Practice Genetics Problems Answer Key

ANSWER KEY - Practice Genetics Problems Instructions: Try and complete the following practice problems without looking at your notes. When you’re finished, check your answers with the key posted on Canvas and review any questions you got wrong. If you are having difficulty completing these practice problems, please visit an instructor or LA office hour for guidance and assistance! Tutorials: Mendel and Modern Genetics & Predicting Phenotypes and Genotypes 1. In humans, albinism is caused by a mutation at a single gene. A homozygous recessive individual (aa) will not have skin pigmentation. Ann is albino. Her husband Mike is not albino. They have two children, Sarah (who has skin pigmentation) and Martha (who is albino). a. Write out the genotypes of Ann, Mike, Sarah and Martha. Ann – aa, Mike – Aa, Sarah – Aa, Martha - aa b. What is the probability that their next child will be an albino? Ann produces only aa gametes, while Mike produces 1/2 A and 1/2 a gametes, so the probability that their next child will be an albino is 1 X 1/2 = 1/2 c. If Ann remarries and has five more children, all with normal pigmentation, what is most likely the genotype of her new husband? Why? He is most likely AA, and all of their children are Aa, so would have normal pigment. If he were Aa, half of their children would be expected to be aa. d. If Ann remarries a man who is heterozygous for albinism, what is the probability that her second family will have three children that are pigmented and one child that is albino? For this problem, you use the rules of multiplication and addition. Ann is genotype aa and her husband is Aa. The probability of having a child with normal pigment (Aa) is 1/2 and of having an albino child (aa) is 1/2. There four possible birth orders that would give you a family with three pigmented children and one albino, each has the same probability: PPPA = 1/2 X 1/2 X 1/2 X 1/2 = 1/16 PPAP = 1/16 PAPP = 1/16 APPP = 1/16 Add these together – 1/16 + 1/16 + 1/16 + 1/16 = 4/16 or 1/4 probability 1 2. Given two separate, independently assorting genes, A and B, with A completely dominant to a and B completely dominant to b, answer the following questions: a. What is the probability of getting an AB gamete from an individual who is heterozygous at both genes? The genotype of the individual is AaBb – the probability of producing a gamete with the A allele is 1/2, a gamete with a allele = 1/2, B = 1/2 , b = ½ The probability of producing an AB gamete is then the product of producing an A gamete and a B gamete, so 1/2 X 1/2 = 1/4 b. What is the probability of getting an Ab gamete from an individual who is heterozygous at both genes? Using the same information in 4a, the probability is 1/2 X 1/2 = 1/4 c. What is the probability of getting an AB gamete from an individual who is AABb? This individual can only produce A gametes, so the probability is 1 X 1/2 = 1/2 d. What is the probability of getting an AABB child from a cross of two people heterozygous at both genes? This requires the information in 4a. The probability an AaBb individual producing a AB gamete is 1/4. The probability of two AB gametes undergoing fertilization is then 1/4 X 1/4 = 1/16. e. From a cross of two people heterozygous at both genes, what is the probability of having a child that expresses the dominant phenotype for both genes? For the dominant phenotype, since A and B are both inherited as a complete dominant, the genotypes AA and Aa, and BB and Bb will have the same phenotype. Therefore, you need to determine the probability of a child having any genotype that results in A and B phenotypes. If you do the cross AaBb X AaBb, the probability of producing a child with at least on A allele is the probability of a AA plus the probability of Aa, so 1/4 + 1/2 = 3/4, and the same is true for the B allele. So, the probability of a child with an A and B phenotype is 3/4 X 3/4 = 9/16. If this is confusing, go back and review the figure for the dihybrid cross in Tutorial 27. 3. If a plant of genotype AaBbCc is allowed to self-fertilize and the three genes (all inherited as a complete dominant) assort independently, how many different genotypes will be found among the offspring? To solve this, consider each gene separately. If you look at Aa X Aa, this is a monohybrid cross of two heterozygotes, so the result of this cross is three genotypes, AA, Aa and aa. The same is true of the remaining two genes. 2 For each gene, the number of different possible genotypes is 3, so for three genes: 3 X 3 X 3 = 27 4. If 2 dogs of genotype KkLlMmNn are repeatedly mated, and the four genes (all inherited as a complete dominant) assort independently, how many different phenotypes will be found in the offspring? This is similar to question 5, but now you are considering phenotypes rather than genotypes. If you look a Kk X Kk, this is a monohybrid cross of two heterozygotes, and the result of this cross is two phenotypes, KK/Kk and kk. The same is true of the remaining three genes. For each gene, the number of possible phenotypes is 2, so the number of possible combinations with four genes is: 2 X 2 X 2 X 2 = 16 5. In a cross between a female AaBbccDdee and a male AabbCcDdee, what proportion of the offspring will be phenotypically identical to the female parent? (Assume independent assortment of all genes and complete dominance). The female has a dominant phenotype for genes A, B and D, and a recessive phenotype for c and e. You need to consider each gene separately. For gene A, a cross of Aa X Aa, the probability of having a dominant phenotype is 3/4 (see the answer to Question 4e). For gene B, a cross of Bb X bb, the probability of having a dominant phenotype is ½ For gene C, a cross of cc X Cc , the probability of having a recessive phenotype is ½ For gene D, a cross of Dd X Dd, the probability of having a dominant phenotype is 3/4 (same as for gene A) For gene E, a cross of ee X ee, the probability of having a recessive phenotype is 1 You then multiply these probabilities: 3/4 X 1/2 X 1/2 X 3/4 X 1 = 9/64 Tutorials: The Complex Expression Patterns of Multiple Alleles and Pedigree Analysis 6. What type of dominance best explains a phenotypic ratio of 1:2:1 in the F2 generation of a monohybrid cross for a character with two traits (two alleles)? If you have two answers to this question, how would you determine which type of dominance relationship exists between the two alleles? You would see this phenotypic ratio for both incomplete dominance and codominance. To determine which type of dominance relationship exists, you would look at the heterozygote phenotype. If both alleles are expressed, it is codominance. If you see a phenotype intermediate between the two alleles, it is incomplete dominance. 3 7. In humans there are many blood group genes – you have been introduced to two of them the MN blood group and the ABO blood group. Another blood group gene is the Bombay phenotype gene, which shows a complete dominant pattern of inheritance. Individuals who are homozygous recessive at this gene do not express (produce) the H antigen, the antigen that is modified by the gene that encodes the ABO blood group, but instead have only the h antigen. Within the ABO blood group, the i allele has an unmodified H antigen, while the IA and IB alleles further modify the H antigen with additional sugars (see the figure). Figure for Question 9 a. What type of modified gene expression is this interaction an example of? Why? It is an example of epistasis, because the genotype at the Bombay (H) gene determines the expression of the ABO gene. b. If an individual is genotype hhIAIB, what is their blood type? It is Bombay and they do not produce an H antigen, so their AB blood type is not expressed (although if a standard blood test is performed the result will indicate that they have Type O blood because a standard blood test tests for the presence of A and/or B antigens, not the H antigen). c. What would be the blood type of an individual with the genotype HhIAIB? Because they have one copy of the H allele, they will produce H antigen so their blood type would be AB. d. Would there be any difference between an individual with the genotype hhii and the genotype HHii? Why or why not? Yes, there is a difference. While both have the genotype ii, which produced bloodtype O, individuals with hh don’t produce H antigen, so they would not have bloodtype O. 4 e. What does having the genotype hh mean if the person needs a blood transfusion? They must get blood from another hh individual who does not have any H antigen. f. If a man who has the genotype hhIAi has children with a woman who has the genotype HhIBi, which genotypes and blood type phenotypes would you see in their children? hhIAIB, hhIAi, hhIBi, hhii – All of these individuals would have the Bombay blood type, regardless of their blood type alleles for the ABO blood group gene HhIAIB – Type AB HhIAi – Type A HhIBi – Type B Hhii – Type O 8. This pedigree shows the pattern of inheritance for a particular trait in a family. You will need to examine this carefully to determine the mode of inheritance of this trait; only affected individuals are indicated so if there are carriers that information is not available. Table for Question 10 a) What is the genotype of individual I-1? How do you know this? This individual must be bb. He does not have any affected children, so this cannot be inherited as a dominant genetic disorder. b) What is the genotype of individual I-3? How do you know this? This individual must be Bb, a carrier, because he has one affected child. 5 c) Do the individuals II-8 and II-9 have the genotype? Why or why not? Yes, they must both carriers (Bb) because they have an affected child but neither has the disorder. 9. This pedigree shows the pattern of inheritance for a particular trait in a family. Figure for Question 11 a) Using this information, what is the mode of inheritance? Why? This trait is inherited as a sex linked recessive. Males are affected much more frequently than females, while females who have the allele are almost always carriers. b) What is the genotype of the affected male in Generation I? He would be hemizygous, do would have d allele and a Y chromosome, dY genotype. He passes that allele to all of his daughters. c) What is the genotype of the female in Generation I? She is most likely DD because she does not have any sons with the trait. d) What is the genotype of the affected female in Generation IV? Why is she the only affected female in this pedigree? She is dd. To be affected, a woman must have an affected father and a carrier (or affected) mother, which describes the genotypes of her parents but not the parents of any other females in the pedigree except her sister. If her mother was affected instead of a carrier, what would her sister’s genotype be? 6 Tutorial: Chromosome Behavior and Sex Chromosomes 10. In humans, sex is determined by chromosome complement. Which of these individuals will not be a phenotypic male? Why? a. XYY b. XX c. XY d. XXY 11. A man with colorblindness will pass the allele for the disease to only some of his offspring. Which of his offspring will receive this allele from their father? Only his daughters, because his sons will receive his Y chromosome. 12. A man who does not have hemophilia and a woman who has hemophilia have two children with hemophilia. What is the probability that their next son will have hemophilia? What is the probability that their next daughter will have hemophilia? What do you know about the gender of their first two children? The genotype of the man would be HY and the woman would be hh. The probability of having a son with hemophilia is 100% - that is the only allele that the mother can contribute. None of their daughters will have hemophilia because they get the H allele from their father, but all will be carriers. Since their first two children have hemophilia, they must both be males. Tutorial: Gene Linkage Make sure you understand what’s being asked in each of the following questions. Do not simply memorize mathematical patterns! Use the information below to answer the next two questions: The recessive mutations (cut wings) and (vermillion eyes) identify two autosomal genes on the second chromosome of . When females heterozygous for these genes were crossed with cut winged, vermillion eyed males, the following classes and numbers of progeny (out of 1000) were obtained: wildtype wings, wildtype eyes 60 cut wings, wildtype eyes 427 wildtype wings, vermillion eyes 419 cut wings, vermillion eyes 94 7 13. Based upon these results, the map distance between the and genes is estimated to be: a. 6.0 map units b. 42.7 map units c. 15.4 map units d. greater than 50 map units because all four classes of offspring were observed 14. Which homozygous parent stocks could be used to produce the heterozygous females used in the cross described above (circle one)? a. wildtype wings, wildtype eyes X cut wings, vermillion eyes b. wildtype wings, vermillion eyes X cut wings, wildtype eyes c. wildtype wings, wildtype eyes X wildtype wings, wildtype eyes d. cut wings, vermillion eyes X cut wings, vermillion eye 15. In the following cross (P generation) the genes are inherited independently (not linked) except for genes G and H, which are completely linked and show no recombination: EEffggHH x eeFFGGhh a. What would be the genotype of their offspring (F1)? EeFfGgHh b. What type of gametes (genotypes) could their offspring produce? How many types of gametes would you expect if all four genes assorted independently? EFgH, EFGh, EfgH, EfGh, eFgH, eFGh, efgH, efGh If all underwent independent assortment, you would expect 24 = 16 types of gametes. Since G and H are completely linked, you can treat them as one gene - 23 = 8 types of gametes. c. What is the probability that they will have an EeFFggHH offspring? Prob of Ee = ½ Prob of FF = ¼ Prob of gg = ¼ Prob of HH = 1 because g and H are completely linked and are inherited as a unit. Overall probability = 1/32 d. What is the probability that they will have an eeffgghh offspring? Prob = 0 because gghh cannot happen 8 16. The recessive mutations (red flower color) and (round pollen) identify two linked genes in (sweet pea). The normal phenotypes are a purple colored flowers ( ) and long pollen ( ). When a plant heterozygous for these alleles at both genes was crossed with a plant that had red flowers and round pollen, the following classes and numbers of progeny (out of 1000) were obtained: Red flower, round pollen 444 Purple flower, round pollen 59 Purple flower, long pollen 440 Red flower, long pollen 57 The autosomal recessive mutation gives yellow green seedlings rather than the normal green seedlings ( ). You calculate that the gene is 22 map units away from the gene and 34 units away from the gene. What is the most likely order of these three genes? a. b. c. d. None of the above give the correct order of these three genes. 17. A wild-type allele is: a. always beneficial to the organism. b. expressed in the heterozygous state. c. expressed only in the homozygous state. d. the most common allele for a gene in a population. e. always completely dominant. 18. In the following cross (P generation) the genes are inherited independently (not linked) except for genes C and D, which are completely linked and show no recombination: AAbbccDDeeFF x aaBBCCddEEff In the F2 generation, what proportion of the progeny will be: A. True breeding? A½ B½ CD ½ - you have to treat this as a single gene because they don’t undergo crossing over. Think of the arrangement of the genes on the chromosomes - c D and C d. So, ½ of the time you will get offspring that get two cD chromosomes or two Cd chromosomes, so would be homozygous for both (either ccDD or CCdd) 9 E½ F½ B. Heterozygous for all loci? The same answer as #1. For the C and D genes, you have ½ probability that the offspring will get one cD chromosome and one Cd chromosome, so would be heterozygous for both genes (CcDd). C. Homozygous recessive for all loci? Zero. It is not possible for an F2 to be either ccdd or CCDD. 10