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Unformatted text preview: © Prep101 www.prep101.com/freestuff Solutions to HMB265 Practice Midterm 1. In a diploid cell in the metaphase of mitosis there are 100 chromatids. The haploid number of chromosomes for this organism would be: a) b) c) d) e) 200 100 50 25 cannot accurately decide based on data given Solution: Since there are 2 sister chromatid for every chromosome, there would be 50 chromosomes in this cell. And, given that it is a diploid cell, there are two sets of chromosomes, or homologues, thus if 2n=50, n=25. 2. Which of the following is NOT a feature of chromosomes? a) b) c) d) Homologous chromosomes carry the same set of genes Metacentric chromosomes are comprised of two sister chromatids Nonhomologous chromosomes have unrelated sets of genes The definition of karyotype is the entire chromosome complement of an individual organism or cell, as seen during metaphase e) All of the above are correct 3. There are two types of variation. These are continuous and discontinuous variation. Which of the following statements is FALSE? a) Weight is an example of continuous variation b) Continuous variation produces a spread of variation within a population c) Discontinuous variation is often reflective of a 1:1 relationship between genotype:phenotype d) An example of discontinuous variation is height e) All of the above are correct Solution: Discontinuous variation is variation having distinct classes of phenotypes for a particular character. Examples of discontinuous variation are blood type or eye colour. Page 1 of 22 © Prep101 www.prep101.com/freestuff 4. During what stage of mitosis are centromeric connections severed? a) b) c) d) e) anaphase metaphase prometaphase prophase telophase Solution: Anaphase is characterized by centromeres dividing, 2 kinetochores splitting and sister chromatid (now daughter chromosomes) moving to opposite poles. 5. During what stage of mitosis does the nuclear membrane reform? a) b) c) d) e) anaphase metaphase prometaphase prophase telophase Solution: Telophase is characterized by the nuclear envelope reforming, spindle fibers disappearing, chromosomes de-condensing and nucleoli reappearing. 6. Arabidopsis thaliana is a model organism studied by many of the world’s plant biologists. It has a diploid chromosome number of eight. How many chromosomes are present in the cell nucleus at the beginning of meiosis II? a) b) c) d) e) 4 16 8 2 32 Solution: Meiosis (2N to N) takes place in 2 successive nuclear divisions called Meiosis I (reductional division) and Meiosis II (equational division) in special cells called meiocytes (germ cells). At the end of meiosis I, homologous chromosomes have split and thus, the cell now has a haploid number of chromosomes that are still comprised of two sister chromatid, which are divided at the end of Meiosis II. Page 2 of 22 © Prep101 www.prep101.com/freestuff 7. The following is a list of mitotic events. What is the correct order? A. chromosomes align on the midplate of the cell B. kinetochores being attaching to spindle fibers C. nuclear membrane reforms, chromosomes, decondense D. chromosomes condense, centrosomes migrate to the opposite poles E. pairs of sister chromatids separate, one of a pair moves to each pole a) b) c) d) e) BDACE DABEC DBAEC ABDCE EDBAC Solution: The order reflects the order of mitotic events corresponding to the details of that particular stage as outlined in the following chart. Phases of Mitosis Prophase Metaphase Prometaphase Metaphase Anaphase Telophase Cytokinesis Important details chromatin condense (n # of chromosomes, but 2n chromatid) and become visible-even though chromosomes have replicated, because they are still attached by the centromere, the number of chromosomes has NOT increased! sister chromatid attach at centromere spindle apparatus forms outside of nucleus centrosomes form the spindle poles nucleoli disappear nuclear envelope breaks down (vesiculates) microtubules attach to kinetochore polar microtubules span centrosomes astral microtubules extend from centrosomes sister chromatid attach to opposite poles chromosomes align along the “equator” sister chromatid face OPPOSITE poles-this is an important difference from meiosis!! centromeres divide; 2 kinetochores split sister chromatid (now daughter chromosomes) move to opposite poles nuclear envelope reforms spindle fibers disappear chromosomes de-condense nucleoli reappear cytoplasm divides at cleavage furrow via a contractile ring produce 2 equal daughter cells Page 3 of 22 © Prep101 www.prep101.com/freestuff 8. At the beginning of meiosis II a cell has 36 chromatid, how many chromosomes will each cell have at the anaphase II? a) 36 b) 72 c) 18 d) 9 e) 16 Solution: At the beginning of meiosis II, homologous chromosomes have already split, thus, there are only the haploid number of chromosomes remaining which are each comprised of two sister chromatid. Although sister chromatid divides and move to the opposite poles during anaphase II. Thus, each chromatid is considered a separate chromosome and so the number of chromosomes at the end of anaphase II equals the number of chromosomes at the beginning of meiosis II. 9. Before the S phase of mitosis, a diploid cell has 20 chromosomes, how many chromosomes will it have in late anaphase? a) 20 b) 40 c) 10 d) 80 e) 30 Solution: In late anaphase centromeres divide; 2 kinetochores split, sister chromatid (now daughter chromosomes) move to opposite poles. Thus, the number doubles from before mitosis to the end as the 2 daughter cells need to have the exact chromosome # as the parent cell. 10. How many chromosomes does a diploid organism have at the end of anaphase I, if n represents the haploid number of chromosomes? a) n/2 b) n c) 2n d) 4n e) 8n Solution: Because cells go through the S phase prior to meiosis, a diploid cell would have 2n chromosomes, but 4n chromatid. Because homologous chromosomes split in meiosis I, the chromosome number is the same as the original 2n number. Page 4 of 22 © Prep101 www.prep101.com/freestuff 11. To identify the genotype of yellow-seeded pea plants as either homozygous dominant (YY) or heterozygous (Yy), you could do a testcross with plants of genotype _______. a) b) c) d) e) y Y yy YY Yy Solution: By definition, a testcross is a cross to an animal that is homozygous recessive for the query genes. 12. A pea plant is heterozygous for both seed shape and seed color. S is the allele for the dominant, spherical shape characteristic; s is the allele for the recessive, dented shape characteristic. Y is the allele for the dominant, yellow color characteristic; y is the allele for the recessive, green color characteristic. What will be the distribution of these two alleles in this plant's gametes assuming that these genes are on different chromosomes? a) 50% of gametes are Sy; 50% of gametes are sY b) 25% of gametes are SY; 25% of gametes are Sy; 25% of gametes are sY; 25% of gametes are sy. c) 50% of gametes are sy; 50% of gametes are SY d) 100% of the gametes are SsYy e) 50% of gametes are SsYy; 50% of gametes are SSYY. Solution: Because of independent assortment, the dominant alleles of the different genes on different chromosomes have an equal chance of sorting together or with the recessive alleles and likewise for the recessive alleles. Mendel’s Second Law: Independent Assortment Genes on different chromosomes segregate independently of each other. So, an equal chance of SY, Sy, sY, sy is 1:1:1:1 or a 25% (equal) chance of any of the 4 combinations. Page 5 of 22 © Prep101 www.prep101.com/freestuff 13. Which of the following genetic crosses would be predicted to give a phenotypic ratio of 9:3:3:1? a) b) c) d) e) SSYY x ssyy SsYY x SSYy SsYy x SsYy SSyy x ssYY ssYY x ssyy Solution: The 9:3:3:1 ratio is unique to a dihybrid cross (two heterozygotes mating). 14. In a dihybrid cross, what fraction of the offspring will be homozygous for both recessive traits? a) b) c) d) e) 1/16 1/8 3/16 1/4 ¾ Solution: Homozygous recessive offspring represent the “1” in the 9:3:3:1 dihybrid cross outcome. Thus, 1/(9+3+3+1=16). 15. In a trihybrid cross, how many different phenotypes would you observe in the resulting offspring? a) b) c) d) e) 3 8 9 12 27 Solution: Using the branching method, there are two possible phenotypes for any of the three genes = 23 = 8. Page 6 of 22 © Prep101 www.prep101.com/freestuff 16. For the following cross, A/a; B/B; c/c; D/d; E/e x A/a; b/b; C/c; D/D; e/e what would be the chance of having a completely heterozygous offspring with the genotype: A/a; B/b; C/c; D/d; E/e? a) b) c) d) e) 1/2 ¼ 1/8 1/16 1/32 Solution: Using the product rule, there is a ½ chance that the parental gametes will combine to produce Aa x 1/1 for B/b (each parent only has a B or b to give, so all offspring will have B/b) x ½ for C/c; ½ for D/d and ½ for E/e. ½ x 1 x ½ x ½ x ½ = 1/16 17. Hemophilia in humans is recessive and is due to an X-chromosome mutation. What will be the results of mating between a normal (non-carrier) female and a hemophilic male? a) b) c) d) e) half of daughters are normal and half of sons are hemophilic all sons are normal and all daughters are carriers half of sons are normal and half are hemophilic; all daughters are carriers all daughters are normal and all sons are carriers half of daughters are hemophilic and half of daughters are carriers; all sons are normal Solution: Because the father will only contribute the disease and it is on the X chromosome, the father will pass on the affected X chromosome to all of the daughters and none of the sons. Because this is a sex-linked recessive disease, the daughters will only be carriers. Page 7 of 22 © Prep101 www.prep101.com/freestuff 18. What is the inheritance pattern in this pedigree? a) b) c) d) e) Autosomal recessive Autosomal dominant X-linked recessive X-linked dominant Recessive lethal Solution: Because the disease is rare and there is no bias towards one gender, it is an autosomal recessive disease. Consanguineous marriages often results in the manifestation of disease in their offspring due to the increased chance of rare recessive alleles combining. 19. According to the above pedigree, what is the probability that the offspring between individuals 1 and 2 will have the trait? a) b) c) d) e) 1/16 1/8 1/2 0 2/3 Solution: Focus on individual 1. There is a 100% chance that his dad is A/a and so there is ½ chance that he is A/a and ½ chance that he will pass it on to his offspring. The father of individual 2 has the trait and so there is 100% chance that individual 2 is heterozygous and then a ½ chance that if she is a heterozygote, she will pass it on. Multiply these chances together to give you: ½ * ½ * ½ = 1/8 Page 8 of 22 © Prep101 www.prep101.com/freestuff 20. Below is a pedigree of a human genetic disease in which stricken individuals are solid-colored. Apply the laws of probability and calculate the probability the offspring of the cousin marriage 1 x 4 will have the disease. a) 2/3 b) 1/2 c) 1/6 d) 1/4 e) 1/8 Solution: Because the affected individuals had unaffected parents and there is no gender-differences, this disease is an autosomal recessive disease. Cousin 1 has a 100% chance that he is a carrier and ½ chance that he will pass the recessive allele on to his offspring. Cousin 4 has the trait and so there is a 100% chance that she will pass the trait on to her offspring. Thus, there is a ½ chance that the offspring will inherit the disease. 21. A man and a woman are having a baby. Both the man and woman have siblings with sickle-cell anemia, a debilitating blood disease caused by a recessive allele, but neither the man nor the woman nor any of their parents suffer from the disease themselves. What is the probability that the baby will have sickle-cell anemia? a) 1/8 b) 1/16 c) 1/9 d) 2/3 e) 1/3 Solution: Because the siblings of these individuals have the disease, we know that their parents are A/a. A/a x A/a = A/A, A/a, A/a, a/a. Because these individuals do not have the disease, there is a 2/3 chance that they are carriers (eliminate a/a from the possibility and you are left with A/A, A/a, A/a = 2/3 = A/a). If they are A/a, there is a ½ chance Page 9 of 22 © Prep101 www.prep101.com/freestuff that they will pass the disease allele on to their offspring. Thus the chance of having a child with a/a is 2/3 * ½ * 2/3 * ½ = 1/9 22. Linked loci are loci that a) Are on the same chromosome b) Determine sex c) Govern traits (such as hair texture and hair color) that are functionally related d) Have the same alleles residing on them e) Govern traits that have nothing to do with one another 23. How many map units is a recombination frequency of 10% equal to? a) 10% b) 15% c) 10 centimorgans d) 90 anticentimorgans e) Both c and d Solution: 1% recombination = 1 map unit (m.u.) and 1m.u. = 1 centiMorgan (cM). So, 10% rf = 10 cM. 24. The pairwise map distances for four linked genes are as follows: A-B = 9 m.u., B-C = 4 m.u., C-D = 6 m.u., A-C = 5 m.u., A-D= 1 m.u., B-D = 10 m.u. What is the order of these four genes? a) A-B-C-D b) B-D-C-A c) B-C-A-D d) C-A-B-D e) B-C-D-A Solution: Because B and D have the greatest map distance they are the furthest apart. C is closest to B and A is closest to D. So far, you know that the order is B, C, A, D. Check by making sure that these all add up. Page 10 of 22 © Prep101 www.prep101.com/freestuff Use the following information for the next four questions: An experiment was done to determine the linkage relationship between three genes in Drosophila (d, e, and f). Homozygous females phenotypically d,f were crossed with homozygous males phenotypically e. The F1 progeny were all wild type. The F1 females were then crossed to a male tester and the resulting F2 offspring are listed in the following table. Genotype Total offspring d+f 290 d++ 7 def 139 ++f 81 +++ 145 +ef 10 +e+ 265 de+ 63 25. What was the gene order and allele combination in the original parents? a) b) c) d) e) def and +++ d+f and +e+ df+ and ++e +df and e++ df; + and ++; e Solution: Genotype Total offspring d+f 290 d++ 7 def 139 ++f 81 +++ 145 +ef 10 +e+ 265 de+ 63 Total 1000 Parental or recombinant P R R R R R P R m.u. (#/total*100) Page 11 of 22 de 139 81 145 63 428 42.8 ef df 7 139 7 81 145 10 10 301 30.1 63 161 16.1 © Prep101 www.prep101.com/freestuff The offspring genotype with the largest number of offspring is the Parental and any other combination arose due to recombinations. Go down the list and make three columns to represent the recombinations between the three genes based on those that deviate from the parental combinations. Add these up and divide by the total number of progeny, multiplied by 100 for the m.u. (recall 1% rf = 1 m.u.). From these numbers, you will determine that d and e are the furthest apart and that f is closer to d than e. 26. How many double recombinant offspring are there? a) b) c) d) e) 17 34 48 144 70 Solution: Since the parentals are df+ and ++e, the double recombinants are +fe and d++ which are 10 and 7, respectively, and 17 in total. 27. What is the distance from d to e? a) b) c) d) e) 42.8 m.u. 46.2 m.u. 30.1 m.u. 16.1 m.u. 10.3 m.u. Solution: The distance in the de column from question 25 is 42.8, but this is not accounting for double recombinant. Add (428 + 2 x 17)/1000 = 46.2 m.u. 28. Calculate the interference value I a) b) c) d) e) 1 0 0.35 0.52 0.65 Page 12 of 22 © Prep101 www.prep101.com/freestuff Solution: I = 1 – Observed number of double-recombinant/expected number of double recombinants. I = 1 – (17)/((161*301)/1000) = 0.65. The following 3 questions are pertaining to the pedigree below: i i IB i i i i i IB i i i IB i i i i i i i i i i i IBi IB i IB i IB i IB i i i i i IB i i i IB IB i iii i i IB i i i IB i IB i IB i i i The above pedigree shows the inheritance of a dominant phenotype (D) and blood type (type O = ii and type B = IBi). Assume that those marrying into the family are homozygous recessive (ddii), while those with the dominant phenotype are heterozygous (D/d). 29. How many different recombinant offspring are in this family? a) 3 b) 6 c) 12 d) 0 e) 8 Solution: Determine the parental genotype with regards to D and blood type-look at oldest person in the pedigree that had the phenotype (is D linked to IB or is D linked to i)? Given that most of the people in the second generation that have the phenotype are of blood types IBi, their mom must be DIB/di. Anytime someone has the disease but isn’t IB (thus, Di), or someone does not have the dominant phenotype but is IB (dIB), must have resulted from a recombination. There are 6 in total. 30. What is the RF for these two genes in this family? a) 0.21 b) 0.19 c) 0.17 d) 0.25 e) 0.3 Page 13 of 22 © Prep101 www.prep101.com/freestuff Solution: RF = recombinants/total = 6/24 (only include progeny, not people who married into the family) = 0.25 31. What is the Lod score for this family? a) 1.4 b) 1.9 c) -1.4 d) -1.9 e) none of the above Solution: Let’s calculate the probabilities under both hypotheses, linked (RF=0.25) and unlinked (0.5)-independent assortment. RF P P R R 0.5 0.25 0.25 0.25 0.25 0.2727 0.375 0.375 0.125 0.125 The probability of obtaining the result under independent assortment (RF of 50%) will be equal to 0.2524=3.55 x 10-15 x B (the number of possible birth orders for 18 parental and 6 recombinant individuals) For an RF of 0.25 the probability is 0.37518 x 0.1256 = 8.20 x 10-14 x B The ratio of the odds is 8.2 x 10-14 x B/3.55 x 10-15 x B (B’s cancel out) = 23.1 Log (23.1) = 1.4 32. If a striped cat from a pure-breeding stock was mated to a spotted cat from another pure-breeding stock, and the resultant progeny are all both spotted and stripped, these alleles for coat pattern are. . . a) b) c) d) e) Recessive Dominant Epistatic Codominant Incompletely dominant Page 14 of 22 © Prep101 www.prep101.com/freestuff Solution: The definition of Codominance is when both traits become visible in an F1 hybrid, but not as a mix. 33. In certain breed of plant, a single gene is responsible for flower colour. If a blue flowered plant from a pure-breeding stock is crossed to a yellow flowered plant from a pure-breeding stock and the resultant progeny all have orange flowers, these alleles for flower colour are: a) b) c) d) e) Dominant Recessive Epistatic Codominant Incompletely dominant Solution: The definition of incomplete dominant is when the F1 hybrid resembles neither purebred parent, but appears as a combination of the parents. 34. Suppose you are studying eye size in a species of spider. Eye size is controlled by two genes (each with two alleles: Q/q and R/r). If a spider carries at least one Q allele, it has large eyes no matter what alleles it has at the R locus. If it is qq, its phenotype depends on the R locus—if it is qqRr, it has medium eyes; if it is qqrr, it has small eyes. What ratio of (large: medium: small) phenotypes do you expect to see resulting from a QqRr X qqrr cross? a) 9:3:3:1 b) 9:4:3 c) 2:1:1 d) 3:1 e) 1:2:1 Solution: Given that the first parent will contribute Q ½ of the time, ½ of the progeny will have large eyes. The other have will have a split between medium and small, which is determined by the fact that R locus will be R/r half the time and r/r the other half. Thus, 2:1:1 Page 15 of 22 © Prep101 www.prep101.com/freestuff 35. Two unlinked loci effect mouse hair color. AA or Aa mice are agouti. Mice with genotype aa are albino because all pigment production is blocked, regardless of the phenotype at the second locus. At the second locus, the B allele (agouti coat) is dominant to the b allele (black coat). What would be the result of a cross between two agouti mice of genotype AaBb? a) b) c) d) e) 4 agouti: 4 black: 8 albino 9 agouti: 3 black: 3 albino: 1 grey 9 agouti: 3 black: 4 albino 8 agouti: 4 black: 4 albino 12 agouti: 3 black: 1 albino Solution: In order to get agouti, you need A/-; B/-, which happens 9/16; and black when you have A/-; b/b 3/16 and a/a; -/- is 4/16. Thus, albino is epistatic to black. 36. Symptoms of Marfan syndrome include skeletal, optical, and cardiovascular abnormalities. Skeletal abnormalities include lengthening of the long bones, scoliosis and others. Optical abnormalities can include dislocation of the lens into the anterior chamber of the eye. Cardiovascular abnormalities are numerous (ex. dissecting aneurysms), which attributes to the shorter life span of Marfan syndrome patients as a group. Each person with this genetic makeup display a range of the symptoms (from no symptoms to severe). This is an example of a) b) c) d) e) Pleiotropy Expressivity Penetrance Incomplete dominance Expressivity and pleiotropy Solution: Because people with the disease genotype don’t all have symptoms, this is an example of penetrance. However, within the population of people that have symptoms, there is a range, which is an example of expressivity. Page 16 of 22 © Prep101 www.prep101.com/freestuff 37. Determine the order of the compounds (letters) and genes (numbers), respectively, acting in this biosynthetic pathway given the following information that evaluates for the presence of the end product in the pathway: Compound added Mutant A B C D E F 1 - -++-+ 2 -+++ -+ 3 --+--+ 4 -+++++ 5 -----+ a) b) c) d) e) A,E,B,D,C,F; 4,2,1,3,5 F,C,D,B,E,A; 5,3,1,2,4 F,C,D,B,E,A; 4,2,1,3,5 A,E,B,D,C,F; 5,3,1,2,4 None of the above Solution: The first enzyme in the pathway can be recognized by the fact that it is the one that is rescued by the most number of compounds. The compound that doesn’t rescue the mutant is before the first enzyme in the pathway. Thus, gene 4 is first and compound A is first, then 2/E, 1/B, 3/D, 5/C, F (or to determine the first compound, look at the one that rescues no mutants to all mutants; conversely, to ID the first mutant, look for the enzyme that is rescued by the most compounds). A--------E---------B---------D---------C-----------F 4 2 1 3 5 A cannot be synthesized into E (and ultimately the product at the end of the pathway) without a functioning enzyme 4, likewise, a mutation in enzyme 2 requires that B be added in order to complete the pathway etc. 38. Quantitative genetics is: a) b) c) d) e) The study of genetics of continuously varying characters Influenced by variation in the environment Often involves multifactorial inheritance Uses statistical methods to make predictions about inheritance All of the above Page 17 of 22 © Prep101 www.prep101.com/freestuff 39. The spread in a distribution refers to: a) variation b) dispersion of a measurement c) central tendency d) relation between different variables e) Two of the above are correct Solution: Both variation and dispersion are measures of spread. Central tendency is the mode, while correlation describes the relation between different variables. 40. You have two pure breeding lines of plants (average height is 10 cm and 30 cm for line 1 and 2, respectively). You make a cross of a plant from each line and get F1 progeny with an average height of 22 cm. You cross the F1 progeny and get an average height of 24 cm, but a much greater distribution of heights (12 cm to 29 cm). From the F2 plants, you selectively breed two of the smallest plants together and two of the largest plants together. What are the expected outcomes from these crosses? a) Means in each cross is close to the size of the F2 parents b) An average in each cross is close to the size of the parents, but a much larger distribution than in F2 c) An average in each cross is different from the parental heights d) The variation in F3 distributions will be overlapping e) Two of the above are correct Solution: It is not likely that the distributions in each cross will be larger than F2 rather they will likely be smaller due to the combination of genotypes that are more similar (in comparison to the cross of the F1s generating the F2s). 41. Which of the following examples reflects covariance that is calculated in broad-sense heritability? a) Tall parents having tall children b) Overweight parents that have children who are overweight c) Siblings inheriting the same genotype d) The survival of a butterfly with dark wings in the forest e) Different allelic combinations in parents Solution: Covariance in broad-sense heritability is the covariance between genotype and environment (populations variation is influenced by the environment of individuals). Page 18 of 22 © Prep101 www.prep101.com/freestuff 42. A heritability value of 0.8 indicates which of the following? a) b) c) d) e) The majority of phenotypic variation is due to environmental variation The phenotype is the result of 3 genes interacting The majority of the phenotypic variation has a genetic basis There is no developmental noise occurring No answers are correct 43. You are studying the heritability of height in a given population at two altitudes. You calculate the genotypic variance to be 33.4 and the variance of the population to be 49.8. What is the environmental variance and heritability (assume zero covariance) of this trait in this study? a) b) c) d) e) 16.4, 33% 16.4, 49% 16.4, 67% 83.2, 40% 83.2, 67% Solution: se2 = sp2 - sg2 se2 = 49.8 – 33.4 se2 = 16.4 Heritability = H2 = sg2 / sp2 H2 = 33.4/49.8 H2 =67% 44. Which of the following is NOT an assumption of the Hardy-Weinberg equilibrium? a) b) c) d) e) The population in infinitely large Individuals are mating randomly Endogamy is taking place within subpopulations There are no new mutations introduced into the population There is no migration into or out of the population Solution: Endogamy is the mating of individuals within a group or subgroup rather than at random in a population. Page 19 of 22 © Prep101 www.prep101.com/freestuff Use the following table to answer the next 3 questions. Gametic type Population AB Ab aB ab 1 0.25 0.3 0.15 0.3 2 0.4 0.2 0.3 0.1 45. What is the heterozygosity of the gametes for population 1? a) b) c) d) e) 0.265 0.735 0.4875 0.2438 0.975 Solution: Gamete heterozygosity = 1-homozygosity 1- (f AB/AB + f Ab/Ab + f aB/aB + f ab/ab) = 1- (0.25*0.25 + 0.3*0.3 + 0.15*0.15 + 0.3*0.3) = 0.735 46. What is the allelic heterozygosity for population 1? a) b) c) d) e) 0.265 0.735 0.4875 0.2438 0.975 Solution: Allelic heterozygosity = ( f A/a + f B/b )/2 = (2pq + 2rs)/2 = (2(0.25 + 0.3)(0.15+0.3) + 2(0.25 + 0.15)(0.3 + 0.3))/2= 0.4875 Page 20 of 22 © Prep101 www.prep101.com/freestuff 47. What is the heterozygosity of the gametes for population 2? a) b) c) d) e) 0.7 0.3 0.45 0.9 0.225 Solution: Gamete heterozygosity = 1-homozygosity 1- (f AB/AB + f Ab/Ab + f aB/aB + f ab/ab) = 1- (0.4*0.4 + 0.2*0.2 + 0.3*0.3 + 0.1*0.1) = 0.7 48. Humans with the genotype DD and Dd show the Rh+ blood phenotype, whereas those with the genotype dd show the Rh- blood phenotype. In a sample of 400 Basques from Spain, 230 people were Rh+ and 170 people were Rh-. Assuming that this population is in Hardy-Weinberg proportions, what is the allele frequency of the allele D? a) b) c) d) e) 0.348 0.652 0.425 0.575 0.288 Solution: q2 = 170/400 = 0.425 q = √(0.425) = 0.6519 p = 1 – 0.6519 = 0.348 49. Which of the following statements about inbreeding is FALSE? a) There is a 1/2N loss of heterozygosity in each generation b) In each generation, there is an increase towards recessive homozygosity c) Inbreeding results in non-random mating which alters Hardy-Weinberg frequencies d) An inbreeding coefficient F needs to be calculated to adjust for the proportion of shared genes between related individuals e) Two of the above Page 21 of 22 © Prep101 www.prep101.com/freestuff Solution: There is an increase towards either homozygous recessive or homozygous dominant 50. If in a population of 1 million people, 200 blind (homozygous recessives, aa) were found, how many normal (homozygous dominants, AA) individuals would be found in the next generation under equilibrium conditions? a) b) c) d) e) 749,850 999,800 985,858 27,888 971,916 Solution: q2 = 200/1,000,000 = 0.0002 q = √(0.0002) = 0.01414 p = 1 – 0.01414 = 0.985858 p2 = (0.985858)2 = 0.971916 * 1,000,000 f(AA)* 1e6 =971,916 Page 22 of 22 ...
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