Genetics Quiz: Interpreting Recombination Data

What this quiz covers

This quiz focuses on Interpreting Recombination Data, giving you a quick way to practice the rules, question types, and explanations that matter most for Genetics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. The genes are unlinked, and the chi-square test would likely fail to reject the null hypothesis (p > 0.05).
2. The genes are linked with a map distance of 87 cM, and the chi-square test would reject the null hypothesis.
3. The genes are linked, but the recombination frequency cannot be estimated from test cross data.
4. The genes are linked with a map distance of 13 cM, and the chi-square test would reject the null hypothesis. (correct answer)

Explanation: When you encounter test cross data with unequal phenotype frequencies, you're looking at evidence for gene linkage and need to calculate recombination frequency to determine map distance. In this test cross, the heterozygous parent (AB/ab) can produce four types of gametes. The data shows AB (880) and ab (860) are the most frequent classes, totaling 1,740 offspring. These represent the parental types - the original gene combinations that were linked together on the same chromosome. The less frequent classes, Ab (140) and aB (120), totaling 260 offspring, are recombinant types produced by crossing over between the two gene loci. The recombination frequency equals the number of recombinants divided by total offspring: $\frac{260}{2000} = 0.13 = 13\%$. Since 1 map unit (centimorgan) equals 1% recombination frequency, the map distance is 13 cM. Answer A is incorrect because the unequal distribution of phenotypes clearly indicates linkage, not independent assortment. If genes were unlinked, you'd expect roughly equal frequencies (500 each). Answer B miscalculates by using the parental frequency (87%) instead of recombination frequency (13%) - a common error. Answer C is wrong because test cross data is actually the ideal method for estimating recombination frequency, as it directly reveals the gamete types produced by the heterozygous parent. The chi-square test would reject the null hypothesis of independent assortment because the observed ratios deviate significantly from the expected 1:1:1:1 ratio. Study tip: Always identify parental vs. recombinant classes first - the least frequent classes are usually the recombinants, and their proportion gives you the map distance directly.

Question 2

1. 3.5 cM
2. 7.0 cM (correct answer)
3. 14.0 cM
4. 28.0 cM

Explanation: In ordered tetrad analysis, the map distance between a gene and its centromere is calculated based on the frequency of second-division segregation (MII) patterns. A second-division segregation occurs when there is a crossover between the gene and the centromere. The formula for the map distance is: Distance (cM) = (1/2) × (% of MII asci). First, calculate the percentage of MII asci: (140 / 1000) × 100% = 14%. Then, apply the formula: Distance = (1/2) × 14% = 7.0 cM. The 1/2 factor is included because only half of the chromatids in an MII ascus are actually recombinant.

Question 3

1. 0%
2. 3.75%
3. 7.50% (correct answer)
4. 15.0%

Explanation: First, determine how the 'pqr' gamete is formed from the parental chromosomes (PqR and pQr). Comparing 'pqr' to the parental chromosome 'pQr', we see that the alleles for Q and R have been exchanged. This requires a single crossover (SCO) event between genes Q and R. Complete interference (I=1) means that no double crossovers (DCOs) occur. The map distance between Q and R (15 cM) represents the frequency of all crossovers in that region (SCOs + DCOs). Since DCOs = 0, the frequency of SCOs between Q and R is 0.15. This frequency accounts for two reciprocal gametes: pqr (from the pQr chromosome) and PQR (from the PqR chromosome). Therefore, the frequency of the 'pqr' gamete alone is half of the total SCO frequency, which is 0.15 / 2 = 0.075, or 7.50%. A common trap is to assume the parents were PQR/pqr, in which case pqr would be parental.

Question 4

1. The coefficient of coincidence between the genes.
2. The recombination frequency between X and Z. (correct answer)
3. The physical distance in kilobases between Y and Z.
4. The results from a cross involving a fourth linked gene.

Explanation: The given data allows for two possible gene orders. If Y is in the middle, the order is X--25--Y--30--Z, and the distance between X and Z would be 55 cM. If X is in the middle, the order is Y--25--X--5--Z, requiring the distance between Y and Z to be 30 cM (which it is) and the distance between X and Z to be 5 cM. A third possibility is Z in the middle: X--25--Y and X--??-Z--??--Y. To distinguish between these possibilities (e.g., Y in the middle vs. X in the middle), the most direct and crucial piece of information is the recombination frequency between the remaining pair of genes, X and Z. If RF(X-Z) is approximately 50% (the maximum observable for a 55 cM distance), then Y is in the middle. If RF(X-Z) is approximately 5%, then X is in the middle.

Question 5

1. The LOD score is 5.0, indicating the genes are 5 cM apart.
2. The LOD score is 100,000, providing conclusive evidence of linkage.
3. The LOD score is 2.0, providing suggestive but not conclusive evidence of linkage.
4. The LOD score is 5.0, providing very strong evidence of linkage. (correct answer)

Explanation: When you encounter LOD score questions in genetics, you're dealing with linkage analysis - a statistical method to determine if genes are inherited together more often than expected by chance. The LOD score (logarithm of the odds) is calculated as: \text{LOD} = \log_{10}\left(\frac{\text{likelihood at θ}}{\text{likelihood at θ = 0.5}\right) Here, the likelihood is 100,000 times higher at 5% recombination than at 50% recombination (independent assortment). So: $\text{LOD} = \log_{10}(100,000) = \log_{10}(10^5) = 5.0$ This LOD score of 5.0 provides very strong evidence of linkage. In genetics, LOD scores ≥ 3.0 indicate strong evidence for linkage (odds of at least 1000:1 in favor), while scores ≥ 2.0 but < 3.0 suggest linkage but aren't considered conclusive. Answer A incorrectly assumes the LOD score directly equals the map distance in centimorgans - while both happen to be 5.0 here, this is coincidental. The recombination frequency (5%) approximates the map distance (5 cM), but the LOD score measures statistical confidence, not distance. Answer B confuses the likelihood ratio (100,000) with the LOD score itself. The LOD score is the logarithm of this ratio. Answer C correctly calculates a logarithm but uses the wrong base or misreads the data, arriving at 2.0 instead of 5.0. Remember: LOD scores measure statistical confidence in linkage, not genetic distance. Always convert likelihood ratios to logarithms, and memorize the interpretation thresholds (≥3.0 = strong evidence, ≥2.0 = suggestive).

Question 6

1. C = 0.67; it indicates positive interference, where one crossover suppresses nearby crossovers. (correct answer)
2. C = 1.50; it indicates negative interference, where one crossover promotes nearby crossovers.
3. C = 0.01; it indicates very strong positive interference.
4. C = 0.33; it indicates positive interference, where one crossover suppresses nearby crossovers.

Explanation: First, calculate the expected frequency of double crossovers (DCOs) assuming no interference. This is the product of the recombination frequencies in the two adjacent regions: Expected DCO = RF(A-B) × RF(B-C) = 0.10 × 0.15 = 0.015. The observed frequency of DCOs is given as 1.0% or 0.01. The coefficient of coincidence (C) is the ratio of the observed DCO frequency to the expected DCO frequency: C = Observed DCO / Expected DCO = 0.01 / 0.015 ≈ 0.67. Interference (I) is calculated as I = 1 - C. In this case, I = 1 - 0.67 = 0.33. Since C is less than 1 (and I is positive), this indicates positive interference, meaning that a crossover event in one region reduces the probability of a second crossover event in the adjacent region.

Question 7

1. 15.0 cM
2. 17.5 cM (correct answer)
3. 30.0 cM
4. 35.0 cM

Explanation: For unordered tetrads, the presence of far more Parental Ditype (PD) tetrads than Non-Parental Ditype (NPD) tetrads (700 >> 50) indicates that the two genes are linked. The map distance can be estimated using the formula: Distance (cM) = [ (1/2 * T) + NPD ] / Total Tetrads * 100. Plugging in the given values: Distance = [ (1/2 * 250) + 50 ] / 1000 * 100 = [ 125 + 50 ] / 1000 * 100 = 175 / 1000 * 100 = 17.5 cM. Other formulas are incorrect; for example, simply summing the recombinant tetrads (T+NPD) or failing to multiply T by 1/2 are common errors.

Question 8

1. High positive interference is suppressing crossovers in that region.
2. The gene order was determined incorrectly.
3. One of the recombinant genotypes has reduced viability or penetrance. (correct answer)
4. The parental F1 generation had an unexpected linkage phase.

Explanation: Reciprocal crossover events are generally expected to produce progeny in equal frequencies. A significant, consistent deviation from a 1:1 ratio for a pair of reciprocal classes (e.g., one SCO class and its counterpart) strongly suggests that the resulting genotypes are not equally viable or that the phenotypes are not equally expressed (penetrance). The genotype corresponding to the lower count (45) likely has a negative effect on survival. Interference affects the frequency of double crossovers relative to single crossovers, but not the balance between reciprocal classes. Incorrect gene order or linkage phase would re-categorize which classes are parental, SCO, and DCO, but it would not explain why one specific recombinant class is much rarer than its reciprocal counterpart.

Question 9

1. 0%
2. 3.75%
3. 7.50% (correct answer)
4. 15.0%

Explanation: First, determine how the 'pqr' gamete is formed from the parental chromosomes (PqR and pQr). Comparing 'pqr' to the parental chromosome 'pQr', we see that the alleles for Q and R have been exchanged. This requires a single crossover (SCO) event between genes Q and R. Complete interference (I=1) means that no double crossovers (DCOs) occur. The map distance between Q and R (15 cM) represents the frequency of all crossovers in that region (SCOs + DCOs). Since DCOs = 0, the frequency of SCOs between Q and R is 0.15. This frequency accounts for two reciprocal gametes: pqr (from the pQr chromosome) and PQR (from the PqR chromosome). Therefore, the frequency of the 'pqr' gamete alone is half of the total SCO frequency, which is 0.15 / 2 = 0.075, or 7.50%. A common trap is to assume the parents were PQR/pqr, in which case pqr would be parental.

Question 10

1. 10%
2. Approximately 50% (correct answer)
3. 70%
4. It cannot be determined without knowing the interference value.

Explanation: Map distances between linked genes are additive. The map distance between A and C is 30 cM + 40 cM = 70 cM. However, recombination frequency is the observed frequency of recombinant progeny and is capped at 50%. This is because as the distance between genes increases, the likelihood of multiple crossover events (double, triple, etc.) also increases. These multiple crossovers can cancel each other out, resulting in a parental combination of alleles. Consequently, for genes that are far apart on the same chromosome (typically >50 cM), the observed recombination frequency approaches that of unlinked genes, which is 50%. A map distance of 70 cM will result in an observed recombination frequency of approximately 50%.

Question 11

1. 0.085 (correct answer)
2. 0.100
3. 0.170
4. 0.200

Explanation: The gamete d E f is a single crossover (SCO) between genes D and E. The map distance of 20 cM represents the total frequency of all crossovers in that region, which includes both SCO events and double crossover (DCO) events. First, calculate the expected frequency of DCOs: Expected DCO = RF(D-E) × RF(E-F) = 0.20 × 0.30 = 0.06. Next, calculate the observed DCO frequency using the coefficient of coincidence (C): Observed DCO = C × Expected DCO = 0.5 × 0.06 = 0.03. The total recombination frequency in the D-E region (20% or 0.20) is the sum of SCOs in that region (SCO1) and DCOs. So, Freq(SCO1) + Freq(DCO) = 0.20. Therefore, Freq(SCO1) = 0.20 - Freq(DCO) = 0.20 - 0.03 = 0.17. This 0.17 frequency represents both reciprocal SCO1 gametes (d E f and D e F). The frequency of the specific d E f gamete is half of this value: 0.17 / 2 = 0.085.

Question 12

1. The LOD score is 5.0, indicating the genes are 5 cM apart.
2. The LOD score is 100,000, providing conclusive evidence of linkage.
3. The LOD score is 2.0, providing suggestive but not conclusive evidence of linkage.
4. The LOD score is 5.0, providing very strong evidence of linkage. (correct answer)

Explanation: When you encounter LOD score questions in genetics, you're dealing with linkage analysis - a statistical method to determine if genes are inherited together more often than expected by chance. The LOD score (logarithm of the odds) is calculated as: \text{LOD} = \log_{10}\left(\frac{\text{likelihood at θ}}{\text{likelihood at θ = 0.5}\right) Here, the likelihood is 100,000 times higher at 5% recombination than at 50% recombination (independent assortment). So: $\text{LOD} = \log_{10}(100,000) = \log_{10}(10^5) = 5.0$ This LOD score of 5.0 provides very strong evidence of linkage. In genetics, LOD scores ≥ 3.0 indicate strong evidence for linkage (odds of at least 1000:1 in favor), while scores ≥ 2.0 but < 3.0 suggest linkage but aren't considered conclusive. Answer A incorrectly assumes the LOD score directly equals the map distance in centimorgans - while both happen to be 5.0 here, this is coincidental. The recombination frequency (5%) approximates the map distance (5 cM), but the LOD score measures statistical confidence, not distance. Answer B confuses the likelihood ratio (100,000) with the LOD score itself. The LOD score is the logarithm of this ratio. Answer C correctly calculates a logarithm but uses the wrong base or misreads the data, arriving at 2.0 instead of 5.0. Remember: LOD scores measure statistical confidence in linkage, not genetic distance. Always convert likelihood ratios to logarithms, and memorize the interpretation thresholds (≥3.0 = strong evidence, ≥2.0 = suggestive).

Question 13

1. 0.12
2. 0.24
3. 0.38 (correct answer)
4. 0.76

Explanation: The F1 parent has genotype Jk/jK, meaning the parental gametes are Jk and jK. Progeny with the Jk phenotype receive a Jk gamete from the F1 parent. Since Jk is a parental gamete, its frequency will be part of the non-recombinant fraction. The total recombination frequency is 24% (0.24). This means the total frequency of parental (non-recombinant) gametes is 100% - 24% = 76% (0.76). This 76% is split equally between the two parental gametes, Jk and jK. Therefore, the expected frequency of the Jk gamete is 0.76 / 2 = 0.38. The progeny from a test cross directly reflect the gamete frequencies of the heterozygous parent.

Question 14

1. 15.0 cM
2. 17.5 cM (correct answer)
3. 30.0 cM
4. 35.0 cM

Explanation: For unordered tetrads, the presence of far more Parental Ditype (PD) tetrads than Non-Parental Ditype (NPD) tetrads (700 >> 50) indicates that the two genes are linked. The map distance can be estimated using the formula: Distance (cM) = [ (1/2 * T) + NPD ] / Total Tetrads * 100. Plugging in the given values: Distance = [ (1/2 * 250) + 50 ] / 1000 * 100 = [ 125 + 50 ] / 1000 * 100 = 175 / 1000 * 100 = 17.5 cM. Other formulas are incorrect; for example, simply summing the recombinant tetrads (T+NPD) or failing to multiply T by 1/2 are common errors.

Question 15

1. The coefficient of coincidence between the genes.
2. The recombination frequency between X and Z. (correct answer)
3. The physical distance in kilobases between Y and Z.
4. The results from a cross involving a fourth linked gene.

Explanation: The given data allows for two possible gene orders. If Y is in the middle, the order is X--25--Y--30--Z, and the distance between X and Z would be 55 cM. If X is in the middle, the order is Y--25--X--5--Z, requiring the distance between Y and Z to be 30 cM (which it is) and the distance between X and Z to be 5 cM. A third possibility is Z in the middle: X--25--Y and X--??-Z--??--Y. To distinguish between these possibilities (e.g., Y in the middle vs. X in the middle), the most direct and crucial piece of information is the recombination frequency between the remaining pair of genes, X and Z. If RF(X-Z) is approximately 50% (the maximum observable for a 55 cM distance), then Y is in the middle. If RF(X-Z) is approximately 5%, then X is in the middle.

Question 16

1. High positive interference is suppressing crossovers in that region.
2. The gene order was determined incorrectly.
3. One of the recombinant genotypes has reduced viability or penetrance. (correct answer)
4. The parental F1 generation had an unexpected linkage phase.

Explanation: Reciprocal crossover events are generally expected to produce progeny in equal frequencies. A significant, consistent deviation from a 1:1 ratio for a pair of reciprocal classes (e.g., one SCO class and its counterpart) strongly suggests that the resulting genotypes are not equally viable or that the phenotypes are not equally expressed (penetrance). The genotype corresponding to the lower count (45) likely has a negative effect on survival. Interference affects the frequency of double crossovers relative to single crossovers, but not the balance between reciprocal classes. Incorrect gene order or linkage phase would re-categorize which classes are parental, SCO, and DCO, but it would not explain why one specific recombinant class is much rarer than its reciprocal counterpart.

Question 17

1. The genes are unlinked, and the chi-square test would likely fail to reject the null hypothesis (p > 0.05).
2. The genes are linked with a map distance of 87 cM, and the chi-square test would reject the null hypothesis.
3. The genes are linked, but the recombination frequency cannot be estimated from test cross data.
4. The genes are linked with a map distance of 13 cM, and the chi-square test would reject the null hypothesis. (correct answer)

Explanation: When you encounter test cross data with unequal phenotype frequencies, you're looking at evidence for gene linkage and need to calculate recombination frequency to determine map distance. In this test cross, the heterozygous parent (AB/ab) can produce four types of gametes. The data shows AB (880) and ab (860) are the most frequent classes, totaling 1,740 offspring. These represent the parental types - the original gene combinations that were linked together on the same chromosome. The less frequent classes, Ab (140) and aB (120), totaling 260 offspring, are recombinant types produced by crossing over between the two gene loci. The recombination frequency equals the number of recombinants divided by total offspring: $\frac{260}{2000} = 0.13 = 13\%$. Since 1 map unit (centimorgan) equals 1% recombination frequency, the map distance is 13 cM. Answer A is incorrect because the unequal distribution of phenotypes clearly indicates linkage, not independent assortment. If genes were unlinked, you'd expect roughly equal frequencies (500 each). Answer B miscalculates by using the parental frequency (87%) instead of recombination frequency (13%) - a common error. Answer C is wrong because test cross data is actually the ideal method for estimating recombination frequency, as it directly reveals the gamete types produced by the heterozygous parent. The chi-square test would reject the null hypothesis of independent assortment because the observed ratios deviate significantly from the expected 1:1:1:1 ratio. Study tip: Always identify parental vs. recombinant classes first - the least frequent classes are usually the recombinants, and their proportion gives you the map distance directly.

Question 18

1. 3.5 cM
2. 7.0 cM (correct answer)
3. 14.0 cM
4. 28.0 cM

Explanation: In ordered tetrad analysis, the map distance between a gene and its centromere is calculated based on the frequency of second-division segregation (MII) patterns. A second-division segregation occurs when there is a crossover between the gene and the centromere. The formula for the map distance is: Distance (cM) = (1/2) × (% of MII asci). First, calculate the percentage of MII asci: (140 / 1000) × 100% = 14%. Then, apply the formula: Distance = (1/2) × 14% = 7.0 cM. The 1/2 factor is included because only half of the chromatids in an MII ascus are actually recombinant.

Question 19

1. C = 0.67; it indicates positive interference, where one crossover suppresses nearby crossovers. (correct answer)
2. C = 1.50; it indicates negative interference, where one crossover promotes nearby crossovers.
3. C = 0.01; it indicates very strong positive interference.
4. C = 0.33; it indicates positive interference, where one crossover suppresses nearby crossovers.

Explanation: First, calculate the expected frequency of double crossovers (DCOs) assuming no interference. This is the product of the recombination frequencies in the two adjacent regions: Expected DCO = RF(A-B) × RF(B-C) = 0.10 × 0.15 = 0.015. The observed frequency of DCOs is given as 1.0% or 0.01. The coefficient of coincidence (C) is the ratio of the observed DCO frequency to the expected DCO frequency: C = Observed DCO / Expected DCO = 0.01 / 0.015 ≈ 0.67. Interference (I) is calculated as I = 1 - C. In this case, I = 1 - 0.67 = 0.33. Since C is less than 1 (and I is positive), this indicates positive interference, meaning that a crossover event in one region reduces the probability of a second crossover event in the adjacent region.

Question 20

1. 10%
2. Approximately 50% (correct answer)
3. 70%
4. It cannot be determined without knowing the interference value.

Explanation: Map distances between linked genes are additive. The map distance between A and C is 30 cM + 40 cM = 70 cM. However, recombination frequency is the observed frequency of recombinant progeny and is capped at 50%. This is because as the distance between genes increases, the likelihood of multiple crossover events (double, triple, etc.) also increases. These multiple crossovers can cancel each other out, resulting in a parental combination of alleles. Consequently, for genes that are far apart on the same chromosome (typically >50 cM), the observed recombination frequency approaches that of unlinked genes, which is 50%. A map distance of 70 cM will result in an observed recombination frequency of approximately 50%.