2. A chromosome with a centromere at the very end is called telocentric.

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1 Problem Set 1A Due August A diploid somatic cell from a rat has a total of 42 chromosomes (2n = 42). As in humans, sex chromosomes determine sex: XX in females and XY in males. i. What is the total number of DNA molecules in a rat cell in G 2? 84 ii. What is the total number of telomeres in a rat cell in G 2? 168 iii. What is the total number of chromosomes present in the cell during metaphase I of meiosis? 42 iv. What is the total number of chromosomes in a polar body cell from a rat? A chromosome with a centromere at the very end is called telocentric. 3. Use the following choices for this questions: a. Meiosis I prophase b. Meiosis I anaphase c. Meiosis II prophase d. Meiosis II anaphase e. Mitosis prophase f. Mitosis anaphase i. Chromosomes are in unseparated, sister-chromatid form, at the end of the phase(s) a, b, c, e. ii. The first stage after which a dividing cell that started as a diploid would be haploid b. iii. Sister chromatids separate during d, f. iv. Chromosomes are randomly partitioned during b, contributing to genetic diversity. v. Crossing over (genetic recombination) occurs in a. 4. Write a calculation, with an explanation, for the number of different combinations of chromosomes in the gametes of a rat (2n = 42). 2 n = 2 21 where n = number of homologous pairs, of which there are 21 in a (female) rat.

2 Chapter 2 5. Microscopy to look at a cell's chromosomes is often done when the cell is in mitotic metaphase. For example, karyotypes that extract chromosomes from a single cell and photograph them to look for abnormalities are done on metaphase, rather than interphase, cells. Why? In metaphase, chromosomes are condensed and are more easily visualized. 6. The cells illustrated below belong to a species with a diploid chromosome number of four. Each of the cells below is in which stage of mitosis or meiosis? a. meiosis I anaphase b. mitosis metaphase c. mitosis anaphase d. meiosis I anaphase e. meiosis II metaphase

3 Chromosomes and Cellular Reproduction 7. Find and describe at least four errors in the drawing below of mitosis anaphase. a. Chromosomes that are separating are still duplicated. b. Spindles are not coming from a common spindle pole body. c. Sister chromatids do not have identical alleles for the B gene. d. Two alleles of the D gene are on one chromosome. e. No alleles of the A gene are on the homologous chromosome. f. Homologous chromosomes appear to have paired and to be segregating. A - -A B - -b B - -b d - D - d - D -

4 Chapter 2 8. Two gene loci, A and B, are unlinked (and thus assort independently), and alleles A and B are dominant over alleles a and b. Indicate the probabilities of producing the following. a. An AB gamete from an AaBb individual? a. 1/4 b. An AB gamete from an AABb individual? b. 1/2 c. An AABB zygote from a cross AaBb AaBb? c. 1/16 d. An AaBb zygote from a cross AaBb AABB? d. 1/4 e. An Aabb zygote from a cross AaBb AAbb? e. 1/4 f. An AB phenotype from a cross AaBb AaBb? f. 9/16 g. An AB phenotype from a cross aabb AABB? g. 1 (100%) h. An ab phenotype from a cross AaBb AaBB? h. 1/4 9. While doing field work in Madagascar, you discover a new dragonfly species that has either red (R) or clear (r) wings. Initial crosses indicate that R is dominant to r. You perform three crosses using three different sets of redwinged parents with unknown genotype and observe the following data: Cross Phenotypes 1 72 red-winged, 24 clear-winged 2 4 red-winged 3 96 red-winged a. What is the most likely genotype for each pair of parents? Cross 1 results in a 3:1 ratio of red-winged to clear-winged progeny, therefore the parents are most likely both Rr. Crosses 2 and 3 result in only red-winged progeny, therefore the parents are most likely all RR. b. Do you think there are a sufficient number of progeny to support each of your answers in the previous question? Crosses 1 and 3 have a sufficient number of progeny, but the low number of progeny from cross 2 precludes making any conclusions. 10. In snapdragons, the allele for red flowers is incompletely dominant over the allele for white flowers, and thus heterozygotes have pink flowers. What ratios of snapdragon flower colors would you expect to see among progeny generated from the following crosses? a. red white 100% pink b. red pink 50% pink; 50% red c. white pink 50% pink; 50% white e. white white 100% white f. pink pink 25% red; 50% pink; 25% white g. red red 100% red

5 Chromosomes and Cellular Reproduction 11. The red kernel color in wheat is caused by the presence of at least one dominant allele from each of two independently segregating gene pairs (e.g., R-B-). Wheat plants with rrbb genotypes have white kernels, and plants with genotypes R-bb and rrb- have yellow kernels. You cross a plant true breeding for red kernels with a plant true breeding for white kernels. a. What is the expected phenotype(s) and ratios of the F 1 plants? Using P for the leaf color locus and S for the stem type locus: PpSs ppss. b. What are the relative proportions of the phenotypic classes expected in the F 2 progeny after selfing the F 1 progeny? A possible hypothesis is that each trait is controlled by independently assorting single gene pairs, and that one allele in each pair exhibits complete dominance over the other. Class 1 = 3/8, class 2 = 3/8, class 3 = 1/8, class 4 = 1/ List at least four phenomena that can alter expected Mendelian phenotypic ratios in genetic crosses. (1) Linkage (2) Epistasis (3) X-linked genes (4) Lethal recessive alleles (5) Environmental effects (sex-influenced traits, etc.) (6) Continuous traits (7) Variable expressivity 13. Albinism is a somatic recessive condition resulting from the inability to produce the dark pigment melanin in skin and hair. A man and woman with normal skin pigmentation have two children. The man has one albino parent; the woman has parents with normal pigmentation, but an albino brother. a. What is the probability that at least one of the children is albino? Using alleles A and a, since one of the man s parents is albino, he must be a carrier (Aa) of the recessive allele (probability = 1). The woman has an albino brother, which means both her parents must be carriers (Aa). However, the woman (who is not albino) could have either an AA or Aa genotype. In the woman s case the aa (albino) genotype must be excluded as a possibility, therefore the final probability of the woman having either AA or Aa genotypes is 1/3 and 2/3, respectively. Case 1: If the woman is Aa, the probability of at least one of the two children being albino from the cross (Aa Aa) equals the probability that one child is albino (1/4) the probability that the other child is normal (3/4) 2 (must multiply by 2 because either birth order is possible for this outcome), plus the probability that both children are

6 Chapter 2 albino, (1/4) (1/4) = 1/16. The probability that at least one child is albino equals 2(3/4)( 1/4) + (1/16) = 6/16 + 1/16 = 7/16 (43.75%). Because the probability of the woman having the Aa genotype is 2/3, the final probability of her having albino children is (7/16)(2/3) = 29.2%. Case 2: If the woman is AA, the probability of at least one of the two children being albino from the cross (Aa AA) is zero. b. What is the probability of both children being albino? As noted above, if the woman is Aa, then the probability of both children being albino (aa) equals (1/4) (1/4) = 1/16 (6.25%). Because the probability of the woman having the Aa genotype is 2/3, the final probability of both her children being albino is (1/16)(2/3) = 4.2%. 14. In order to put yourself through school, you decide to start commercially breeding parrots. In one of your lines you discover two birds (a male and a female) with brilliant purple feathers. You cross these parents several times and observe 20 progeny with brilliant purple feathers and 10 with wild-type color. Offer an explanation for this ratio. If the allele for purple feathers (P) is dominant to wild-type color (p) and the parental birds are heterozygous you would expect to see a 3:1 ratio of purple (PP and Pp) and wild-type (pp) progeny. If the P allele also acts as a recessive lethal (i.e. lethal in homozygous state) then the PP class would never occur in the progeny resulting in a 2:1 purple:wildtype ratio. The observed progeny ratio is consistent with 2:1 segregation.

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