Genetics - Problem Drill 05: Genetic Mapping: Linkage and Recombination No. 1 of 10 1. A corn geneticist crossed a crinkly dwarf (cr) and male sterile (ms) plant; The F1 are male fertile with normal height. He observed the F2 progeny from F1 intercross segregate as: Cr_Ms_ 181, Cr_msms 34, crcrms_ 36, and crcrmsms 28, which obviously deviate from the 9:3:3:1 ratio. The possible reason for this phenomenon could be. (A) Abnormal segregation of cr gene. (B) Abnormal segregation of ms gene. (C) Linkage between cr and ms. (D) The cr gene interferes with the phenotype of male sterile, or vice versa. (E) None of the above A. Incorrect! Ignore the ms gene; just look at the cr gene: in F2 the ratio is (181+34)/(36+28) = 215 : 64, Chi square to fit 3:1 segregation ratio. Therefore, there is no significant abnormal segregation at cr locus. Analyze the ms locus just as we did for cr locus. Ms : ms = 217 : 62, Chi square to fit 3:1 segregation. Therefore, there is no significant abnormal segregation at cr locus. C. Correct! Again, use Chi square method to test. Assume there is no linkage and the data would fit into 9:3:3:1 mode. This explanation is possible but unlikely to occur; it is very rare in eukaryotic genetics that linked genes interfere with each other s phenotype. There is one right answer above. Linkage refers to the proximity of a gene on the same chromosome and how that impacts inheritance patterns. The simple way to detect linkage is through a chi square test. (C) Linkage between cr and ms.
No. 2 of 10 2. Two genes are linked on the same chromosome, A and B; the parent genotypes are both Ab/aB. The crossover ratio is 30%. The progeny wild type A_B_ and aabb ratio would be. (A) 3:1 (B) 16:1 (C) 209:9 (D) 175:9 (E) 64:9 A. Incorrect! 3:1 ratio is for single gene heterozygous self-cross. If the two genes are not linked, the aabb would be counted as 1/16 of total progeny; since they are linked, this ratio is obviously off. C. Correct! To solve this problem, first list all possible gametes from both parents: Ab (35%), ab (35%), AB (15%) and ab (15%). Make a Punnet Table (see below) and add up all combinations that can bring up A_B_ genotypes, resulting 0.5225; and then count the aabb type which is easier because, in each parent, the aabb gamete count for 15%. Therefore, in total, it is 0.15*0.15 = 0.0225. A_B_:aabb = 0.5225/0.0225 = 209:9. There is no way to get number 175; it is a wild guess. There is also no way to have 64:9 ratio. The key to solve this problem is to list all gamete types and calculate their amount. Ab 0.35 ab 0.35 AB 0.15 ab 0.15 Ab 0.35 AAbb 0.1225 AaBb 0.1225 AABb 0.0525 Aabb 0.0525 ab 0.35 AaBb 0.1225 aabb 0.1225 AaBB 0.0525 aabb 0.0525 AB 0.15 AABb 0.0525 AaBB 0.0525 AABB 0.0225 AaBb 0.0225 ab 0.15 AAbb 0.0525 aabb 0.0525 AaBb 0.0525 aabb 0.0225 (C) 209:9
No. 3 of 10 3. In a plant, leaf color and leaf shape are controlled by two linked genes. Leaves of the wild-type plant are red. A recessive mutation in this gene causes white leaves. Wild-type leaves are pointed, and a recessive mutation in this gene causes them to be smooth. The following crosses were performed: Cross 1: pure breeding white, smooth X pure breeding wild type gives F1: all red, pointed Cross 2: red, pointed F1 X pure breeding white, smooth (test cross) gives F2: 40 white, smooth; 36 red, pointed; 10 white, pointed; 14 red, smooth. What is the recombination frequency between the gene for color and for shape? (A) 0.1 (B) 0.14 (C) 0.5 (D) 0.24 (E) 0.12 A. Incorrect! White point is the recombined type but is half of the recombinants; the other half is 14 of red and smooth. Same as A, only half of the recombinants are counted. When the recombination frequency reach 0.5 (50%), these genes are considered not to be linked; they assort independently. D. Correct! The total recombinants are 10 + 14 and the total progeny is 100. Therefore, the recombination frequency is 0.24. By definition, the recombination frequency is the total recombinant number divided by total progeny number; it includes both types of recombinant but does not divide into half. Recombination frequency = number of recombinants/number of total progeny. (D) 0.24
No. 4 of 10 4. In a testcross of a corn plant heterozygous for three linked recessive genes (a,b, and d), the following phenotypes and numbers of progeny were obtained (total is 1000): ABd 460; AbD 12; Abd 19; abd 476; abd 12; abd 20; abd 1 The coefficiency of coincidence is. (A) 1.0 (B) 0.5 (C) 0.75 (D) 1.5 (E) Not enough information to determine. First, we need to determine which gene is in the middle. The smallest class is abd type; this is the double crossover. Comparing to the parental type abd, d is the one whose position is changed; therefore, d is located in the middle. Then, we determine the recombination frequency between A and D. Since the parental types are Ad and ad, ignore b for now and only count the Ad and ad type (parental, 975), as well as the sum of AD and ad type (recombinants, 25). The frequency between A and d is 0.025; similarly, we can calculate the recombination frequency between d and B is 0.040. The expected double crossover is 0.025*0.040*1000 = 1. The observed double crossover is also 1. Therefore, the coefficiency of coincidence is 1.0. If coefficiency of coincidence is 0.5, then the expected abd should be 0.5. If coefficiency of coincidence is 0.75, then the expected abd should be 0.75. If coefficiency of coincidence is 1.5, then the expected abd should be 1.5. There is sufficient information to pick the answer. Follow the step by step technique, first determine the gene order, then count two genes at a time and calculate the recombination frequency (i.e., the crossover frequency). (A) 1.0
No. 5 of 10 5. For the tetrads recovered below, the genetic distance between mat-ura and leuura is? (A) 8.07 and 5.55 (B) 15.2 and 10.2 (C) 16.1 and 11.1 (D) 7.4 and 5.1 (E) None of the above The first step is to determine which is PD, T and NPD. Here, class I is the most abundant type and, therefore, it is PD. Class II is double crossover, the rarest one, NPD; class III and VI are NPD. For each pair of genes, PD>T>NPD, so all three genes are linked. In the double crossover, ura switched the position; therefore, ura is in the middle. The genetic distance is calculated by ½ (NPD + T) / total. For mat and ura, cm = ½ (106+6)/(511+6+106+71) = 8.07; for ura and leu, cm = ½ (71+6)/(511+6+106+71) = 5.55. Getting this result is because the double crossover class is ignored, and ½ is ignored in the formula. The formula includes ½. This doubles the answer because ½ is ignored. This result also ignored the double crossover. There is one correct answer above. The key to solve the tetrad analysis problem is to determine the type of tetrads, and then follow rules about double crossover and the formula. (A) 8.07 and 5.55
No. 6 of 10 6. Which of the following statements about recombination is true? (A) A centi morgan is a unit of genetic distance that is equal to a 1% probability of recombination during meiosis. (B) A centi morgan is a unit of genetic distance that is equal to a 10% probability of recombination during meiosis. (C) One centi morgan is equal, on average, to one centi base in the human genome. (D) A centi morgan is always equal to one mega base; this amount never changes. (E) If two genes are one centi morgan apart, there is a 10% chance they will separate during meiosis. A centi morgan is a unit of genetic distance that is equal to a 1% probability of recombination during meiosis. A centi organ is a unit of genetic distance that is equal to a 1% probability of recombination during meiosis. One centi morgan is equal, on average, to one mega base in the human genome. One centi morgan is equal, on average, to one mega base in the human genome. If two genes are one centi morgan apart, there is a 1% chance they will separate during meiosis. A centi morgan is a unit of genetic distance that is equal to a 1% probability of recombination during meiosis. One cm is equivalent, on average, to a physical distance of about 1 megabase in the human genome. This is an average because recombination rates are different at different parts of the chromosomes. (A) A centi morgan is a unit of genetic distance that is equal to a 1% probability of recombination during meiosis.
No. 7 of 10 7. What affects crossover events during meiosis? (A) Chiasma interference, because of the presence of a crossover, increases the likelihood of another crossover event nearby. (B) Chiasma interference, because of the presence of a crossover, decreases the likelihood of another crossover event nearby. (C) The number of centi morgans between 2 genes. (D) The location of the somatic cell which is undergoing meiosis. (E) Whether the somatic cell is in meiosis I or meiosis II. A. Incorrect! Chiasma interference, because of the presence of a crossover, decreases the likelihood of another crossover event nearby. B. Correct! Chiasma interference, because of the presence of a crossover, decreases the likelihood of another crossover event nearby. The number of centi morgans between 2 genes affects their likelihood of genetic recombination but not whether crossover will take place. Meiosis typically does not take place in somatic cells but instead produces gametes. Meiosis typically does not take place in somatic cells but instead produces gametes. The chiasma is the point where two homologous non-sister chromatids undergo crossing over and genetic exchange. Its presence affects crossing over in the local vicinity. (B) Chiasma interference, because of the presence of a crossover, decreases the likelihood of another crossover event nearby.
No. 8 of 10 8. In genetics, what would an 18p14 identification mean? (A) It would identify the location of a gene as chromosome 18, short arm, and a position of 14 from the centromere. (B) It would identify the location of a gene as chromosome 18, and a position of 12 from the centromere. (C) It would identify whether the 2 genes, designated 18 and 14, would separate during meiosis. (D) The gene would be closer to the chiasma than a gene at 18p15. (E) The gene would be farther away from the chiasma than a gene at 18p15. It would identify the location of a gene as chromosome 18, short arm, and a position of 14 from the centromere. It would identify the location of a gene as chromosome 18, short arm, and a position of 14 from the centromere. The identification is describing the location of a single gene on the chromosome. The designation is in reference to the centromere of the chromosome, 12 being closer than 14. The designation is in reference to the centromere of the chromosome, 12 being closer than 14. Example: For 17q12, the number 17 is the chromosome number. The letter q means it is on the q arm. The final number 12 shows its relationship in terms of distance from the centromere. A gene with the #12 is closer to the centromere than a gene with the #14. (A)It would identify the location of a gene as chromosome 18, short arm, and a position of 14 from the centromere.
No. 9 of 10 9. Which of the following statements about genetic linkage is true? (A) Linkage was discovered by Bateson, Saunders, and Punnett. (B) It was Thomas Hunt that first discovered genetic linkage. (C) Linkage was inferred from experimental data, in which the parental phenotypes were identical to the expected numbers in the F2. (D) Linkage was inferred as a result of data that showed that all phenotypes that transmit together to the offspring occur in equal frequency as expected. (E) Genes within 5-10 centi morgans are always linked and transmitted together to the offspring. Linkage was discovered by Bateson, Saunders, and Punnett. Linkage was discovered by Bateson, Saunders, and Punnett. Linkage was inferred from experimental data, in which the parental phenotypes were over-represented in the F2. Linkage was inferred from experimental data, in which the parental phenotypes were over-represented in the F2. Although, in development of the linkage theory, it was observed that some phenotypes couple and are transmitted together, there is not an exact distance over which this occurs. Using the following data: Observed: 284, 21, 21, 55 Expected: 215, 71, 71, 24 Linkage is inferred because the parental phenotypes are over-represented in the F2. Yes, these two factors directly violate the Law of Independent Assortment. Noteworthy is that these two factors are coupled then they are transmitted. Also, the new traits (phenotypes) are under-represented. (A)Linkage was discovered by Bateson, Saunders, and Punnett.
No. 10 of 10 10. The coefficient of coincidence. (A) Is defined as the observed number of double crossovers divided by the expected. (B) Is defined as the expected number of double crossovers subtracted from the observed. (C) Can be calculated from the data resulting from a single test cross. (D) Can be calculated by dividing the number of observed double crossovers by the expected number, in the data from a single test cross. (E) Is independent of interference. The coefficient of coincidence is defined as the observed number of double crossovers divided by the expected. The coefficient of coincidence is defined as the observed number of double crossovers divided by the expected. A single test cross could not reveal double crossover events. A single test cross could not reveal double crossover events. The coefficient of coincidence is part of the calculation of interference (Interference = 1-coeffencient of coincidence). The coefficient of coincidence is the observed number divided by the expected number of double crossovers. The interference is defined as: Interference = 1 coefficient of coincidence. (A)Is defined as the observed number of double crossovers divided by the expected.