Genetics Quiz: Interference And Coefficient Of Coincidence
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Interference And Coefficient Of CoincidenceQuestion 1 of 20

In a three-point test cross analyzing genes P, Q, and R, the recombination frequency between P and Q is 12%, and between Q and R is 18%. The observed frequency of double crossovers is 1.512%. What is the value of the interference?

0.20
0.30
0.70
1.43
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Genetics Quiz

Genetics Quiz: Interference And Coefficient Of Coincidence

Practice Interference And Coefficient Of Coincidence in Genetics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Interference And Coefficient Of Coincidence, giving you a quick way to practice the rules, question types, and explanations that matter most for Genetics.

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Question 1

In a three-point test cross analyzing genes P, Q, and R, the recombination frequency between P and Q is 12%, and between Q and R is 18%. The observed frequency of double crossovers is 1.512%. What is the value of the interference?

  1. 0.20
  2. 0.30 (correct answer)
  3. 0.70
  4. 1.43

Explanation: Interference (I) is calculated as 1 - Coefficient of Coincidence (C). The Coefficient of Coincidence is the ratio of observed double crossover (DCO) frequency to expected DCO frequency.

  1. Calculate the expected DCO frequency: Expected DCO = RF(P-Q) × RF(Q-R) = 0.12 × 0.18 = 0.0216.
  2. The observed DCO frequency is given as 1.512%, or 0.01512.
  3. Calculate the Coefficient of Coincidence (C): C = Observed DCO / Expected DCO = 0.01512 / 0.0216 = 0.70.
  4. Calculate Interference (I): I = 1 - C = 1 - 0.70 = 0.30.

Question 2

If a geneticist observes complete (or total) positive interference for a set of three very closely linked genes, what is the expected coefficient of coincidence (C), and which progeny class from a test cross would be entirely absent?

  1. C = 0; the double-crossover class would be absent. (correct answer)
  2. C = 1; the parental class would be absent.
  3. C = 0; the single-crossover classes would be absent.
  4. C = 1; the double-crossover class would be absent.

Explanation: Complete positive interference means that the occurrence of one crossover completely prevents the occurrence of another crossover nearby. This is represented by an interference value (I) of 1. The coefficient of coincidence (C) is calculated as C = 1 - I. Therefore, C = 1 - 1 = 0. A C value of 0 means that the observed number of double crossovers is zero, so this progeny class would be entirely absent from the results of a cross.

Question 3

The map distances for three linked genes are A-15cM-B and B-25cM-C. If the coefficient of coincidence is 0.8 in this region, what is the expected frequency of double-crossover gametes produced by an individual with genotype ABC/abc?

  1. 3.75%
  2. 4.69%
  3. 3.00% (correct answer)
  4. 0.80%

Explanation: The frequency of double-crossover gametes is the observed double crossover frequency.

  1. Calculate the expected double crossover (DCO) frequency by multiplying the recombination frequencies of the two intervals: Expected DCO = 0.15 × 0.25 = 0.0375 (or 3.75%).
  2. Use the coefficient of coincidence (C) to find the observed DCO frequency: Observed DCO = C × Expected DCO = 0.8 × 0.0375 = 0.030 (or 3.00%).

Question 4

A coefficient of coincidence of 0.65 is calculated for three linked genes in Neurospora. Which of the following statements provides the most accurate interpretation of this value?

  1. The number of observed double crossovers was 65% of the number expected based on the individual crossover frequencies. (correct answer)
  2. A crossover in one region prevents a crossover in the adjacent region with 65% efficiency.
  3. The observed frequency of double crossovers was 35% lower than the frequency of single crossovers.
  4. The genetic distance between the outer two genes is 65 map units, indicating a high degree of recombination.

Explanation: The coefficient of coincidence (C) is the ratio of the observed frequency of double crossovers (DCOs) to the expected frequency of DCOs. A value of 0.65 means that the actual number of DCOs observed is 65% of the number that would be expected if the two crossover events occurred independently of each other. Choice B describes interference (I), which would be 1 - 0.65 = 0.35 or 35%. Choice C makes an irrelevant comparison to single crossovers. Choice D incorrectly equates the coefficient of coincidence with genetic distance.

Question 5

A student analyzes data from a three-point cross. They calculate the recombination frequency for region I as 20% and for region II as 30%. They correctly calculate the expected double crossover (DCO) frequency as 6%. However, upon observing only 3% DCOs in the progeny, they conclude that I = 1 - (0.06 / 0.03) = -1.0. What was the student's primary mistake?

  1. They should have added the recombination frequencies, not multiplied them.
  2. They inverted the ratio for the coefficient of coincidence calculation. (correct answer)
  3. They misidentified the parental and double-crossover classes in the raw data.
  4. They should have subtracted the observed DCO frequency from the expected DCO frequency.

Explanation: The student's calculation for interference (I) was I = 1 - C. Their calculation for C was Expected DCO / Observed DCO (0.06 / 0.03 = 2.0), which is incorrect. The correct formula for the coefficient of coincidence (C) is Observed DCO / Expected DCO. The correct calculation would be C = 0.03 / 0.06 = 0.5, leading to an interference of I = 1 - 0.5 = 0.5. The student inverted the ratio in the C calculation.

Question 6

A plant breeder is performing a trihybrid test cross. The three genes are linked in the order R-S-T. The map distance between R and S is 20 cM and between S and T is 20 cM. Interference in this region is 0.5. What is the expected frequency of parental gametes (R S T and r s t) from the trihybrid parent?

  1. 40%
  2. 34%
  3. 68%
  4. 62% (correct answer)

Explanation: When you encounter linked gene problems with interference, you need to account for how crossovers in one region affect crossovers in adjacent regions. This requires calculating both single and double crossover frequencies. Start by finding the recombination frequencies: R-S distance is 20 cM (20% recombination) and S-T distance is 20 cM (20% recombination). If genes were unlinked, you'd expect 4% double crossovers (0.20 × 0.20 = 0.04). However, interference = 0.5 means only half the expected double crossovers actually occur, so the actual double crossover frequency is 2%. Now calculate the gamete frequencies:

  • Single crossover between R-S: 20% - 2% = 18%
  • Single crossover between S-T: 20% - 2% = 18%
  • Double crossovers: 2%
  • Parental types: 100% - 18% - 18% - 2% = 62%
Therefore, parental gametes (RST and rst) appear at 62% frequency. Answer choice A (40%) ignores interference entirely and miscalculates recombinant frequencies. Answer choice B (34%) appears to incorrectly subtract total recombination from 100% without properly accounting for double crossovers. Answer choice C (68%) likely represents a calculation error where someone added instead of subtracted recombination frequencies. Study tip: Always remember that interference reduces double crossover frequency below the expected value (distance₁ × distance₂). The coefficient of coincidence equals (1 - interference), so multiply expected double crossovers by this value to get the actual frequency.

Question 7

Two genes, A and B, are 10 map units apart, and two other genes, C and D, are also 10 map units apart, but they are in different regions of the genome. A cross involving A and B shows an interference of 0.8, while a cross involving C and D shows an interference of 0.3 when measured with a third gene. What is a valid conclusion from this comparison?

  1. The A-B region must be located in heterochromatin, while the C-D region is in euchromatin.
  2. The measurement for the C-D region is more accurate because its interference value is lower.
  3. Fewer double crossovers are observed than expected in the A-B region compared to the C-D region. (correct answer)
  4. The actual physical distance between C and D must be greater than the physical distance between A and B.

Explanation: Interference (I) reflects the reduction in observed double crossovers (DCOs) compared to expected. A higher interference value means a greater reduction. An interference of 0.8 in the A-B region means the observed DCO frequency is only 1 - 0.8 = 0.2 (20%) of the expected frequency. An interference of 0.3 in the C-D region means the observed DCO frequency is 1 - 0.3 = 0.7 (70%) of the expected frequency. Therefore, there is a much stronger suppression of double crossovers in the A-B region than in the C-D region.

Question 8

The genetic map for three genes in tomato is d - 20 cM - m - 10 cM - p. In a test cross of a trihybrid plant (D M P / d m p), interference is 0.4. Out of 5,000 progeny scored, what is the expected number of individuals with the genotype d M p or D m P?

  1. 40
  2. 60 (correct answer)
  3. 80
  4. 100

Explanation: The genotypes d M p and D m P represent the double crossover (DCO) progeny. The number of DCO progeny depends on the expected DCO frequency and the interference.

  1. Calculate the expected DCO frequency: RF(d-m) × RF(m-p) = 0.20 × 0.10 = 0.02 (or 2%).
  2. Calculate the coefficient of coincidence (C) from interference (I): C = 1 - I = 1 - 0.4 = 0.6.
  3. Calculate the observed DCO frequency: Observed DCO = C × Expected DCO = 0.6 × 0.02 = 0.012 (or 1.2%).
  4. Calculate the number of DCO progeny: Number = Observed DCO frequency × Total progeny = 0.012 × 5,000 = 60.

Question 9

A geneticist calculates a coefficient of coincidence of 0.80 and an interference of 0.20. Which value is a more direct measure of the reduction in expected double crossovers, and which value reflects the proportion of expected double crossovers that actually occur?

  1. Reduction: 0.80; Proportion: 0.20
  2. Reduction: 0.20; Proportion: 0.80 (correct answer)
  3. Both measure reduction.
  4. Both measure proportion.

Explanation: This question tests the precise definitions of interference and coefficient of coincidence. The coefficient of coincidence (C = Obs/Exp) directly represents the fraction or proportion of expected double crossovers that are observed (in this case, 0.80 or 80%). Interference (I = 1 - C) represents the degree to which expected double crossovers are not observed, i.e., the reduction or inhibition (in this case, 0.20 or 20%).

Question 10

In a three-point test cross analyzing genes P, Q, and R, the recombination frequency between P and Q is 12%, and between Q and R is 18%. The observed frequency of double crossovers is 1.512%. What is the value of the interference?

  1. 0.20
  2. 0.30 (correct answer)
  3. 0.70
  4. 1.43

Explanation: Interference (I) is calculated as 1 - Coefficient of Coincidence (C). The Coefficient of Coincidence is the ratio of observed double crossover (DCO) frequency to expected DCO frequency.

  1. Calculate the expected DCO frequency: Expected DCO = RF(P-Q) × RF(Q-R) = 0.12 × 0.18 = 0.0216.
  2. The observed DCO frequency is given as 1.512%, or 0.01512.
  3. Calculate the Coefficient of Coincidence (C): C = Observed DCO / Expected DCO = 0.01512 / 0.0216 = 0.70.
  4. Calculate Interference (I): I = 1 - C = 1 - 0.70 = 0.30.

Question 11

In some organisms, such as certain fungi, a crossover event can sometimes increase the probability of a second crossover event nearby. If this phenomenon, known as negative interference, is occurring, what would be the expected relationship between the observed and expected double crossover (DCO) frequencies?

  1. Observed DCO frequency would be less than the expected DCO frequency, resulting in I > 0.
  2. Observed DCO frequency would be greater than the expected DCO frequency, resulting in I < 0. (correct answer)
  3. Observed DCO frequency would be equal to the expected DCO frequency, resulting in I = 0.
  4. Observed DCO frequency would be zero, resulting in I = 1.

Explanation: Negative interference describes the situation where one crossover enhances the chance of a second crossover. This means more double crossovers (DCOs) will be observed than expected if the events were independent. Therefore, the observed DCO frequency is greater than the expected DCO frequency. This leads to a coefficient of coincidence (C = Obs/Exp) greater than 1, and an interference (I = 1 - C) that is negative (less than 0).

Question 12

The genetic map for three genes in tomato is d - 20 cM - m - 10 cM - p. In a test cross of a trihybrid plant (D M P / d m p), interference is 0.4. Out of 5,000 progeny scored, what is the expected number of individuals with the genotype d M p or D m P?

  1. 40
  2. 60 (correct answer)
  3. 80
  4. 100

Explanation: The genotypes d M p and D m P represent the double crossover (DCO) progeny. The number of DCO progeny depends on the expected DCO frequency and the interference.

  1. Calculate the expected DCO frequency: RF(d-m) × RF(m-p) = 0.20 × 0.10 = 0.02 (or 2%).
  2. Calculate the coefficient of coincidence (C) from interference (I): C = 1 - I = 1 - 0.4 = 0.6.
  3. Calculate the observed DCO frequency: Observed DCO = C × Expected DCO = 0.6 × 0.02 = 0.012 (or 1.2%).
  4. Calculate the number of DCO progeny: Number = Observed DCO frequency × Total progeny = 0.012 × 5,000 = 60.

Question 13

A three-point test cross yields an interference value of 1.0. Which statement is the least likely to be true about the genes being studied?

  1. The three genes are located very close to one another on the chromosome.
  2. The observed number of double-crossover progeny is zero.
  3. The coefficient of coincidence is zero.
  4. The genes are on different chromosomes. (correct answer)

Explanation: Interference is a phenomenon that applies to linked genes on the same chromosome. If the genes were on different chromosomes, they would assort independently, and the concepts of linkage, recombination frequency between them, and interference would not be applicable. An interference of 1.0 (complete interference) means C=0 and no double crossovers are observed, a situation that occurs when genes are very tightly linked.

Question 14

A student analyzes data from a three-point cross. They calculate the recombination frequency for region I as 20% and for region II as 30%. They correctly calculate the expected double crossover (DCO) frequency as 6%. However, upon observing only 3% DCOs in the progeny, they conclude that I = 1 - (0.06 / 0.03) = -1.0. What was the student's primary mistake?

  1. They should have added the recombination frequencies, not multiplied them.
  2. They inverted the ratio for the coefficient of coincidence calculation. (correct answer)
  3. They misidentified the parental and double-crossover classes in the raw data.
  4. They should have subtracted the observed DCO frequency from the expected DCO frequency.

Explanation: The student's calculation for interference (I) was I = 1 - C. Their calculation for C was Expected DCO / Observed DCO (0.06 / 0.03 = 2.0), which is incorrect. The correct formula for the coefficient of coincidence (C) is Observed DCO / Expected DCO. The correct calculation would be C = 0.03 / 0.06 = 0.5, leading to an interference of I = 1 - 0.5 = 0.5. The student inverted the ratio in the C calculation.

Question 15

Positive interference is thought to be a consequence of the physical nature of chromosomes and crossing over. Which molecular mechanism is the most accepted explanation for this phenomenon?

  1. Mechanical rigidity of the chromosome and synaptonemal complex, which makes it difficult to form two chiasmata close together. (correct answer)
  2. Localized depletion of the Spo11 enzyme after it initiates the first double-strand break.
  3. A feedback mechanism where the first chiasma signals the cell to enter anaphase I, limiting time for a second crossover.
  4. The limited number of specific DNA hotspot sequences where recombination can be initiated.

Explanation: When you encounter questions about positive interference in genetics, focus on the physical constraints that limit where crossovers can occur along chromosomes. Positive interference refers to the observation that when one crossover forms, it reduces the probability of another crossover forming nearby. The most widely accepted explanation involves the mechanical properties of chromosomes during meiosis. Answer A correctly identifies that the physical rigidity of the synaptonemal complex and chromosome structure creates spatial constraints that prevent chiasmata from forming too close together. Think of it like trying to tie two knots very close together on a stiff rope—the physical properties of the material make this mechanically difficult. Answer B incorrectly focuses on Spo11 depletion. While Spo11 does initiate double-strand breaks, its localized depletion isn't the primary mechanism explaining positive interference. Multiple breaks can still occur nearby even with varying Spo11 concentrations. Answer C misrepresents the timing of meiotic progression. The first chiasma doesn't trigger immediate entry into anaphase I—cells remain in prophase I for extended periods, giving ample time for additional crossovers to form if interference weren't operating. Answer D incorrectly attributes interference to limited hotspots. While recombination hotspots exist, positive interference occurs even in regions with abundant potential crossover sites. The phenomenon isn't about hotspot availability but about physical spacing constraints. Remember: positive interference questions typically test your understanding of the physical, mechanical aspects of chromosome behavior during meiosis, not just the biochemical processes involved in recombination.

Question 16

Two genes, A and B, are 10 map units apart, and two other genes, C and D, are also 10 map units apart, but they are in different regions of the genome. A cross involving A and B shows an interference of 0.8, while a cross involving C and D shows an interference of 0.3 when measured with a third gene. What is a valid conclusion from this comparison?

  1. The A-B region must be located in heterochromatin, while the C-D region is in euchromatin.
  2. The measurement for the C-D region is more accurate because its interference value is lower.
  3. Fewer double crossovers are observed than expected in the A-B region compared to the C-D region. (correct answer)
  4. The actual physical distance between C and D must be greater than the physical distance between A and B.

Explanation: Interference (I) reflects the reduction in observed double crossovers (DCOs) compared to expected. A higher interference value means a greater reduction. An interference of 0.8 in the A-B region means the observed DCO frequency is only 1 - 0.8 = 0.2 (20%) of the expected frequency. An interference of 0.3 in the C-D region means the observed DCO frequency is 1 - 0.3 = 0.7 (70%) of the expected frequency. Therefore, there is a much stronger suppression of double crossovers in the A-B region than in the C-D region.

Question 17

A geneticist calculates a coefficient of coincidence of 0.80 and an interference of 0.20. Which value is a more direct measure of the reduction in expected double crossovers, and which value reflects the proportion of expected double crossovers that actually occur?

  1. Reduction: 0.80; Proportion: 0.20
  2. Reduction: 0.20; Proportion: 0.80 (correct answer)
  3. Both measure reduction.
  4. Both measure proportion.

Explanation: This question tests the precise definitions of interference and coefficient of coincidence. The coefficient of coincidence (C = Obs/Exp) directly represents the fraction or proportion of expected double crossovers that are observed (in this case, 0.80 or 80%). Interference (I = 1 - C) represents the degree to which expected double crossovers are not observed, i.e., the reduction or inhibition (in this case, 0.20 or 20%).

Question 18

A plant breeder is performing a trihybrid test cross. The three genes are linked in the order R-S-T. The map distance between R and S is 20 cM and between S and T is 20 cM. Interference in this region is 0.5. What is the expected frequency of parental gametes (R S T and r s t) from the trihybrid parent?

  1. 40%
  2. 34%
  3. 68%
  4. 62% (correct answer)

Explanation: When you encounter linked gene problems with interference, you need to account for how crossovers in one region affect crossovers in adjacent regions. This requires calculating both single and double crossover frequencies. Start by finding the recombination frequencies: R-S distance is 20 cM (20% recombination) and S-T distance is 20 cM (20% recombination). If genes were unlinked, you'd expect 4% double crossovers (0.20 × 0.20 = 0.04). However, interference = 0.5 means only half the expected double crossovers actually occur, so the actual double crossover frequency is 2%. Now calculate the gamete frequencies:

  • Single crossover between R-S: 20% - 2% = 18%
  • Single crossover between S-T: 20% - 2% = 18%
  • Double crossovers: 2%
  • Parental types: 100% - 18% - 18% - 2% = 62%
Therefore, parental gametes (RST and rst) appear at 62% frequency. Answer choice A (40%) ignores interference entirely and miscalculates recombinant frequencies. Answer choice B (34%) appears to incorrectly subtract total recombination from 100% without properly accounting for double crossovers. Answer choice C (68%) likely represents a calculation error where someone added instead of subtracted recombination frequencies. Study tip: Always remember that interference reduces double crossover frequency below the expected value (distance₁ × distance₂). The coefficient of coincidence equals (1 - interference), so multiply expected double crossovers by this value to get the actual frequency.

Question 19

In some organisms, such as certain fungi, a crossover event can sometimes increase the probability of a second crossover event nearby. If this phenomenon, known as negative interference, is occurring, what would be the expected relationship between the observed and expected double crossover (DCO) frequencies?

  1. Observed DCO frequency would be less than the expected DCO frequency, resulting in I > 0.
  2. Observed DCO frequency would be greater than the expected DCO frequency, resulting in I < 0. (correct answer)
  3. Observed DCO frequency would be equal to the expected DCO frequency, resulting in I = 0.
  4. Observed DCO frequency would be zero, resulting in I = 1.

Explanation: Negative interference describes the situation where one crossover enhances the chance of a second crossover. This means more double crossovers (DCOs) will be observed than expected if the events were independent. Therefore, the observed DCO frequency is greater than the expected DCO frequency. This leads to a coefficient of coincidence (C = Obs/Exp) greater than 1, and an interference (I = 1 - C) that is negative (less than 0).

Question 20

The map distances for three linked genes are A-15cM-B and B-25cM-C. If the coefficient of coincidence is 0.8 in this region, what is the expected frequency of double-crossover gametes produced by an individual with genotype ABC/abc?

  1. 3.75%
  2. 4.69%
  3. 3.00% (correct answer)
  4. 0.80%

Explanation: The frequency of double-crossover gametes is the observed double crossover frequency.

  1. Calculate the expected double crossover (DCO) frequency by multiplying the recombination frequencies of the two intervals: Expected DCO = 0.15 × 0.25 = 0.0375 (or 3.75%).
  2. Use the coefficient of coincidence (C) to find the observed DCO frequency: Observed DCO = C × Expected DCO = 0.8 × 0.0375 = 0.030 (or 3.00%).