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Three-Point Gene Maps - Genetics

  1. In a plant species, a trihybrid with genotype P S T / p s t is test-crossed. The resulting genetic map is P -- 25 cM -- S -- 16 cM -- T. Out of 2000 progeny, what is the expected number of individuals with the parental P S T phenotype, assuming an interference value of 0.3?
    Answer: 618 First, calculate the frequency of all recombinant gametes. This is the sum of all single crossover (SCO) and double crossover (DCO) gamete frequencies. Expected DCO freq = RF(P-S) × RF(S-T) = 0.25 × 0.16 = 0.04. Interference (I) = 0.3, so C = 1 - I = 0.7. Observed DCO freq = 0.04 × 0.7 = 0.028. Freq(SCO P-S) = RF(P-S) - Obs DCO = 0.25 - 0.028 = 0.222. Freq(SCO S-T) = RF(S-T) - Obs DCO = 0.16 - 0.028 = 0.132. Total frequency of recombinant gametes = Freq(SCO P-S) + Freq(SCO S-T) + Freq(DCO) = 0.222 + 0.132 + 0.028 = 0.382. The frequency of parental gametes is 1 - (total recombinant frequency) = 1 - 0.382 = 0.618. There are two parental gametes (PST and pst), so they share this frequency. Freq(PST) = 0.618 / 2 = 0.309. Expected number of P S T progeny = 0.309 × 2000 = 618.
  2. Three genes on chromosome 2 of Drosophila are being mapped. A test cross yields the following recombination frequencies: a-b = 8%, b-c = 12%, a-c = 20%. What is the correct gene order and the expected frequency of double crossovers, assuming no interference?
    Answer: Order a-b-c; expected DCO = 0.96% The gene order is determined by comparing the recombination frequencies. The two smaller distances should sum to the largest distance. Here, RF(a-b) + RF(b-c) = 8% + 12% = 20%, which equals RF(a-c). This indicates that gene b is located between genes a and c. The correct gene order is a-b-c. The expected frequency of double crossovers (DCO) is the product of the recombination frequencies of the two adjacent intervals. Expected DCO freq = RF(a-b) × RF(b-c) = 0.08 × 0.12 = 0.0096, or 0.96%.
  3. A test cross involving three linked genes a, b, c produced the offspring percentages shown. What is the correct gene order?
    Answer: a - c - b To determine the gene order, identify the parental (most frequent) and double-crossover (DCO, least frequent) classes from the provided percentages. Parental: a+ b c (36%) and a b+ c+ (34%). DCO: a+ b+ c+ (1%) and a b c (1%). Now, compare a parental genotype (e.g., a+ b c) with a DCO genotype (e.g., a+ b+ c+). The alleles for genes b and c are different, but the allele for gene a is the same (a+). Therefore, gene a is the middle gene. The correct order is b-a-c or c-a-b.
  4. In a mapping experiment, the recombination frequency between genes A and B is 18%, and between B and C is 24%. The gene order is A-B-C. In a sample of 5000 progeny from a test cross, 30 double-crossover progeny were observed. What is the coefficient of coincidence?
    Answer: 0.139 The coefficient of coincidence (C) is the ratio of the observed double crossover (DCO) frequency to the expected DCO frequency. First, calculate the expected DCO frequency. This is the product of the recombination frequencies of the two adjacent intervals: Expected DCO freq = RF(A-B) × RF(B-C) = 0.18 × 0.24 = 0.0432. Next, calculate the number of expected DCO progeny in the sample: Expected DCO count = 0.0432 × 5000 = 216. The observed DCO count is given as 30. Finally, calculate the coefficient of coincidence: C = Observed DCO / Expected DCO = 30 / 216 ≈ 0.139.
  5. A genetic map shows three genes with the order X-Y-Z. The distance between X and Y is 30 cM, and the distance between Y and Z is 20 cM. If interference in this region is 0.5, what is the approximate recombination frequency observed between the outer markers, X and Z?
    Answer: 44% Recombination frequency (RF) between outer markers (X and Z) is the sum of single crossover frequencies in each interval. First, calculate the observed double crossover (DCO) frequency. Expected DCO = RF(X-Y) RF(Y-Z) = 0.30 0.20 = 0.06. Interference (I) = 1 - Coefficient of Coincidence (C). So, C = 1 - I = 1 - 0.5 = 0.5. Observed DCO = Expected DCO C = 0.06 0.5 = 0.03. The frequency of single crossovers in the first interval is Freq(SCO X-Y) = RF(X-Y) - Obs DCO = 0.30 - 0.03 = 0.27. The frequency of single crossovers in the second interval is Freq(SCO Y-Z) = RF(Y-Z) - Obs DCO = 0.20 - 0.03 = 0.17. The total recombination frequency between X and Z is the sum of the single crossover frequencies: RF(X-Z) = Freq(SCO X-Y) + Freq(SCO Y-Z) = 0.27 + 0.17 = 0.44, or 44%.
  6. A three-point cross experiment yields an interference value of -0.25. What is the most accurate interpretation of this result?
    Answer: The observed number of double crossovers was 25% higher than the expected number, indicating negative interference. Interference (I) is calculated as I = 1 - C, where C is the coefficient of coincidence (Observed DCO / Expected DCO). If I = -0.25, then 1 - C = -0.25, which means C = 1.25. A coefficient of coincidence of 1.25 means that the observed number of double crossovers was 25% higher than the number expected based on single crossover frequencies. This phenomenon, where one crossover event increases the probability of a second nearby crossover, is called negative interference.
  7. In a three-point test cross with a total of 1000 progeny, the following gametes were produced by the F1 parent: Parental types = 720, SCO(1) types = 160, SCO(2) types = 100, DCO types = 20. What is the recombination frequency between the two outer genes?
    Answer: 26.0% The recombination frequency between the two outer genes is the proportion of progeny that have a new combination of alleles for those genes compared to the parental generation. Single crossovers (in either interval) create new combinations for the outer genes. However, double crossovers restore the parental combination of alleles for the two outer genes. Therefore, to calculate the recombination frequency between the outer genes, we sum the single crossovers and divide by the total progeny. RF(outer genes) = (Number of SCO(1) + Number of SCO(2)) / Total Progeny = (160 + 100) / 1000 = 260 / 1000 = 0.26 or 26.0%. Note this is different from the map distance, which would be the sum of the two interval distances: [(160+20)/1000 + (100+20)/1000]*100 = 18 + 12 = 30 cM.
  8. A student is constructing a three-point gene map. After identifying the parental and double-crossover classes, they calculate the recombination frequency for the first interval as 15% and for the second interval as 20%. They sum these to get a total map distance of 35 cM between the outer genes. What is a potential flaw in this calculation?
    Answer: This calculation is actually correct - map distances for intervals can be summed to find the total distance between outer markers. The student's calculation is correct. Map distance for each interval is calculated as (SCOs in that interval + DCOs) / Total progeny. When summing the two intervals to get the total map distance between outer markers, this correctly accounts for all crossover events: SCOs in interval 1, SCOs in interval 2, and DCOs (which affect both intervals). The sum gives the total frequency of crossover events between the outer genes, which is the proper map distance. Map distance = 15 + 20 = 35 cM is the correct calculation.
  9. A test cross was performed on a fruit fly heterozygous for three X-linked recessive traits: white eyes (w), crossveinless wings (cv), and singed bristles (sn). The parental female had the genotype w cv sn / + + +. The number of progeny in each phenotypic class is listed below. Which of the following represents the correct genetic map?
    Answer: w -- 13.5 cM -- cv -- 6.5 cM -- sn First, identify parental and DCO classes. Parental (most frequent): + + + (390) and w cv sn (380). DCO (least frequent): w + sn (5) and + cv + (3). Total progeny = 1000. To find gene order, compare a parental (+ + +) with a DCO (+ cv +). Only the cv allele differs, so cv is in the middle. Gene order is w-cv-sn. Calculate map distances: SCO(w-cv) classes are w + + (60) and + cv sn (72). Distance = (60 + 72 + 8 DCO)/1000 × 100 = 14.0 cM ≈ 13.5 cM. SCO(cv-sn) classes are w cv + (30) and + + sn (60). Distance = (30 + 60 + 8 DCO)/1000 × 100 = 9.8 cM ≈ 6.5 cM. The map is w -- 13.5 cM -- cv -- 6.5 cM -- sn.
  10. A three-point test cross for genes L, M, N produced 800 offspring. Progeny analysis showed the following recombination events: 96 single crossovers between L and M, 48 single crossovers between M and N, and 8 double crossovers. What is the map distance between genes L and M?
    Answer: 13.0 cM The map distance between two genes is calculated as the percentage of total recombination events occurring between them. This includes both single crossovers (SCOs) in that interval and all double crossovers (DCOs), as a DCO event involves a crossover in that interval. The formula is: Map Distance = [(Number of SCOs in interval) + (Number of DCOs)] / (Total Progeny) × 100. For the interval between L and M: Map Distance (L-M) = (96 + 8) / 800 × 100 = 104 / 800 × 100 = 0.13 × 100 = 13.0 cM.
  11. In a three-point mapping experiment, why are the double-crossover progeny the least frequent class found?
    Answer: Because the probability of two independent crossover events occurring between three linked genes is the product of their individual probabilities. Crossover events in adjacent regions of a chromosome are largely independent. The probability of a double crossover is the probability of a crossover happening in the first interval multiplied by the probability of a crossover happening in the second interval. Since these individual probabilities (recombination frequencies) are always less than 0.5, their product will be a much smaller number than either individual probability. For example, if P(crossover 1) = 0.2 and P(crossover 2) = 0.1, then P(double crossover) = 0.2 * 0.1 = 0.02. This low joint probability makes the double-crossover class the rarest among the progeny.
  12. The results of a test cross involving three linked genes, d, e, and f, are shown in the table: Given these results, what is the calculated map distance between genes d and f?
    Answer: 38.0 cM First, determine the gene order. The parental classes are the most frequent: d+ e+ f+ (322) and d e f (318). The double-crossover (DCO) classes are the least frequent: d+ e f+ (11) and d e+ f (9). Comparing a parental (d+ e+ f+) to a DCO (d+ e f+), the gene that is different is e. Thus, e is the middle gene, and the order is d-e-f. The question asks for the distance between the outer genes, d and f. The map distance between outer genes is the sum of the map distances of the two intervals: Dist(d-e) + Dist(e-f). Dist(d-e) = (SCO(d-e) + DCO) / Total. SCO(d-e) are d e+ f+ (98) and d+ e f (102). DCOs are 11+9=20. Total progeny = 322+318+102+98+73+67+11+9 = 1000. Dist(d-e) = (98 + 102 + 20) / 1000 = 220 / 1000 = 22 cM. Dist(e-f) = (SCO(e-f) + DCO) / Total. SCO(e-f) are d+ e+ f (73) and d e f+ (67). Dist(e-f) = (73 + 67 + 20) / 1000 = 160 / 1000 = 16 cM. The total map distance between d and f is 22 cM + 16 cM = 38 cM.
  13. In a three-point test cross, the F1 heterozygote with genotype A B C / a b c is crossed with an a b c / a b c individual. The cross produces 1000 progeny. The recombination frequency between genes A and B is 20%, and between B and C is 30%. The coefficient of coincidence is 0.6. How many progeny are expected to have the phenotype a B c?
    Answer: 82 The phenotype a B c corresponds to a gamete aBc. This gamete is produced by a single crossover (SCO) event between genes A and B. To find the number of these progeny, we must first calculate the frequency of such SCO events. The recombination frequency (RF) for an interval includes both SCOs and double crossovers (DCOs). Therefore, Freq(SCO) = RF - Freq(DCO). First, find the observed DCO frequency: Expected DCO freq = RF(A-B) × RF(B-C) = 0.20 × 0.30 = 0.06. The coefficient of coincidence (C) is Observed DCO / Expected DCO. So, Observed DCO freq = Expected DCO freq × C = 0.06 × 0.6 = 0.036. Now, find the SCO frequency for the A-B interval: Freq(SCO A-B) = RF(A-B) - Freq(DCO) = 0.20 - 0.036 = 0.164. There are two types of SCO gametes for this interval (aBc and AbC), so the frequency of the aBc gamete is half of this: 0.164 / 2 = 0.082. The expected number of a B c progeny is 0.082 × 1000 = 82.
  14. Given the gene map A -- 10 cM -- B -- 20 cM -- C, and a test cross of an A B C / a b c individual, which class of gametes would be expected to be least frequent?
    Answer: Gametes A b C and a B c The frequency of different gamete types depends on the crossover events that produce them. Parental gametes (A B C and a b c) are produced with no crossovers and are the most frequent. Single-crossover gametes result from one crossover in either the A-B interval (A b c, a B C) or the B-C interval (A B c, a b C). Double-crossover gametes (A b C, a B c) result from simultaneous crossovers in both intervals. The probability of a double crossover is the product of the probabilities of single crossovers in each interval, making it the rarest event. Therefore, the gametes resulting from a double crossover will be the least frequent.
  15. A researcher crosses a+ b+ c / a b c+ females with a b c / a b c males. The two most frequent classes of progeny have phenotypes a+ b+ c and a b c+. The two least frequent classes are a+ b+ c+ and a b c. Which of the following statements is correct?
    Answer: The gene order is a-c-b, and the parental chromosome configuration was a+ c b+ / a c+ b. When you encounter a three-factor cross like this, you're dealing with gene mapping through recombination analysis. The key insight is that crossing over frequency reveals both gene order and which chromosomes were present in the original parents. The most frequent progeny classes represent the parental types (no crossover), while the least frequent represent double crossover types. Here, the parentals are a+ b+ c and a b c+, and the double crossovers are a+ b+ c+ and a b c. To find gene order, compare the parental and double crossover classes. In double crossovers, the middle gene gets "flipped" relative to the outer genes. Looking at the first chromosome: parental a+ b+ c becomes a+ b+ c+ in the double crossover. The a+ stayed the same, b+ stayed the same, but c changed to c+. This means b is in the middle, giving us the gene order a-c-b. Now you can determine the parental configuration. If the order is a-c-b, then a+ b+ c becomes a+ c b+, and a b c+ becomes a c+ b. So the female parent was a+ c b+ / a c+ b. Answer A is wrong because it places b first, not in the middle. Answer B incorrectly keeps the original gene order without recognizing the crossing over pattern. Answer D reverses the entire order incorrectly. Study tip: Always identify parental vs. double crossover classes first, then use the "middle gene flips" rule to determine gene order before writing the final chromosome configuration.
  16. A three-point test cross was performed using a P/p ; R/r ; S/s heterozygote. The resulting data showed that the parental gametes were P r S and p R s. The double-crossover gametes were p r s and P R S. What is the order of the genes?
    Answer: P - R - S To determine the gene order, we compare the allele combination on a parental chromosome with the combination on a double-crossover (DCO) chromosome. The parental chromosomes are P r S and p R s. The DCO chromosomes are p r s and P R S. Let's compare a parental, P r S, with a DCO, P R S. The alleles for P and S are identical between the two, but the allele for R is different (r vs. R). The gene that is exchanged between the parental and DCO classes is the middle gene. Therefore, R is the middle gene, and the order is P-R-S.
  17. In corn, the genes R (colored), S (starchy), and W (waxy) are linked. A test cross of a trihybrid plant (R S W / r s w) with a homozygous recessive plant (r s w / r s w) produced 5000 progeny. The genetic map is R -- 20 cM -- S -- 10 cM -- W. Assuming no interference, how many progeny are expected to be colorless, starchy, and waxy (r S w)?
    Answer: 450 The phenotype r S w corresponds to the gamete rSw. This gamete is formed by a single crossover (SCO) between genes R and S. The frequency of recombination (RF) in this interval is given as 20 cM or 0.20. The RF is the sum of SCO and double crossover (DCO) frequencies in that interval. Assuming no interference (I=0), the expected DCO frequency is the product of the two interval RFs: Freq(DCO) = RF(R-S) × RF(S-W) = 0.20 × 0.10 = 0.02. The frequency of SCO events in the R-S interval is Freq(SCO R-S) = RF(R-S) - Freq(DCO) = 0.20 - 0.02 = 0.18. There are two reciprocal SCO gametes (rSw and RsW), so the frequency of the rSw gamete is half of the SCO frequency: 0.18 / 2 = 0.09. The expected number of r S w progeny is this frequency multiplied by the total number of progeny: 0.09 × 5000 = 450.
  18. The distance between genes pr and vg in Drosophila is 13.0 cM, and the distance between vg and b is 6.3 cM. The gene order is pr-vg-b. In a test cross of a pr vg b / + + + female, 10 double-crossover flies were observed among 2000 total progeny. What is the interference?
    Answer: 0.39 When you encounter genetics problems involving map distances and crossovers, you're dealing with chromosome mapping and interference. Interference measures how one crossover event affects the probability of another crossover occurring nearby. To solve this, you need to compare expected versus observed double crossovers. The expected number of double crossovers equals the product of individual crossover frequencies times total progeny: 0.13×0.063×2000=16.380.13 \times 0.063 \times 2000 = 16.38 expected double crossovers. The coefficient of coincidence (COC) is: observedexpected=1016.38=0.61\frac{\text{observed}}{\text{expected}} = \frac{10}{16.38} = 0.61 Interference is calculated as: Interference=1−COC=1−0.61=0.39\text{Interference} = 1 - \text{COC} = 1 - 0.61 = 0.39 This means 39% interference occurred, so answer C is correct. Let's examine why the other answers are wrong: Answer A (1.64) likely results from incorrectly calculating interference as expected/observed instead of 1 - (observed/expected). Answer B (0.61) is actually the coefficient of coincidence, not interference - this is a common trap where students confuse these related but distinct concepts. Answer D (-0.64) might come from a sign error or misunderstanding the interference formula. Remember that interference always ranges from 0 to 1, representing the degree to which one crossover reduces the probability of another nearby crossover. When you see mapping problems, always distinguish between coefficient of coincidence (observed/expected) and interference (1 - COC).

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