BISC403 Genetic and Evolutionary Biology Spring, Summary of requirements for Exam 2 (to be given on March 24) plus exam 2 from Fall, 2010.

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1 BISC403 Genetic and Evolutionary Biology Spring, 2011 March 17, 2011 Summary of requirements for Exam 2 (to be given on March 24) plus exam 2 from Fall, The primary responsibility is for any topic covered in lecture. This will include those portions of the textbook which relate to the lecture. Reading: Chapter 13 as follows: Assigned: Chromosome banding, pages Heterochromatin and position effect, pages X-inactivation, Polytene chromosomes, pages Aneuploidy and polyploidy, pages , Table 13.1 NOTE that most of pages was not discussed in lecture, especially The sections on polyploidy. You are responsible for the material through Figure Rearrangements, pages , with general features of transposons. Supplement with solution to problem Chapter 14 as follows: 1. Review: Pages , be sure to understand: auxotroph plasmid These will not be tested specifically, but it will be assumed that students are familiar with them. Disregard penicillin enrichment (Figure 14.4 and associated text) and transposable elements in bacteria (Figure 14.8 and associated text). 2. Assigned: Conjugation as mechanism of gene transfer in bacteria, pages (Figures to very important) 3. Not Assigned (will not be on this exam, but will be on the 3rd exam): a) Transformation (Figure 14.14) b) Transduction (pages )

2 Last year s second exam is included below with answers. Also included are the first 9 questions and question 31 from exam 3 which address topics that were covered this spring. Aside from the specific topics listed above, any topics which were not covered in lecture will not be on the examination. Exam 2 from fall, 2010 BISC403 Genetic and Evolutionary Biology Examination II, October 19, 2010 Instructions: 1. Grid in your name on the answer sheet, write your name on page Do NOT enter or grid in your student number. Multiple choice questions are of equal value worth a total of 85% of the exam. Short answer questions are of equal value and together they are worth 15% of the exam. 1. High resolution G-banding analysis of human chromosomes: a) provides cytological markers for the human genome. b) is valuable because each band contains one gene. c) cannot be done with chromosomes from gametes, since the haploid cells are missing half of the bands. d) has allowed the precise localization of active genes. e) none of the above. 2. Aneuploidy: a) arises from nondisjunction in either meiotic division. b) usually affects only some tissues in humans, such as skin. c) does not occur with sex chromosomes in humans due to X-inactivation. d) usually has very mild phenotypic effects in humans. e) all of the above. 3) The inactivation of X chromosomes: a) produces equal expression of almost all X-chromosome genes in both genders of mammals. b) results in the formation of Barr bodies in male mammals with karyotype XY. c) does not occur in male mammals. d) causes sterility in male mammals with karyotype XXY. e) none of the above. 4) Male cats can have the calico coat color pattern: a) if they have a karyotype with multiple X chromosomes.

3 b) when they are homozygous for the X-linked coat color alleles. c) only if their mothers had the calico coat pattern. d) if they are heterozygous for the X-linked coat color genes. e) a) and d). 5) Consider a mammal with multiple X chromosomes and a fly with multiple X chromosomes. Which is healthier (and why)? a) the mammal because all X chromosomes are active. b) the mammal because the activity of X chromosome genes is averaged to give the proper (balanced) activity. c) the fly because all X chromosomes are active. d) the fly because genetic imbalance does not harm flies. e) none of the above.

4 6) Heterochromatin: a) is chromosomal material which functions much the same way as euchromatin. b) becomes euchromatin when it is moved via an inversion or translocation. c) may diminish if euchromatin "invades" it and causes activation of genes. d) is generally seen in the vicinity of centromeres. 7) Position effect variegation in fruit flies: a) occurs when heterochromatin enhances the activity of genes in nearby euchromatic regions. b) only occurs for genes on the X chromosome, such as white eye color. c) increases the activity of genes in inverted regions. d) was detected as the variable inactivation of genes in inverted regions. 8) The polytene chromosomes of Drosophila: a) arise when mitosis occurs without chromosomal duplication. b) are valuable as cytological markers. c) are inactive meaning they are not being transcribed. d) b) and c). 9) The polytene chromosomes of Drosophila: a) show crossing over very clearly since the chromosomes are so large. b) allow visual distinction between the dominant and recessive alleles for a gene, such as white eye color. c) do not undergo transcription because the chromatids are packed too tightly. d) contain puffs which appear in different places as development proceeds. 10) Chromosome deletions: a) can remove several genes and still allow survival, depending on which information is deleted. b) are easily visible in mitosis as the pairing of homologues is made more difficult. c) are usually detected in the homozygous condition, since the deletion will be masked in heterozygotes. d) generally cause sterility, since meiosis cannot proceed normally. 11) Gene mapping with deletions: a) uses organisms heterozygous for deletions. b) cannot be done when crossing over occurs, therefore male flies are very useful. c) involves the masking of recessive alleles. d) allows the localization of the chromosomal position of a gene. e) a) and d)

5 Questions 12, 13, 14, 15, 16, and 17 relate to the following information. A series of chromosomal rearrangements are isolated in the X chromosome of Drosophila following X-irradiation. Five of the new strains contain deletions and two have inversions. When the polytene chromosomes of the salivary glands of individuals heterozygous for these changes are examined, the following results are obtained. Deletion Missing chromomeres (a chromomere is a single band in a polytene chromosome) 1 3, 4, 5, 6, 7 2 2, 3, 4, 5 3 6, 7, 8, 9 4 8, 9, 10, , 11, 12, 13 Inversion Breakpoints A chromomere 3 and chromomere 11 B chromomere 7 and chromomere 13 Seven different female flies are prepared, so that one X chromosome contains one of the rearrangements, but has the dominant allele of all genes that are present and the other chromosome contains the Bar eye mutation. In this case, all the dominant alleles are the wild type. Each of these females is crossed with a male that is hemizygous for six different recessive traits on the X chromosome: st, f, car, ec, l, ct (the sequence does not indicate the actual sequence on the chromosome). The following table presents the results of these seven crosses. The only females examined are those which do not have the Bar eye phenotype. Deletion Phenotypes of non Bar-eyed females 1 ec ct st 2 ec st 3 l ct 4 car l 5 f car Inversion A B Phenotypes of non Bar-eyed females st car wild type NOTE carefully that your objective is to assign each gene to the narrowest range of chromomeres, even to a single chromomere when possible. 12) What chromomere(s) probably contain the f gene? a) 12 b) 13

6 c) 12, 13 d) 10, 11 e) 11 13) What chromomere(s) probably contain the st gene? a) 3, 4, 5 b) 4, 5 c) 2 d) 4 e) 3 14) What chromomere(s) probably contain the car gene? a) 11 b) 10 c) 10, 11 d) 9, 10, 11 e) 12, 13 15) What chromomere(s) probably contain the ec gene? a) 3, 4, 5 b) 4, 5 c) 2 d) 3 e) 4 16) What chromomere(s) probably contain the l gene? a) 6, 7, 8, 9 b) 8, 9 c) 9, 10, 11 d) 9 e) 9 17) What chromomere(s) probably contain the ct gene? a) 7 b) 6, 7 c) 6 d) 6, 7, 8, 9 e) 5, 6, 7 18) In the deletion mapping experiment described for questions 12 to 17, female offspring of the 7 different crosses were examined instead of males. Which of the following statements is true? a) Males were not examined, because deletion mapping requires heterozygotes. b) Crossing over does not occur in male fruit flies and without crossing over, deletion mapping cannot be done.

7 c) Deletions are lethal in males which would be hemizygous for these X-linked genes. d) none of the above.

8 19) Duplications: a) are usually less harmful than deletions of the same size. b) generally have minimal phenotypic effects. c) are visible in the polytene chromosomes of Drosophila. d) only cause unbalanced gametes when crossing over occurs in the duplicated region. e) all of the above except d). 20) The Bar eye phenotype in Drosophila: a) is caused by a tandem duplication of region 16A on the X chromosome. b) can lead to the Double-Bar phenotype as a result of recombination after mistaken pairing in meiosis. c) can appear in both males and females. d) all of the above 21) Crossovers within inversion loops: a) cause the production of unbalanced chromosomes in half the resultant gametes. b) lead to the formation of acentric fragments if the inversion is paracentric. c) cause the apparent map distances between genes to decrease. d) all of the above 22) Organisms which are heterozygous for an inversion: a) always produce 50% defective gametes. b) only produce defective gametes if there is a crossover. c) produce defective gametes 50% of the time if there is a crossover within the inversion loop. d) all of the above. 23) Inversions are called crossover suppressors because: a) the inversion loop physically interferes with crossing over in heterozygotes. b) apparent map distances are decreased even in inversion homozygotes. c) inversion heterozygotes are sterile. d) all of the above. 24) Balancer chromosomes: a) have both inversions and translocations. b) must be present in homozygous condition to affect an organism. c) carry a distinctive recessive allele so their presence can be followed. d) completely prevent crossing over with a non-inverted homologue. 25) Translocations usually have only minor effects on phenotype, but exceptions to this general pattern include: a) fusion of genes from the translocated regions, possibly causing cancer as is seen in some forms of leukemia. b) interference with normal disjunction due to damage of the centromere by the exchange between chromosomes.

9 c) duplication of genes adjacent to the translocated chromosomes. d) crossovers which may cause the formation of unbalanced chromosomes.

10 26) During the first meiotic division (meiosis I), the chromosomes in a translocation heterozygote may segregate in three different ways. These are termed: alternate, adjacent-1 and adjacent-2. Which of the following is true? a) Crossovers have no effect on the formation of balanced chromosomes in the gametes. b) All gametes produced after alternate segregation have unbalanced chromosomes. c) About 3/4 of gametes produced will have unbalanced chromosomes. d) Adjacent-1 segegation produces more seriously imbalanced chromosomes than does adjacent-2 segregation. 27) Semisterility due to translocations: a) arises from alternate segregation. b) means that one gender (either male or female) becomes sterile. c) is not observed in translocation homozygotes. d) is associated with high frequencies of nondisjunction. 28) Semisterility due to translocations: a) does not occur in animals. b) is most easily seen after self-fertilization. c) means that half the population cannot reproduce. d) can only be detected if crossing over does not occur. e) none of the above 29) In corn, the wild type leaf color is green and is controlled by a gene that displays complete dominance. Plants homozygous for the recessive allele (yg) have yellow-green leaves. Consider the following two crosses. 1. A semisterile strain of corn homozygous for the yg + allele is crossed with a normally fertile strain that is homozygous for the yg allele. 2. The semisterile F1 from the first cross are test crossed to a plant with normal fertility and yellow-green leaves. Which of the following is true for the offspring of this second cross? a) If the yg gene is not on either translocated chromosome, then there will be two different phenotypes in equal proportions. b) If the yg gene is on one of the translocated chromosomes, then in the absence of crossing over there will be four different phenotypes in equal proportions. c) If the yg gene is on a translocated chromosome, crossing over between the yg gene and the translocation break point will result in complete sterility for all plants that have normal leaf color. d) If the yg gene is on a translocated chromosome, then offspring with normal fertility will have the dominant green leaf color only if there is a crossover between the yg gene and the translocation break point. 30) Transposable elements: a) are small portions of DNA that move from one part of the genome to another.

11 b) contain genes for inactivating the neighboring genetic information. c) are only found in simple organisms and not higher species such as corn and fruit flies. d) are usually lethal when heterozygous. Answers to questions from exam 2 in fall, a 2. a 3. a 4. e 5. e 6. d 7. d 8. b 9. d 10. a 11. e 12. a 13. e 14. a 15. b 16. b 17. c 18. a 19. e 20. d 21. d 22. c 23. a 24. d 25. a 26. a 27. c 28. e 29. d 30. a 31. Which type of chromosomal rearrangement is more likely to cause severe phenotypic effects: deletions or duplications? Explain. Both deletions and duplications cause imbalance with the rest of the genome and therefore have the potential to have phenotypic effects. However it is more likely that deletions will have the more serious effect because deletions are more imbalanced than

12 duplications are when compared to the original chromosomes with neither rearrangement. Specific examples of phenotypic effects were not asked for and are not part of this question. Distinction between homozygotes and heterozygotes was also not part of this question. In order to receive full credit, it was necessary to address the concept of balance as the primary topic of concern. Simply saying that deletions were more harmful without addressing balance was worth only 1 point. 32. In mammals, trisomy for sex chromosomes causes fewer phenotypic effects than does trisomy for autosomes. Explain. Mammals who are trisomic for sex chromosome will inactivate their X chromosomes until only one active one remains. As a result these trisomic individuals will be in normal balance, which is one active X and two copies of each autosome. This will be true for males as well as females. There is no such inactivation mechanism for autosomes. Therefore, an individual who is trisomic for an autosome will express the genes on all three copies of the chromosome. This creates a severe imbalance of gene expression, event for the smallest chromosome. 33. During the first meiotic division (meiosis I), the chromosomes in a translocation heterozygote may segregate in three different ways. These are termed: alternate, adjacent-1 and adjacent-2. Describe alternate and adjacent-1 segregation. Diagrams may be helpful in your answer. The descriptions or diagrams are directly from the book. QUESTIONS FROM EXAM 3, FALL ) Chromosomes in bacterial cells: a) are covalently closed double-stranded circles for most species. b) are usually single-stranded and become double-stranded at cell division. c) undergo frequent translocations with each other. d) segregate via mitosis in those cells with multiple chromosomes. 2) The replication of circular bacterial chromosomes: a) may begin anywhere, since they are circular. b) proceeds in two directions at the same time from a common starting point. c) requires the circular chromosomes to become linear before replication. d) leads to fewer copies of genes near the origin in rapidly growing cells.

13 3) Most bacterial phenotypes studied in genetics experiments: a) are detected by the ability to form colonies on carefully defined media. b) relate to antibiotic resistance and sensitivity because they are so easy to see. c) require bacteria to be prototrophic so they can grow on defined medium. d) change rapidly since they are not passed on to the daughter cells after division. 4) An auxotrophic bacterium: a) is one which uses carbon dioxide from the atmosphere for growth. b) could not survive in nature due to its defects. c) has lost the ability to synthesize some essential molecule such as an amino acid. d) can no longer metabolize glucose for its energy needs. 5) Conjugation in E. coli: a) can occur between any two members of the species. b) does not occur during interrupted mating experiments. c) may transfer chromosomal genes without transferring any part of the F factor. d) does not change the F factor content or arrangement of the donor cell. 6) Conjugation: a) is gene transfer between bacterial cells controlled by genes on a plasmid. b) refers to reciprocal (both direction) genetic exchange in bacteria. c) only happens when chromosomes are sorted properly, as in mitosis. d) requires the recipient cell to have an F pilus. 7) Insertion Sequences (IS): a) allow integration of the F factor into the cellular chromosome. b) cannot be the same on plasmids and on chromosomes in the same cell. c) are deleted when Hfr cells revert back to the F + configuration. d) can lead to the formation of linear chromosomes if two of them pair and them recombine. 8) When Hfr strains are mated with F - strains, it is common to use a recipient strain which has a nutritional defect, such as the inability to make the amino acid threonine. Why? a) F - cells that have not received DNA will not grow. b) Hfr cells do not have this defect and they can be detected after transfer. c) The nutritional defect enhances the killing of Hfr by antibiotics.. d) This will prevent all auxotrophs from growing and confusing the results. 9) When chromosomal DNA is transferred from an Hfr cell to an F - cell during conjugation, it must become a permanent part of the recipient chromosome: a) or else it will not be possible to detect the growth of the exconjugants. b) through homologous alignment and recombination. c) otherwise the linear DNA from the donor will be degraded. d) all of the above

14 Answers to exam 3 questions 1. a 2. b 3. a 4. c 5. d 6. a 7. a 8. a 9. d 31. Describe three ways in which genetic transfer by bacteria during conjugation is different from sex in eucaryotes. There are three primary differences: a) Bacterial conjugation involves a small percentage of the genome, while eucaryotic sex involves the entire genome. b) Eucaryotic sex creates an entirely new individual, whereas bacterial conjugation does not. c) The parents in eucaryotic sex are not changed as a result of the reproduction. In bacterial conjugation, the recipient cell has an altered genome. Variations on these points were accepted. What was not accepted was comments on the physical process (pilus, etc).