Genetics Quiz: Map Distance Calculations

What this quiz covers

This quiz focuses on Map Distance Calculations, 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. 0.20
2. 0.40 (correct answer)
3. 0.60
4. 0.80

Explanation: First, calculate the expected frequency of double crossovers (DCOs) based on the map distances, assuming no interference. The probability of a crossover between X and Y is 0.20, and between Y and Z is 0.10. Expected DCO frequency = (distance X-Y) * (distance Y-Z) = 0.20 * 0.10 = 0.02. Second, calculate the expected number of DCO progeny in a sample of 2000: Expected DCO number = 0.02 * 2000 = 40. Third, determine the observed frequency of DCOs from the data provided: Observed DCO number = 24. Fourth, calculate the coefficient of coincidence (C), which is the ratio of observed to expected DCOs: C = Observed DCOs / Expected DCOs = 24 / 40 = 0.60. Finally, calculate interference (I), which is I = 1 - C: I = 1 - 0.60 = 0.40. Distractors explained: 0.60 is the coefficient of coincidence, not interference. 0.20 could result from a calculation error (e.g., 24/40 = 0.6, then 1-0.2-0.1=0.7 or some other confusion). 0.80 might come from 1 - 0.20 = 0.8.

Question 2

1. 2.75%
2. 47.25% (correct answer)
3. 49.72%
4. 94.50%

Explanation: Male progeny receive their single X chromosome from their mother, so their phenotype directly reflects the genotype of her gametes. We need to find the frequency of the + + + gamete.

1. The + + + gamete is a parental (non-recombinant) type, as the female's genotype is y w ec / + + +.
2. The total frequency of recombinant gametes is the sum of single crossovers (SCOs) and double crossovers (DCOs). The map distance between the outer genes (y and ec) is 1.5 + 4.0 = 5.5 cM. This distance represents the frequency of all crossover events between them.
3. The total frequency of all recombinant gametes is 5.5% (or 0.055). The total frequency of parental gametes is therefore 1 - 0.055 = 0.945 (or 94.5%).
4. There are two parental gametes (y w ec and + + +), which are produced in equal frequency. Therefore, the frequency of the + + + gamete is half of the total parental frequency: 0.945 / 2 = 0.4725, or 47.25%.

Question 3

1. 2.75%
2. 47.25% (correct answer)
3. 49.72%
4. 94.50%

Explanation: Male progeny receive their single X chromosome from their mother, so their phenotype directly reflects the genotype of her gametes. We need to find the frequency of the + + + gamete.

1. The + + + gamete is a parental (non-recombinant) type, as the female's genotype is y w ec / + + +.
2. The total frequency of recombinant gametes is the sum of single crossovers (SCOs) and double crossovers (DCOs). The map distance between the outer genes (y and ec) is 1.5 + 4.0 = 5.5 cM. This distance represents the frequency of all crossover events between them.
3. The total frequency of all recombinant gametes is 5.5% (or 0.055). The total frequency of parental gametes is therefore 1 - 0.055 = 0.945 (or 94.5%).
4. There are two parental gametes (y w ec and + + +), which are produced in equal frequency. Therefore, the frequency of the + + + gamete is half of the total parental frequency: 0.945 / 2 = 0.4725, or 47.25%.

Question 4

1. 7.2%
2. 9.1% (correct answer)
3. 10.0%
4. 18.2%

Explanation: The gamete r S t is formed by a single crossover (SCO) in the interval between S and T. The frequency of SCOs is affected by the occurrence of double crossovers (DCOs).

1. First, calculate the frequency of observed DCOs.
   • Expected DCO frequency = (distance R-S) × (distance S-T) = 0.15 × 0.20 = 0.03.
   • Observed DCO frequency = Expected DCO frequency × Coefficient of Coincidence (C) = 0.03 × 0.6 = 0.018.
2. The map distance (e.g., 20 cM for S-T) represents the total frequency of crossovers in that interval, which includes both SCOs and DCOs. So, Freq(SCO in S-T) + Freq(DCO) = 0.20.
3. The frequency of single crossovers only in the S-T interval is: Freq(SCO_ST only) = map distance(S-T) - Freq(observed DCO) = 0.20 - 0.018 = 0.182.
4. This frequency of 0.182 (or 18.2%) represents both types of SCO gametes in this region (R s T and r S t). Since they are produced in equal numbers, the frequency of the r S t gamete is half of this value. Freq(r S t) = 0.182 / 2 = 0.091, or 9.1%.

Question 5

1. F and G
2. G and H
3. F and H (correct answer)
4. All pairs are equidistant.

Explanation: The pair of genes furthest apart will be the two outer genes on the chromosome. To find them, we must first determine the gene order.

1. Identify parental classes (most frequent): fgh (315) and FGH (305).
2. Identify double crossover classes (least frequent): fgH (16) and FGh (14).
3. Compare a parental genotype (FGH) with a DCO genotype (fgH). The G allele remains the same as the parental, while F and H have recombined. This indicates that G is the middle gene. The order is F-G-H.

• Dist(F-G) = (SCO_FG + DCO) / Total = (fGH(80)+Fgh(70) + fgH(16)+FGh(14)) / 1000 * 100 = 18.0 cM.
• Dist(G-H) = (SCO_GH + DCO) / Total = (fGh(105)+FgH(95) + fgH(16)+FGh(14)) / 1000 * 100 = 23.0 cM.
• The distance between F and H is the sum of these distances: 18.0 + 23.0 = 41.0 cM, which is the largest distance.

Question 6

1. 7.2%
2. 9.1% (correct answer)
3. 10.0%
4. 18.2%

Explanation: The gamete r S t is formed by a single crossover (SCO) in the interval between S and T. The frequency of SCOs is affected by the occurrence of double crossovers (DCOs).

1. First, calculate the frequency of observed DCOs.
   • Expected DCO frequency = (distance R-S) × (distance S-T) = 0.15 × 0.20 = 0.03.
   • Observed DCO frequency = Expected DCO frequency × Coefficient of Coincidence (C) = 0.03 × 0.6 = 0.018.
2. The map distance (e.g., 20 cM for S-T) represents the total frequency of crossovers in that interval, which includes both SCOs and DCOs. So, Freq(SCO in S-T) + Freq(DCO) = 0.20.
3. The frequency of single crossovers only in the S-T interval is: Freq(SCO_ST only) = map distance(S-T) - Freq(observed DCO) = 0.20 - 0.018 = 0.182.
4. This frequency of 0.182 (or 18.2%) represents both types of SCO gametes in this region (R s T and r S t). Since they are produced in equal numbers, the frequency of the r S t gamete is half of this value. Freq(r S t) = 0.182 / 2 = 0.091, or 9.1%.

Question 7

1. 9.0 cM
2. 44.4 cM
3. 36.0 cM
4. 18.0 cM (correct answer)

Explanation: When you encounter ordered asci analysis in fungi like Sordaria, you're dealing with gene mapping relative to the centromere. The key insight is that crossing over between a gene and its centromere creates a distinctive pattern in the ascospores. In this cross between tan spores (tn) and wild-type black spores (tn+), the 4:4 pattern (128 asci) represents no crossing over between the tn gene and centromere. The 2:4:2 and 2:2:2:2 patterns (72 asci) indicate crossing over occurred between the gene and centromere, causing the alleles to segregate differently. To calculate map distance: $\text{Map distance} = \frac{\text{Number of crossover asci}}{Total asci} \times \frac{1}{2} \times 100$ The factor of 1/2 is crucial because only half the chromatids in each tetrad that undergoes crossing over will show recombination. $\text{Map distance} = \frac{72}{200} \times \frac{1}{2} \times 100 = 0.36 \times \frac{1}{2} \times 100 = 18.0 \text{ cM}$ This confirms answer D) 18.0 cM. A) 9.0 cM incorrectly applies an additional factor of 1/2, double-counting the correction. B) 44.4 cM appears to miscount the asci types or use an incorrect formula. C) 36.0 cM calculates the percentage of crossover asci but forgets the essential 1/2 correction factor. Study tip: Always remember the 1/2 factor in fungal mapping—only half the spores in a crossover ascus are actually recombinant for the gene-centromere distance.

Question 8

1. 45%
2. 50% (correct answer)
3. 90%
4. 100%

Explanation: Map distance is calculated by summing recombination frequencies over short intervals to account for double crossovers, and thus can exceed 50 cM. However, the observable recombination frequency between two genes in a single cross has a theoretical maximum of 50%. This value is reached when the genes are on different chromosomes (independent assortment) or when they are so far apart on the same chromosome that at least one crossover event occurs between them in every meiotic event. A map distance of 90 cM indicates the genes are very far apart on the same chromosome. Therefore, they will assort independently, and the recombination frequency observed in the progeny will be 50%.

Question 9

1. 9.0 cM
2. 59.0 cM
3. 36.0 cM
4. 18.0 cM (correct answer)

Explanation: When you encounter test cross problems involving linked genes, you're analyzing recombination frequency to determine map distance. The key insight is that map units (centimorgans) directly equal the percentage of recombinant offspring. In this problem, you have an F1 parent with genotype AB/ab crossed with ab/ab. The parental gamete types are AB and ab, while the recombinant types are Ab and aB. You're told that 410 offspring came from AB gametes, which means another 410 came from ab gametes (since crossovers produce recombinants equally). This accounts for 820 parental types out of 1000 total progeny. The remaining offspring are recombinants: $1000 - 820 = 180$ recombinants. The recombination frequency is $\frac{180}{1000} = 0.18 = 18\%$. Since 1 cM = 1% recombination frequency, the map distance is 18.0 cM. Looking at the wrong answers: A) 9.0 cM represents half the recombination frequency—you might get this if you incorrectly calculated only one type of recombinant instead of both. B) 59.0 cM likely comes from mistakenly using parental types as recombinants (820/1000 ≈ 82%, then subtracting from 100% gives 18%, but somehow arriving at 59%). C) 36.0 cM is double the correct answer—this could result from counting recombinants twice or miscalculating the total. Remember: in linkage problems, always identify parental vs. recombinant classes first, then calculate recombination frequency as (total recombinants ÷ total offspring) × 100. This percentage directly equals map distance in centimorgans.

Question 10

1. 4 cM
2. 12 cM (correct answer)
3. 16 cM
4. 20 cM

Explanation: First, determine the gene order by testing which arrangement makes the distances additive. If the order were x-y-z, then x-z distance should equal x-y + y-z = 12% + 16% = 28%, but the observed x-z distance is only 4%. If the order is z-x-y, then z-y distance should equal z-x + x-y = 4% + 12% = 16%, which matches the observed z-y distance of 16%. Therefore, the gene order is z-x-y, making x the middle gene. The map distance between gene y and the middle gene (x) is 12 cM.

Question 11

1. 8
2. 192 (correct answer)
3. 200
4. 384

Explanation: The genotype Abc results from a single crossover (SCO) between genes A and B.

1. First, calculate the expected and observed rates of double crossovers (DCOs). The coefficient of coincidence C = 1 - I = 1 - 0.6 = 0.4.
   • Expected DCO rate = (Dist A-B) × (Dist B-C) = 0.20 × 0.10 = 0.02.
   • Observed DCO rate = Expected DCO rate × C = 0.02 × 0.4 = 0.008.
2. The map distance for A-B (20 cM) represents all crossovers in that region (SCOs + DCOs). To find the rate of SCOs only in that region, we subtract the observed DCO rate.
   • Rate of SCO in region A-B = (Map distance A-B) - (Observed DCO rate) = 0.20 - 0.008 = 0.192.
3. This rate (0.192) represents both reciprocal products of single crossovers in this region (Abc and aBC). The frequency of the Abc class alone is half of this value: 0.192 / 2 = 0.096.
4. Finally, calculate the expected number of Abc progeny out of 2000:
   • Number = 0.096 × 2000 = 192.

Question 12

1. 18.0 cM (correct answer)
2. 36.0 cM
3. 41.5 cM
4. 82.0 cM

Explanation: First, determine the linkage phase of the PpRr parent. It was produced from a cross of purple, wrinkled (PP rr) and yellow, round (pp RR). Therefore, the F1 parent received a Pr chromosome from one parent and a pR chromosome from the other. Its genotype is in repulsion phase: Pr/pR. Second, identify the parental and recombinant classes in the test cross progeny.

• Parental progeny have phenotypes corresponding to the non-recombinant gametes Pr and pR. These are purple, wrinkled (810) and yellow, round (830).
• Recombinant progeny have phenotypes corresponding to the crossover gametes PR and pr. These are purple, round (185) and yellow, wrinkled (175).

• Total progeny = 810 + 830 + 185 + 175 = 2000.
• Total recombinants = 185 + 175 = 360.
• Recombination Frequency = (Recombinants / Total) * 100 = (360 / 2000) * 100 = 18%. The map distance is 18.0 cM.

Question 13

1. Genes A and B are linked and are 48 cM apart.
2. Genes A and B are assorting independently.
3. Genes A and B are either on different chromosomes or are located far apart on the same chromosome. (correct answer)
4. The experiment had insufficient progeny to distinguish linkage from independent assortment.

Explanation: A recombination frequency of 50% indicates independent assortment. A value that is statistically indistinguishable from 50% (such as 48% in a real experiment) also suggests independent assortment. Independent assortment occurs under two conditions: either the genes are on different (non-homologous) chromosomes, or they are located very far apart on the same chromosome. Therefore, observing 48% recombination does not allow one to distinguish between these two possibilities without further tests. Choice A is too strong a conclusion; while the genes could be 48 cM apart, they could also be 100 cM apart or on different chromosomes, both of which would yield RF values near 50%. Choice B is a possible interpretation, but C is more complete because it states the two physical scenarios that lead to independent assortment. Choice D is incorrect as a high RF is a valid result, not necessarily an artifact of sample size.

Question 14

1. 8 cM
2. 12 cM
3. 20 cM
4. 32 cM (correct answer)

Explanation: There are two possible linear arrangements for the three genes based on the given information. Case 1: Gene B is located between genes A and C. The order is A -- B -- C. In this case, the distance between A and C is additive: Distance(A-C) = Distance(A-B) + Distance(B-C) = 12 cM + 20 cM = 32 cM. Case 2: Gene A is located between genes C and B. The order is C -- A -- B. In this case, the distance between C and B would be the sum of the other two distances: Distance(C-B) = Distance(C-A) + Distance(A-B). We are given Dist(B-C) is 20 cM and Dist(A-B) is 12 cM. So, 20 cM = Distance(C-A) + 12 cM. This gives Distance(C-A) = 8 cM. The two possible map distances between A and C are 32 cM and 8 cM. The question asks for the maximum possible distance, which is 32 cM.

Question 15

1. 50.0%
2. 44.0%
3. 38.0% (correct answer)
4. 6.0%

Explanation: Recombination between the outer genes L and N occurs when there is a single crossover in region I (L-M) or a single crossover in region II (M-N). A double crossover (one in region I and one in region II) results in a parental combination of alleles for L and N, so it does not contribute to the recombination frequency between them.

1. Calculate the frequency of double crossovers (DCOs). With no interference (I=0), the expected frequency is the product of the interval distances.
   • Freq(DCO) = Freq(CO in L-M) * Freq(CO in M-N) = 0.20 * 0.30 = 0.06.
2. The frequency of single crossovers only in region I is Freq(SCO_I) = Dist(L-M) - Freq(DCO) = 0.20 - 0.06 = 0.14.
3. The frequency of single crossovers only in region II is Freq(SCO_II) = Dist(M-N) - Freq(DCO) = 0.30 - 0.06 = 0.24.
4. The recombination frequency between L and N is the sum of the frequencies of single crossovers in the intervening regions.
   • RF(L,N) = Freq(SCO_I) + Freq(SCO_II) = 0.14 + 0.24 = 0.38, or 38.0%.

Question 16

1. 4 cM
2. 12 cM (correct answer)
3. 16 cM
4. 20 cM

Explanation: First, determine the gene order by testing which arrangement makes the distances additive. If the order were x-y-z, then x-z distance should equal x-y + y-z = 12% + 16% = 28%, but the observed x-z distance is only 4%. If the order is z-x-y, then z-y distance should equal z-x + x-y = 4% + 12% = 16%, which matches the observed z-y distance of 16%. Therefore, the gene order is z-x-y, making x the middle gene. The map distance between gene y and the middle gene (x) is 12 cM.

Question 17

1. 9.0 cM
2. 44.4 cM
3. 36.0 cM
4. 18.0 cM (correct answer)

Explanation: When you encounter ordered asci analysis in fungi like Sordaria, you're dealing with gene mapping relative to the centromere. The key insight is that crossing over between a gene and its centromere creates a distinctive pattern in the ascospores. In this cross between tan spores (tn) and wild-type black spores (tn+), the 4:4 pattern (128 asci) represents no crossing over between the tn gene and centromere. The 2:4:2 and 2:2:2:2 patterns (72 asci) indicate crossing over occurred between the gene and centromere, causing the alleles to segregate differently. To calculate map distance: $\text{Map distance} = \frac{\text{Number of crossover asci}}{Total asci} \times \frac{1}{2} \times 100$ The factor of 1/2 is crucial because only half the chromatids in each tetrad that undergoes crossing over will show recombination. $\text{Map distance} = \frac{72}{200} \times \frac{1}{2} \times 100 = 0.36 \times \frac{1}{2} \times 100 = 18.0 \text{ cM}$ This confirms answer D) 18.0 cM. A) 9.0 cM incorrectly applies an additional factor of 1/2, double-counting the correction. B) 44.4 cM appears to miscount the asci types or use an incorrect formula. C) 36.0 cM calculates the percentage of crossover asci but forgets the essential 1/2 correction factor. Study tip: Always remember the 1/2 factor in fungal mapping—only half the spores in a crossover ascus are actually recombinant for the gene-centromere distance.

Question 18

1. 8 cM
2. 12 cM
3. 20 cM
4. 32 cM (correct answer)

Explanation: There are two possible linear arrangements for the three genes based on the given information. Case 1: Gene B is located between genes A and C. The order is A -- B -- C. In this case, the distance between A and C is additive: Distance(A-C) = Distance(A-B) + Distance(B-C) = 12 cM + 20 cM = 32 cM. Case 2: Gene A is located between genes C and B. The order is C -- A -- B. In this case, the distance between C and B would be the sum of the other two distances: Distance(C-B) = Distance(C-A) + Distance(A-B). We are given Dist(B-C) is 20 cM and Dist(A-B) is 12 cM. So, 20 cM = Distance(C-A) + 12 cM. This gives Distance(C-A) = 8 cM. The two possible map distances between A and C are 32 cM and 8 cM. The question asks for the maximum possible distance, which is 32 cM.

Question 19

1. 50.0%
2. 44.0%
3. 38.0% (correct answer)
4. 6.0%

Explanation: Recombination between the outer genes L and N occurs when there is a single crossover in region I (L-M) or a single crossover in region II (M-N). A double crossover (one in region I and one in region II) results in a parental combination of alleles for L and N, so it does not contribute to the recombination frequency between them.

1. Calculate the frequency of double crossovers (DCOs). With no interference (I=0), the expected frequency is the product of the interval distances.
   • Freq(DCO) = Freq(CO in L-M) * Freq(CO in M-N) = 0.20 * 0.30 = 0.06.
2. The frequency of single crossovers only in region I is Freq(SCO_I) = Dist(L-M) - Freq(DCO) = 0.20 - 0.06 = 0.14.
3. The frequency of single crossovers only in region II is Freq(SCO_II) = Dist(M-N) - Freq(DCO) = 0.30 - 0.06 = 0.24.
4. The recombination frequency between L and N is the sum of the frequencies of single crossovers in the intervening regions.
   • RF(L,N) = Freq(SCO_I) + Freq(SCO_II) = 0.14 + 0.24 = 0.38, or 38.0%.

Question 20

1. 10.0 cM (correct answer)
2. 45.0 cM
3. 50.0 cM
4. 90.0 cM

Explanation: First, determine the genotype and linkage phase of the F1 dihybrid. The parents were tall-pubescent (DDpp) and dwarf-smooth (ddPP). The F1 plant is thus DdPp, with the alleles in repulsion phase (genotype Dp/dP). Second, identify the parental and recombinant progeny from the test cross (Dp/dP x dp/dp). Parental progeny from the Dp/dP F1 are tall-pubescent (445) and dwarf-smooth (455). Recombinant progeny result from crossover gametes (DP and dp). These are tall-smooth (52) and dwarf-pubescent (48). Third, calculate the map distance. Total progeny = 445 + 455 + 52 + 48 = 1000. Total recombinants = 52 + 48 = 100. Recombination Frequency (RF) = (100/1000) × 100 = 10%. The map distance in centiMorgans equals the recombination frequency, thus 10.0 cM.