BISC403 Genetic and Evolutionary Biology Spring 2011

Transcription

1 BISC403 Genetic and Evolutionary Biology Spring 2011 February 22, 2011 Summary of requirements for Exam 1 (to be given on March 1) plus first exam from fall of 2010 The primary responsibility is for any topic covered in lecture. This will include those portions of the textbook which relate to the lecture. Note that the supplemental lecture on chapter 4, linked from the web site, is to be treated as though it were a lecture that was delivered in class. Reading: Chapter 1 is an overview and no questions will be taken from this chapter. Chapter 2 (including supplemental web page on Mendel) Chapter 3 Note that several topics were not covered in lecture, but are assigned: penetrance expressivity phenocopy the genetics of apparently continuous traits (see Figures 3.21 and 3.22) This (so-called continuous traits) is an especially important topic, since it explains that the fundamental Mendelian principles apply in all cases and that blending inheritance does not occur. See also the summary on epistasis Chapter 4 as follows: 1. Review pages to be sure this is familiar material. It will be assumed in later lectures that this material is understood, but it will not be tested directly on the examination. Include Figures and 4.12 in this review. 2. Assigned: pages with emphasis on Figures 4-18, 4-19 and 4.20 plus Table 4.3 on page Not assigned: "Fast Forward" on page 91 and "checkpoints" on pages and in Figure 4.11 Chapter 5 as follows: 1. Assigned: Understand following figures and associated text: 5.2, 5.3, 5.4, 5.7, 5.8, 5.9, 5.10,

2 5.11, Note that Figure 5.6 is very important but will not be covered in lecture nor be on the exam. 2. Not Assigned: a) Chi-Square (pages and Figure 5.5) b) "Fast Forward" on page 137. c) pages (correspondence of physical and genetic map) d) pages (tetrad analysis and mitotic recomination) Aside from the Mendel supplement covering chapter 2 and the specific topics listed above, any topics which were not covered in lecture will not be on the examination. Be sure to answer the question that is asked. Each question on the exam will be very focused. Do not spend your time ineffectively, especially on the short answer questions, by addressing the wrong topic. Examples are given with the short answer questions below. Note: So that all students have an equal opportunity on the exam, there will be no questions permitted during the exam. EXAM 1, FALL, 2010 BISC403 Genetic and Evolutionary Biology Examination I, September 23, 2010 Instructions: 1. Grid in your name on the answer sheet. Grid in nothing else. 2. Write your name on page There are blank sheets of paper between questions 20 and At the end of exam, turn in your answer sheet and last page. You may keep the rest of the exam. All multiple choice questions are of equal value and together they are worth 85% of the exam. The three short answer/problems are of equal value and together they are worth 15% of the exam. Answer all questions. 1. Flower color in peas shows complete dominance, with the P allele (for purple) being dominant over the p allele (flowers with pp genotype are white). Flower position in this organism also shows complete dominance with the A allele (for axial) being dominant over the a allele (flowers with aa genotype are terminal, that is, at the end of the stem).

3 Assume that you have a pea plant which has the flower phenotypes of purple and axial. This plant is allowed to self-fertilize. Which of the following is true concerning the plants produced as offspring of this self-fertilization? a) If the original plant were a double heterozygote, then there will be 4 different phenotypes. b) If the original plant were homozygous for both of the genes, then there all the offspring will have the same phenotype. c) If there are 2 different phenotypes in the offspring, they will be present in a ratio of 3:1. d) There will either be 1, 2, or 4 different phenotypes in the offspring. It would not be possible for there to be 3 different phenotypes. e) All of the above. 2. Use the same example as in question 1. Which of the following is true concerning the flowers produced as offspring of the self-fertilization? a) If the original plant were a double homozygote, then there will be 3 different genotypes. b) If the original plant were homozygous for one of the genes and heterozygous for the other, then there will be 9 different genotypes. c) If there are two 2 different phenotypes in the offspring, then there will be 3 different genotypes. d) If there are 2 different phenotypes in the offspring, then there will be 2 different genotypes. e) None of the above. 3. Consider the following three dihybrid crosses that began with pure-breeding strains of plants that can self-fertilize. Assume that one of these pure-breeding strains has the genotype AAbb and the other has the genotype aabb. Cross I. both genes show complete dominance Cross II. both genes show incomplete dominance. Cross III. one gene shows complete dominance and the other shows incomplete dominance. The F1 plants which arise from these crosses are allowed to self-fertilize. Which of the following is true about the F2 offspring of these crosses? a) The genotype ratios will be 1:2:1:2:4:2:1:2:1 in the offspring from all three crosses. b) The genotype ratios will be 1:1:1:1 in the offspring of all three crosses. c) The phenotype ratios will be the same in the offspring of all three crosses. d) Cross I will have the smallest number of different phenotypes in the offspring. e) a and d. 4. For genes with multiple alleles: a) one allele is always recessive to all the others. b) codominance is not possible, since a dominance series will exist. c) an allele may be dominant to one allele and recessive to a third allele. d) polymorphic means that the recessive is harmful compared to the wild type. d) monomorphic means that only one allele is recessive.

4 5) When dihybrid crosses are performed, many different phenotype ratios may be observed in the F2 generation. Which of the following phenotype ratios are consistent with independence (non-linkage) of the genes involved? a) 9:3:3:1 b) 9:7 c) 9:3:4 d) 13:3 e) all of the above 6. The S allele of the β-globin gene causes sickle-cell anemia when homozygous (all other alleles for the production of this protein are collectively identified as the A allele). The β-globin gene shows remarkable pleiotropy. Which of the following is true? a) The A allele is completely dominant to the S allele in the production of properly functioning β-globin at all altitudes. b) The S allele is completely dominant to the A allele in causing malaria resistance. c) Heterozygotes are disadvantaged in all environments due to the synthesis of abnormal hemoglobin by the S allele. d) The A allele is more common in regions of the U.S. with a high occurrence of malaria. e) a and b.

5 7) When dihybrid crosses are performed, epistasis is sometimes observed. Which of the following is true for the F2 offspring in these situations? a) The phenotype ratios are no longer 9:3:3:1. b) Epistasis may cause the appearance of a new phenotype that was not displayed by either parent. c) Fewer genotypes are present than there are in the absence of epistasis. d) Lethality is a common consequence of dominant epistasis. e) a and b. 8) Complementary gene action (as seen with flower color production in sweet peas): a) causes the same genotypic ratios in the F2 as recessive epistasis does. b) is generally associated with a 13:3 phenotype ratio in the F2. c) refers to cases in which the product of one gene interferes with the formation of product by the other gene. d) is seen most easily with genes that show incomplete dominance. e) all of the above. 9) In mice, the A Y allele of the agouti gene is a recessive lethal allele, but it is dominant for yellow coat color. This gene is not on the X chromosome. Mice homozygous for the A allele display the agouti phenotype. The albino gene shows complete dominance with the homozygous recessive genotype (cc) being epistatic to the agouti gene. What phenotypes and ratios of offspring would you expect from the cross of two mice with the following genotypes: Mouse 1: A Y A Cc Mouse 2: A Y A cc a) 1 yellow: 2 agouti: 3 albino b) 3 yellow: 0 agouti: 3 albino c) 2 yellow: 1 agouti: 3 albino d) 2 yellow: 2 agouti: 2 albino e) none of the above 10) Assume that colorblindness in humans is a sex-linked trait controlled by one gene with two alleles (CB dominant to cb, individuals homozygous or hemizygous for the recessive allele being colorblind). Consider the following: a colorblind man has children with a woman who has normal vision, but whose father was colorblind. Which of the following is true? a) The probability that their daughters will be carriers is ¼ for each daughter. b) All of their daughters will have at least one recessive allele. c) None of their sons will be colorblind. d) The probability that their sons will be colorblind is ¼ for each son. e) a) and d).

6 11) Cystic fibrosis is an autosomal recessive human disease caused by one gene with two alleles (CF dominant to cf). A man who has a sibling with cystic fibrosis has children with a woman who also has a sibling with cystic fibrosis. Neither the man nor the woman has cystic fibrosis, nor do any of their parents. Which of the following is true? a) The probability that the man is a carrier and that the woman is not a carrier is 1/9. b) The probability that they will have a child who has cystic fibrosis is 4/9. c) The probability that they will have a child who is homozygous dominant is 4/9. d) The probability that they will have a child who is homozygous is 4/9. e) c) and d) 12) Assume the absence of the Bombay phenotype in answering this question concerning human ABO blood types (alleles I A, I B, and i). If a man with type A blood has children with a woman who has type AB blood, then: a) Their children may have either type A, type B, or type O blood. b) All of their children with type A blood will be homozygous. c) They could have children who were homozygous for one of the codominant alleles. d) If one of the parents of the man were type O, then they could have a type O child. e) a), b) and d) 13. Penetrance is a term to describe the observation that: a) some genes show incomplete dominance. b) the genotype determines the phenotype in the same way each time. c) not everyone with the same genotype necessarily displays the same phenotype. d) heterozygotes usually have lower viability than those who are homozygous dominant. e) the environment is unimportant in determining the phenotype. 14. A female fruit fly from a pure-breeding strain with carnation eyes (a mutant color) and normal body color (brown) is crossed to a male from a pure-breeding strain with normal red eyes and ebony body. The F1 progeny consist of females with wild type red eyes and normal brown body color plus males with carnation eyes and normal brown body color. When the F1 progeny are mated to each other, the F2 consists of four types of females and four types of males with the proportion of phenotypes the same in both genders: carnation eyes, normal body (3/8 of total) red eyes, normal body (3/8 of total) carnation eyes, ebony body (1/8 of total) red eyes, ebony body (1/8 of total) Are the genes for eye color and body color X-linked or autosomal? a) The eye-color gene is X-linked and the body color gene is autosomal. b) The eye-color gene is autosomal and the body color gene is X-linked. c) Both genes are X-linked d) Both genes are autosomal e) More than one of the above is consistent with the data.

7 15. Genes Q and R are 20 map units apart. Both genes display complete dominance. If a plant of genotype Qr/qR is test-crossed, what percentage of the progeny will display the dominant phenotype for both genes? a) 0% b) 10% c) 20% d) 50% e) 80% 16. Colorblindness and hemophilia are both recessive sex-linked traits. A woman whose mother is colorblind and whose father has hemophilia is pregnant with a boy and wants to know the probability that her son will be either colorblind or a hemophiliac but not both. What is the probability? Assume that there is no history of hemophilia in the family of the woman s mother and no history of colorblindness in the family of the woman s father. The genes for colorblindness and hemophilia are three map units apart on the X chromosome. a).03 b).15 c).485 d).97 e).015 The following information applies to questions 17 to 19. In fruit flies, the recessive pr and vg alleles (these are two separate genes) cause purple eyes and vestigial wings, respectively. A female who displays the wild type phenotype for both traits is test-crossed to a male with purple eyes and vestigial wings. Their progeny (both genders) have the following distribution of phenotypes: wild type 430 purple vestigial 445 purple 61 vestigial total 17. What is the genotype of the original mother? (Note: the alleles listed before the slash in each answer are on one of the mother's chromosomes and the alleles listed after the slash are on the other chromosome). a) pr vg / pr + vg + b) pr + vg / pr + vg c) pr + vg / pr vg + d) pr vg + / pr vg + e) pr vg / pr vg

8 18. The original mother resulted from a cross between two flies from true breeding lines. What are the genotypes of these two lines? (Note that the chromosomes are NOT indicated this time). a) pr + pr + vg vg and pr pr vg + vg + b) pr + pr + vg + vg + and pr pr vg vg c) pr + pr + vg vg and pr pr vg vg d) pr pr vg + vg + and pr + pr + vg vg e) either a) or d) could be true. 19. What is the map distance between the pr and vg genes in map units? a) 6.25 b) 12.5 c) d) 75.0 e) Three point crosses can give map positions for three genes in a single experiment. Which of the following is true? a) Parental types are always the most numerous class, even with double crossovers. b) Double crossovers can make measured map distances appear to be larger than they really are. c) There will be the same number of wild type and triple recessive phenotypes only if all the recessive alleles were together on the same chromosome to begin with. d) Double crossovers are easier to detect in short regions (small numbers of map units). e) All of the above. 21. Below is a pedigree for a human trait. Shaded symbols are for individuals showing the trait. Females are shown by circles and males by squares. What is the genetic basis of this trait (that is, dominant or recessive, sex-linked or autosomal)? Explain. Also, if individuals 4 and 5, who are cousins, have a child together, what is the probability that their child will display the trait? If the probabilities are different for sons and daughters, calculate each probability separately. Show your work.

9

10 22. Suppose you discover a new variant in mice in which they have spiky fur instead of the usual soft fur. The new trait is only seen in males. You cross a spiky-fur male with a soft-fur female and all the progeny (F1) of both sexes has soft fur. If an F1 male is mated with an F1 female, then in their progeny (F2) all of the females have soft fur, but ¼ of the males have spiky fur. Explain. 23. In the fruit fly, the genes f (forked bristles), y (yellow body) and v (vermillion colored eyes) are all X-linked and all are completely recessive to the wild type alleles. Females from a pure-breeding strain with normal bristles, yellow body and vermillion eyes were mated with males from a pure-breeding strain with phenotype forked bristles, brown (wild type) body color and red (wild type) eyes. If female offspring from this cross are test-crossed, the male progeny that result are given below. Give the sequence of genes and the map units. y f + v 3210 y + f v y f v 1024 y + f + v y f v y + f + v 690 y f + v + 72 y + f v total Multiple Choice Answers 1. e 2. c 3. e 4. c 5. e 6. b 7. e 8. a 9. c 10. b 11. c 12. c 13. c 14. a 15. b 16. d 17. a 18. b 19. b 20. a

11 Answers to Short Answer Questions Question 21. Grading of this question had three portions: a) identification of the basis of the condition. b) explanation for the conclusion in a) c) probability explanation for the child of the indicated mating. a) This trait is autosomal recessive. b) There are several different parts of the pedigree which provide evidence for this conclusion; I offer one approach here. Others are possible and received credit. We know it is recessive because the original parents were not affected but had a child who was. We know it is not sex-linked because individual 5 is a male born to an affected mother. Since it is a recessive trait, she must be homozygous. If it were sex-linked, then all her sons would display the trait. c) There are two parts to the calculation. The first step is to determine the possible genotypes of individuals 4 and 5 and assign probabilities for the occurrence of each. Individual 5 must be heterozygous, since he is the son of an affected parent, but does not display the trait himself. Individual 4 may either be homozygous dominant or heterozygous, since she has an affected sibling, but her parents are not affected. They must be heterozygous. Since individual 4 is not affected herself, then the probability that she is homozygous is 1/3 and that she is heterozygous is 2/3. Therefore, the chance that individuals 4 and 5 will have an affected child is 2/3 times ¼ (2/3 that she is heterozygous and ¼ that the children she has with a heterozygote will be affected). This = 1/6. Question 22. This is an autosomal trait in which soft fur is dominant to spiky. The biggest challenge is to explain the differences in the results for the two genders. It is not sex-linked, because that would result in ½ the males being spiky and ½ with soft fur. The best explanation is complete dominance, with the ¼ arising from the homozygous recessive genotype. The absence of the spiky phenotype in females indicates that this is a sex-limited trait. Question 23. Calculate the map distance for each pair of genes: Recombinant flies for each pair of genes y-f y-v f-v y f + v 3210 y + f v y f v y + f + v y f v y + f + v y f + v

12 y + f v Totals Therefore, the y-f recombination frequency is 34.36, the y-v is 15.0 and f-v is The v gene must be in the middle. The map distances above do not add up because of double crossovers. It was not necessary to do the calculation to correct for the double crossovers.