Biology 201 (Genetics) Exam #1 21 September 2004

Name KEY Biology 201 (Genetics) Exam #1 21 September 2004 Read the question carefully before answering. Think before you write. Be concise. You will have up to 85 minutes hour to take this exam. After that, you MUST stop no matter where you are in the exam. If I can not read your handwriting, I will count the question wrong. Sign the honor pledge if applicable. Good luck! I pledge that I have neither given nor received unauthorized assistance during the completion of this work. Signature: 1

6 pts. 1. Albinism (lack of skin pigment) was thought to be caused solely by one recessive mutation in the gene encoding tyrosinase. However, a study from 1952 reported that two albino parents produced three normally pigmented children. How would you explain this phenomenon at a genetic level. In your answer, make sure you indicate how pigment formation is inherited based on the data from the study cited above. (Although it may be formally possible, give me an answer other than the albino male was not the biological father of the children!). The best answer is that this is an example of two genes controlling skin color and duplicative recessive epistasis. Gene A is for the tyrosinase and Gene B is for another gene required for pigment formation. Recessive mutations in either gene cause albinism. Parent 1 was aabb and parent 2 was AAbb. Thus, children were AaBb (non-albino). I also accepted that this was an example of a mutation distinct from the earlier described recessive mutation in the tyrosinase gene. In this case, the mutation (A D ) is behaving in a dominant fashion to the wildtype allele A. Both parents are AAD and so the kids can be AA, although the chance of this occurring is (1/4) 3. 6 pts. 2. Explain how one would determine whether an organism expressing a dominant trait is heterozygous or homozygous? You would do a test cross. Cross the individual of unknown genotype with an individual expressing the recessive phenotype. If the progeny all have the dominant phenotype, than the individual of unknown genotype was homozygous dominant. If the ½ of the progeny are have the dominant phenotype and ½ of the progeny are have the recessive phenotype, than the individual of unknown genotype was heterozygous. 8 pts. Imagine an individual with the genotype AaDdgg, where each of these genes is on a different chromosome. a. How many different gametes types can be formed by this individual? 4 b. What are the gametes? ADG, Adg, adg, adg c. Now imagine an individual with the genotype AaDdggRr, where each of these genes is on a different chromosome. How many different gametes types can be formed by this individual? 8 2

9 pts. 3. Consider three independently assorting gene pairs, which control three different characters. A is dominant to a, B is dominant to b, and D is dominant to d. For a cross between parents that are AaBbDD and AABbDd, write the answer to the questions below in the blank to the left of the question. 1/8 a. what is the probability of obtaining an offspring that is AABbDd? 1/16 b. what is the probability of obtaining an offspring that is purebreeding for all three dominant traits? 3/4 c. what is the probability of obtaining an offspring that will express all three dominant traits? 3

8 pts. 4. From a highly watched soap opera: Monica had a baby and there was some question as to whether the father of the child was Monica's husband (Alan) or Monica s fellow doctor friend (Rick). Monica's blood type is A, Alan's is AB, and Rick's is O. The baby s blood type is O. The potential pedigree is shown below, along with the blood types of each individual. Based on what you have learned about blood group genetics, answer the questions below. As a reminder: The I gene controls which sugar gets added to H substance on the red blood cell surface. There are three alleles of the I gene (I A, I B, and i) - where I A and I B are codominant, I A is dominant to i (O blood), and I B is dominant to i (O blood). The H allele (which allows H substance to be made) is dominant to h allele (which is defective in making h substance). The h allele is epistatic to the alleles at the I locus a) Could Alan be the father? Why or why not? Yes. If Alan is heterozygous for the H substance gene (Hh) and Monica is heterozygous for the H substance gene (Hh), they would have a ¼ chance of producing a child that was hh. Since the recessive h allele is epistatic to the I gene, the genotype at the I locus is now irrelevant. The child would not have an h substance and thus would appear to have O blood, since the sugars controlled by the I gene could not be added to the surface of the red blood cell. b) Could Rick be the father? Why or why not? Yes. The most likely answer is that Rick is ii and Monica is I A i. Thus, they have a ¼ chance of having a child that is ii (blood group O). The answer from (a) could also have occurred, but it is less likely than the genotypes given for (b). 4

9 pts. 5. Normal hedgehogs have straight fur and long whiskers. Mutant hedgehogs have been isolated with either kinked fur or short whiskers. The genes for each of these characters are on different chromosomes. You capture some hedgehogs and mate them and get the following data. mating genotypes straight fur and long whiskers Straight, short AaBb straight fur and short whiskers Number of progeny kinked fur and long whiskers kinked fur and short whiskers X Straight, short Kinked, long X Straight, short Kinked, short X straight, short AaBb aabb AaBb aabb AABb 30 90 10 30 40 40 40 40 40 120 0 0 a) For each mutation, indicate whether it is dominant or recessive by filling in the blanks below with the word dominant or recessive. The allele for kinked fur is RECESSIVE to the allele for straight fur. The allele for short whiskers is DOMINANT to the allele for long whiskers. b) Provide a legend for each trait (using A/a for fur type and B/b for whisker type). A = straight a = kink B = short b = long c) In the table above in the column labeled genotype, write the genotypes of the parents for each cross. 5

8 pts. 6. A purebreeding variety of opium poppy having leaves that are irregularly shaped (lacerate) was crossed with a variety that has normal leaves. All the F 1 had normal leaves. Two F 1 plants were interbred to produce the F 2. Of the F 2, 145 had normal leaves and 9 had lacerate leaves. These crosses are shown in the figure below. Figure 1: P generation lacerate aabb X normal AABB F1 generation Normal leaves AaBb Cross F1 plants F2 generation 145 normal leaves A-B-; A-bb; aab- 9 lacerate leaves aabb a) Explain how inheritance of lacerate leaves is determined in the opium poppy? There are two genes that control leaf shape and epistasis is occurring since the F2 atio is in 16ths (15:1). Specifically, a dominant allele in either of two genes is sufficient to get normal shaped leaves. Perhaps the genes encode proteins of similar function, so having the dominant form of either allele is sufficient to get normal leaves. In other words, for lacerate leaves to appear, both genes must be present in the recessive form. b) Give the possible genotypes that correspond to each phenotype for the plants in the P, F 1, and F 2 generations by putting the genotypes in the blanks in the figure above. You can use a dash to indicate either of two alleles is possible for the same phenotype (ie A- = AA or Aa). 6

8 pts. 7. In the pedigrees below (they are all the same), a trait is determined by a single gene and affected individuals are shaded. State whether is would be possible for the trait to be inherited in each of the following ways by writing YES or NO in the blank to the left of each inheritance pattern. a. autosomal recessive YES NO b. X-linked recessive NO c. autosomal dominant NO d. X-linked dominant 7

8 pts. 8. Two genes are located on separate autosomes. One gene pair determines hair color in foxes where AA is lethal before birth, Aa is platinum hair, and aa is silver hair. A second gene pair (B/b) determines hair length where short hair is dominant to long hair. A platinum fox that was pure breeding for short hair length is crossed with a silver fox with long hair. The F1 contained foxes that were platinum and others that were silver. Two of the platinum F1 foxes were mated. Give the phenotypes and expected phenotypic ratios for the foxes in the F2 generation using either a branch diagram or statistics to do the problem. 3/4 short 6/12 platinum, short 2/3 platinum 1/4 short 2/12 platinum, long 3/4 short 3/12 silver, short 1/3 silver 1/4 short 1/12 silver, long Note that the problem asks for ratios of the F2 generation. Since the AA progeny die before birth, they are not counted in the frequencies. 6 pts. 9. Below are chromosomes in prophase of mitosis in an organisms with a diploid number of 4. a. Label the homologous chromosomes by indicating which belong in the same pair. b. Label the sister chromatids. c. How many individual molecules of DNA are present in the cell at prophase of mitosis? 8 homologous chromosomes sister chromatids homologous chromosomes 8

C or E A D E D E Multiple choice section: (24 points total 4 points per question) Write your answer in the blank provided to the left. IF you want to explain your answer, you can do so next to the question 1. Imagine a cell with a diploid number of 6. If one of the chromosomes has a centromere that is absent, the effect on mitosis would be: a. There would be no effect because centromeres are only important in meiosis. b. One daughter cell has 8 chromosomes and the other daughter cell has 4 chromosomes. c. One daughter cell has 7 chromosomes and the other daughter cell has 5 chromosomes. d. One daughter cell has 12 chromosomes and the other daughter cell has 0 chromosomes. e. None of the above (if you consider mitosis would probably arrest at cell cycle checkpoint). 2. The following have entirely identical DNA sequences: a. Sister chromatids b. Homologous chromosomes c. Chromatin d. All X chromosomes e. A and B 3. Which of the types of chromosomes below are found in all human diploid cells? a. Sex chromosomes b. Autosomes c. Centrosomic chromosomes d. Both a and b e. All of the above 4. A baby was karotyped as 47, XX, +21. This genotype could have arisen as a result of ONE nondisjunction event: a. in the egg during meiosis II before fertilization. b. in the sperm during meiosis II before fertilization. c. in the egg during meiosis I before fertilization. d. either a or c. e. either a or b or c. 5. Which of Mendel s Laws is NOT exemplified in a monohybrid cross a. Unit factors occur in pairs b. One unit factor in a pair can be dominant to another c. Law of Equal Segragation d. Law of Independent Assortment e. None of the above (ie all of the above do apply in a monohybrid cross). 6. The most fundamental feature of that two members of a homologous chromosome pair share is that they have a. the same length. b. the same banding pattern. c. the same centromere position. d. alleles of the same genes at the same loci, where the chromosomes are genetically identical. e. alleles of the same genes at the same loci, where the chromosomes are genetically similar but not necessarily genetically identical. 9