done.

A lod score of 3.0, which represents 1000:1 odds that the assumed linkage is valid, is considered to be strong evidence for linkage. Smaller positive values are considered to be suggestive of linkage.

4. (25 points, 5 for each part) A female Drosophila heterozygous for linked autosomal genes a, b, and c is crossed with a male that is homozygous recessive for all three genes. The female progeny exhibit the following phenotypes: +++, 340; abc, 350; ++c, 45; ab+, 55; +b+, 95; a+c, 105; a++, 4; +bc, 6. Total flies counted = 1000.

a. Which is the middle gene?

The middle gene is a

The two most frequent classes of progeny reflect the genotypes of the parental gametes (+++, abc). The two least frequent classes are the double crossovers (a++, +bc). The alleles at b and c have remained together in the double crossover, whereas the alleles at a have exchanged positions. This identifies a as the middle locus.

b. What is the corrected map distance between the two outside genes?

32 centimorgans (map units)

The corrected measure of distance between the outside loci (b and c) consists of the sum of the two single crossover classes plus twice the double crossover class. The SCO frequency between b and a is (95+105)/1000 = 200/1000 = 0.20. The SCO frequency between a anc is (45+55)/1000 = 100/1000 = 0.10. The DCO frequency is (4+6)/1000 = 10/1000 = 0.01. The total corrected map distance derived from this three point cross is 0.20 + 0.10 + (2 x 0.01) = 0.32. Converted to percentage, this is 32%, which corresponds to 32 centimorgans, which are also called map units.

c. Why would a two point cross between the two outside genes not yield the same value as the corrected map distance?

The directly measured value in a two point cross would consist only of those recombinant classes that cause the two outside markers to be present in a different relationship to each other than in the parental class. Double crossovers bring the markers back into their original relationship relative to each other, and thus cannot be seen in a two point cross, despite the fact that they represent two crossover events occurring between the outside markers.

In the example above, the measured distance in a two point cross would only be the sum of the two single crossover classes, yielding an apparent map distance of 30 centimorgans, rather than the corrected value of 32 centimorgans.

d. How would the observed results have differed if the male parent had been heterozygous and the female parent homozygous recessive?

+++, 500; abc, 500.

There is no recombination in a male Drosophila. Thus, only the two parental classes of gametes and test cross progeny would be obtained.

e. Would it have made a difference in the original cross if the genes had been sex-linked and the male was hemizygous for all three recessive alleles? Explain your answer.

No. Crossing over would still occur. Female progeny would receive an X-chromosome carrying only recessive alleles from the hemizygous father. Thus, their phenotypes would exactly reflect the genotypes of the gametes produced by the heterozygous female.

Male progeny could also be examined, since their single X chromosomes would be obtained from their mother and would accurately reflect the recombinaiton that had occurred in her during oogenesis.

For those of you who had problems with this examination, help is available. Please take full advantage of my office hours and Manisha's , as well as review sessions, dorm tutoring and SASC tutoring. If you have schedule problems with my regular office hours, I can meet with you at other times by appointment.