An answer adequate for full credit is shown in bold face type. Additional notes are sometimes added in regular type.

1. In peas, the allele for round seeds (R) is dominant over the allele for wrinkled seeds (r) and the allele for yellow seeds (Y) is dominant over the allele for green seeds (y). There is no linkage between these two genetic loci. A true-breeding strain that produces round green seeds is crossed with a true-breeding strain that produces wrinkled yellow seeds.

a. What will be the phenotype of the F1 progeny?

Round and yellow.

All F1 progeny will have the genotype Rr Yy, and thus will exhibit both dominant phenotypes.
It is important to make a clear distinction between genotype and phenotype. Several answers reported the genotype, rather than the phenotype.

b. What phenotypes will be present and in what ratios in the F2 generation?

9/16 Round yellow : 3/16 wrinkled yellow : 3/16 round green : 1/16 wrinkled green

Ratio may be expressed as 9:3:3:1

c. What phenotypes will be present and in what ratios in the progeny of a test cross of the F1 generation?

1/4 Round yellow : 1/4 wrinkled yellow : 1/4 round green : 1/4 wrinkled green

The phenotypes of the test cross progeny reflect the genotypes of the haploid gametes produced by the F1 hybrids. Ratio may be expressed as 1:1:1:1.

d. What fraction of peas from the F2 generation that exhibit both of the dominant traits will be found to be true-breeding?

1/9

In a full Punnett squate, only one of the double dominant phenotypes will have a genotype of RRYY. It will be the only one of the 9 squares with a double dominant phenotype that does not contain any recessive alleles at either of the loci.

e. What fraction of ALL peas in the F2 generation will be found to be true-breeding? (Think about this one carefully before you answer!)

4/16, which can also be expressed as 1/4

The true breeding genotypes are RRYY, RRyy, rrYY, and rryy. A forked line diagram for genotype gives each of these a probability of 1/4 x 1/4 = 1/16. Together, they are 4/16. In this problem, it is easy to overlook the Rryy and rrYY genotypes, which are homozygous at both loci and thus true breeding.

2. (25 points, 5 for each part). Distinguish between the following pairs in a manner that makes it clear you know what each is and how they differ.

a. Sister chromatids and homologous chromosomes

Sister chromatids are the products of replication of a chromosome before the two halves have separated. Homologous chromosomes are comparable chromosomes of maternal and paternal origin in a diploid individual.

During meiosis, homologous chromosomes that have already replicated and thus each contain two sister chromosomes pair together to form a tetrad. The homologous chromosomes separate during the first meiotic division, with the centromeres of the sister chromatids staying together. The sister chromatids separate during the second meiotic division.

b. Codominance and incomplete dominance

In codominance, each of the alleles at a genetic locus expresses the trait it codes for independently, such that in a heterozygous individual, both traits are expressed. In incomplete (partial) dominance, one copy of the dominant allele is not enough to generate a phenotype that is identical to that produced when two copies of the dominant allele are present.

The A and B or M and N blood types are good examples of codominance. Incomplete dominance is seen in cases where a hybrid between a red flower and a white flower yields a pink flower, as in snapdragons.

c. Prophase and anaphase

Prophase is the stage of mitosis or meiosis during which the already replicated chromosomes are condensing, such that they become visible with the light microscope. Anaphase is the stage of mitosis or meiosis during which the fully condensed chromosomes are moving away from the midline of mitotic or meiotic spindle toward the poles.

Prophase refers to all stages from the first visible condensation until the centromeres of the chromosomes are fully aligned at the midline of the spindle in metaphase. Some authors refer to the final process during which alignment is starting to occur as "prometaphase", but this is still considered by most to be a subdivision of prophase. Anaphase begins when the centromeres separate and continues until movement to the poles is completed and the nuclear membrane begins to be formed around the chromosomes, which are beginning to decondense.

d. RFLP and VNTR

RFLP (restriction fragment length polymorphism) refers to differences in sizes of DNA fragments produced by a specific restriction endonuclease and detected by a specific hybridization probe. VNTR (variable length tandem repeats) refers to DNA sequences that are repeated in tandem different numbers of times in different individuals.

Both are widely used as haplotype-specific DNA markers that can be detected by Southern blotting. RFLPs usually result from the presence or absence of restriction sites associated with the sequences that are detected by the particular probes that are used. VNTRs may be visualized in Southern blots either as restriction fragments of variable length (a special class of RFLPs) or as PCR products of variable length.

e. Complementary gene action and duplicate gene action

Complementary gene action refers to a situation where functional alleles coded at two different genetic loci must both be present to achieve a particular phenotype. Duplicate gene action refers to a situation where a functional allele at either of two separate loci is sufficient to achieve a particular phenotype.

Complementary gene action results in an F2 ratio of 9:7. Absence of a functional allele at either of the loci results in loss of the trait in question. Only individuals that express at least one dominant allele at each of the loci exhibit the trait. A cross between individuals that are each homozygous recessive at one of the loci will result in F1 progeny that express the trait even though neither of the parental strains do. The textbook example was purple flowers in sweet peas (figures 13.16 through 13.18). Duplicate gene action results in an F2 ratio of 15:1. Only individuals that do not express any dominant alleles at either of the loci exhibit the recessive phenotype. The textbook example was Spring versus Winter wheat (figure 13.19)

3. (25 points, 5 for each part) Briefly define each of the following and explain its genetic importance.

a. Epistasis

Epistasis refers to a situation where a particular allele at one genetic locus inhibits (or otherwise alters) expression of alleles at another locus

As an example, the recessive albino state blocks expression of any other coat color alleles in mice in the 1/4 of the F2 progeny that are homozygous for the albino allele. Similarly, dominant white in squash (or chickens) inhibits expression of any other colors in the 3/4 of the F2 progeny that have at least one dominant white allele.

b. Dominant loss-of-function mutation

Dominant loss of function occurs when a mutant gene product interacts in a manner that prevents the corresponding normal gene product in a heterozygous cell from achieving its normal function.

The mutant gene product may either combine with the normal gene product to generate a non-functional dimer or multimer, or it may bind to other proteins that normally interact with the functional gene product. In the latter case, the dominant loss-of-function gene product may bind all of the available target proteins and thus prevent them from forming active complexes with the normal gene product, thus preventing the normal gene product from functioning in a heterozygote.

c. Unordered probability

Unordered probability is the probability that a particular end result will be achieved without any concern about the sequence in which its individual component events occur.

For example, two boys in a family with four children could occur as BBGG, BGBG, BGGB, GBBG, GBGB, or GGBG. The ordered probability for any one of those events is 1/16. However, because there are six different ways to achieve the same end result,the unordered probability of two boys in a family with four children is 6/16.

d. Null hypothesis

The null hypothesis proposes that any differences between observed and expected results are entirely due to random chance events.

The Chi-square test asks how great the probability is that the observed deviation from the predicted results is actually due to random chance events. The null hypothesis is usually rejected if the probabilitythat the deviation is entirely due to change is less than 0.05.

e. Annonymous probe

An annonymous probe is a randomly cloned segment of DNA that is capable of hybridizing somewhere within a genome that is being studied, but not within any sequence that has a known function, such as a protien coding sequence.

The most useful annonymous probes hybridize to unique sequence or minimally repetitive DA sequences. In many cases they are used to identify RFLPs or other polymorphic genomic features that can be used as codominant Mendelian markers.

4. Two parents with normal hearing have a daughter who is deaf. Assume that all deafness in these questions is of genetic origin and that none of the genetic loci are sex-linked. Fractions are acceptable as answers. You may wish to explain the reasoning behind your answers so a mistake in caclulation will not cost you all of the points.

a. (5 points) What can you infer about the genotypes of the parents?

Both parents are heterozygous carriers of the same recessive allele for deafness.

The fact that both parents have normal hearing rules out dominant alleles for deafness. There is no need to invoke more than one genetic locus prior to part e of this question. Feasible alternative answers were accepted even if they made the situation more complicated than it needed to be.

b. (5 points) What is the probability that the next child born into the family will be a boy with normal hearing?

Boy = 1/2, normal hearing = 3/4. These are independent events, such that the product rule can be used. Probability of boy with normal hearing = 1/2 x 3/4 = 3/8.

A number of answers lost part of the credit because they calculated probability of deaf boy rather than probability of boy with normal hearing.

c. What is the probability that one and only one of the next three children born into the family will also be deaf?

9/64

This is an unordered probability. The ordered probability of one deaf and two with normal hearing is 1/4 x 3/4 x 3/4 = 9/64. There are three possibilities, DNN, NDN, and NND. Thus, the unordered probability is 3 x 9/64 = 27/64. This can also be done with the binomial expansion, using p = 1/4, q = 3/4, n = 3, x = 1, y = 2.

d. What is the probability that one and only one of the next three children born into the family will be heterozygous? (Hint, think very carefully about the probability classes that you are dealing with as you work on this problem).

3/8

The probability of heterozygous is 1/2, which means that the probability of not-heterozygous is also 1/2. p = 1/2, q = 1/2, n = 3, x = 1, y = 2. Ordered probability = 1/8. binomial coefficient = 3. Unordered probability = 3/8.

e. The deaf daughter marries a man who is also deaf. They have a child with normal hearing. The man's parents also have normal hearing. Provide a feasible genetic explanation for the ability of the child of the two deaf parents to hear.

ddFF (deaf) x DDff (deaf) --> DdFf (can hear)
(d and f are recessive alleles for deafness at different genetic loci)

The daughter is homozygous recessive for deafness (from part a). Her husband's deafness must originate from a different genetic locus. The simplest solution is that he is homozygous recessive for a different defect in hearing, such that the child is heterozygous at both loci and thus able to hear (complementary gene action).

Some answers said that more than one allele was involved, rather than more than one locus. This does not work because alleles are alternatives at the same locus. The only way introducing a new allele could reverse the deafness would be intracistronic complementation, which is rare and has not yet been covered in detail in the course, although it has been mentioned. Several answers suggested a mutation that restored function. This is possible, but it would not normally be called a "gain-of-function" mutation. That term typically refers to a deviation from wild type caused by excessive expression of a particular gene product. It would be more appropriate to refer to the mutation that restored function as a back mutation or a reversion to functionality.

There were several suggestions that penetrance of deafness was less than 100% in the child. This was acceptable if worded properly. The husband could also be deaf because he is heterozygous for a dominant deafness allele at a different locus. In this case, the child would inherit his normal alleles both from that locus and from the locus that is responsible for his wife's deafness. (This gets more complicated than necessary.) Several answers suggested that the husband was deaf as the result of an accident. That possibility was specifically ruled out in the original description at the top of the page. Several people suggested an epistasis-like phenomenon. It is unlikely that a loss of funciton could be blocked by epistasis, which normally caused by a larger scale event not happening at all (such as total absence of pigment blocks expression of different coat colors).

Grade distribution: Mean = 74.9; Median = 76

  100  ---  3
90-99 ---  8
80-89 --- 14
70-79 --- 15
60-69 ---  9
50-59 ---  7
35-49 ---  3