The class did poorly on simple Mendelian inheritance (#3, #4, #9, and #36) and other simple inheritance questions (#1, #8, #16, and #20, #37, and #39). The class did poorly on the hypothesis testing question (#15), autopolyploidy (#19), linkage (#13), and polygenic inheritances (#38). Expect to see questions on these subjects on the next exam, scheduled for 2 November 1998.

1. Sex-linked traits in humans are transmitted from mother to son. The male receives one and only one sex chromosome (the "X") from his mother. The female can be heterozygous for recessive traits, such as green colorblindness, hemophilia A, and Duchenne type muscular dystrophy. If the female is heterozygous, she is said to be a "carrier," which means that she can transmit the trait, but does not express it herself. Even though sex linkage or "X-linkage" was not recognized by Mendel, his postulate of segregation would predict that one half of the male offspring from a heterozygous mother will receive the recessive trait. (The mother has two X chromosomes, and there's a 50:50 chance that the son will receive the X with the recessive allele.) If the son receives the recessive allele, he always expresses it, because he has only one X. Male humans are "hemizygous" for every gene on the X chromosome, that is they have only half of the "normal" diploid set that would typify most chromosomes.
2. The problem can be solved by setting up a Punnett's square, using as rows and columns the expected gametes for females and males. The genotype of dihybrid dogs (IiBb)would produce four kinds of gametes in equal frequency, IB, Ib, iB, and ib for both females males. So these four symbol pairs are arranged in each row and each column of a 4 x 4 table. The progeny genotypes are combined within each cell, and their phenotypes are determined, then phenotypes are counted across cells. Epistatic genes mask or block the action of one another in a nonreciprocal manner. The easiest explanation is that the genes act together in a series to regulate a metabolic process. Genes act by eventually producing enzymes, and enzymes can synthesize or break down compounds. Typically a homozygous recessive genotype, for any gene acting in series, results in no enzyme, or defective enzyme, for that that step in the chain. Thus the metabolic process stops at the stage of the missing or defective enzyme.
3. The flow of genetic information is from parent to progeny (inheritance) and from genotype to phenotype (expression). While one can always determine the phenotype from the genotype, if one knows the gene action (e.g., dominant, additive, or epistatic, etc.), most other inferences are made by a combination of probability and knowledgeable guesswork based on the principles of inheritance and gene action.
4. The inheritance of dominant alleles is identical to recessive alleles. Mendel showed that the dominant phenotype is expressed in a 3:1 ratio.
5. Use Punnett's square, same routine, different answer.
6. Genotypes which do not breed true are heterozygous, thus only one gene is necessary to explain many instances of variable progeny.
7. Female mammals have 2 X chromosome, and one or the other is deactivated in separate patches of the body.
8. The human ABO blood types exemplify dominance, codominance, and multiple alleles.
9. Because of unit factors in pairs (two copies of each gene), the number of possible genotypes for each gene pair is 2, 6, 10, 15, etc., for 2, 3, 4, 5 alleles, etc. This can be verified by drawing a table with the rows and columns representing all the different alleles. With two alleles, this will be a 2 x 2 table. The two heterozygotes are genotypically identical.
10. AB children must not have an O parent.
11. Chromosomes in meiosis disjoin reductionally, which means there are half the number of chromosomes in the gametes.
12. Heritability is a measure of recoverable variance. This means that for every unit of variability of the parents, only part of that variability is recovered in the offspring. Heritability of 100% does not occur in nature.
13. To solve thee-gene linkage problems, first find the gene order based on the least frequent two classes of progeny representing the double-crossovers. What this means is that (assuming we crossed a triple-heterozygote onto a homozygous recessive tester), there are eight possible genotypes in the progeny. Two are least frequent. Draw the three possible orders for the three genes, and compare these to the original genotype. Which ever order represents the right order is the one in which the least frequent classes would have arisen from a double crossover. If your original heterozygote was in coupling (three dominant alleles on the same chromosome), then it is easiest to figure. The least frequent classes will each have an "odd" gene, one which is in a different state than the other two. So if the other two genes are dominant, the odd gene is recessive, and vice-versa. The odd gene is located on the chromosome between the other two genes. Once you know gene order, find the map distances as a percentage crossing over for each region, adding the similar single crossover classes. The double crossover classes count in each single crossover.
14. A homozygous offspring must have received the Thalassemia allele twice, once from each parent. Therefore, both parents must be heterozygous. According to Mendel's postulates, there is only one chance in four that offspring of heterozygotes are homozygous.
15. Whenever observations are differ greatly from expectations, the null hypothesis is rejected. One looks for a statistically small probability value, because the conventional probability associated with an experiment is the likelihood that results as different from the expectations would be observed due to chance alone. The success or failure of an experiment is determined by how confident one can be in the results, not whether the outcome supported or violated conventional wisdom.
16. See problem 9. This is the same problem except it is a 3 x 3 table, and there are three heterozygotes represented twice, and they may not be double counted.
17. Find the diploid number and subtract 2.
18. The octoploid is four times the diploid.
19. Chromosomes in autotetraploids pair poorly; usually their chromosomes don't pair but form tetravalents, that is, cross-shaped groups of four chromosomes. Such configurations do not separate, that is, disjoin, regularly, thus the gametes may have incomplete sets of sometimes translocated chromsomes, and low seed set.
20. Pattern baldness is sex influenced (see the picture caption on page 104).
    

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36. Mendel was well organized, kept good records, and tabulated his data.
37. Mitosis produces identical daughter cells, so each has the same genes and the same chromosomes, resulting in continuity of the genetic makeup of cell throughout the body. Meiosis and fertilization generally operate on homologous pairs of chromosomes in the zygote, which divide by reduction and unite in fertilization. This keeps a complete set of chromosomes and genes together as a genome, and results in continuity of the basic makeup of individuals within a species. Independent assortment and crossing generate new genetic combinations.
38. Three genes A, B, and C control height in loblolly pine in an additive relationship involving incomplete dominance. Loblolly pines with A1A1B1B1C1C1 are 100 feet tall, and loblollies with A2A2B2B2C2C2 are only 40 feet tall. Pines with A2A2B2B2C2C1, A2A2B2B1C2C2, and A2A1B2B2C2C2 are 50 feet tall. If a triple heterozygote A2A1B2B1C2C1self-fertilizes, it will have a range of offspring from A1A1B1B1C1C1 to A2A2B2B2C2C2, with heights form 40 to 100 feet. Based on the forked line method, among 64 offspring there will be only one A1A1B1B1C1C1 and one A2A2B2B2C2C2 with a lot more in the intermediate height classes (i.e., 50, 60, 70, 80, and 90 feet), resulting in fine distinctions of height that cannot be measured precisely, but will look like continuous variation.
39. The cross is x and the male offspring will have white eyes and the female offspring will have red eyes.