Genetics Quiz: Linkage Disequilibrium

What this quiz covers

This quiz focuses on Linkage Disequilibrium, 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. 0.100
2. 0.180
3. 0.020
4. 0.118 (correct answer)

Explanation: The decay of linkage disequilibrium due to recombination is modeled by the equation D_t = D_0(1-c)^t, where D_t is the disequilibrium at generation t, D_0 is the initial disequilibrium, c is the recombination fraction, and t is the number of generations. Plugging in the values: D_5 = 0.20 × (1 - 0.10)^5 = 0.20 × (0.9)^5 = 0.20 × 0.59049 ≈ 0.118.

Question 2

1. 0.18
2. -0.07
3. 0.07 (correct answer)
4. 0.43

Explanation: The coefficient of linkage disequilibrium (D) is calculated as the difference between the observed frequency of a haplotype and its expected frequency if the loci were independent. The expected frequency of the AB haplotype is the product of the individual allele frequencies: E[P(AB)] = p(A) × p(B) = 0.6 × 0.3 = 0.18. The observed frequency is given as P(AB) = 0.25. Therefore, D = P(AB) - E[P(AB)] = 0.25 - 0.18 = 0.07.

Question 3

1. 0.14 (correct answer)
2. 0.24
3. 0.25
4. 0.56

Explanation: The value of D is constrained by the allele frequencies. For a positive D (excess of coupling haplotypes AB and ab), the maximum value is given by D_max = min[p(A)p(b), p(a)p(B)]. First, we find the frequencies of the other alleles: p(a) = 1 - p(A) = 1 - 0.8 = 0.2, and p(b) = 1 - p(B) = 1 - 0.7 = 0.3. Then, we calculate D_max = min[0.8 × 0.3, 0.2 × 0.7] = min[0.24, 0.14] = 0.14.

Question 4

1. Two SNPs located 2 kb apart on chromosome 3 show an r² of 0.9 in a large, randomly mating population.
2. In an isolated island population, almost all pairs of loci on chromosome 10 show moderate LD.
3. A SNP on chromosome 1 and a SNP on chromosome 5 show a strong statistical association in an admixed population. (correct answer)
4. Two alleles at adjacent loci within the MHC region show strong LD across many human populations.

Explanation: Loci on different chromosomes (chromosome 1 and 5) are, by definition, not physically linked and assort independently during meiosis (recombination fraction = 0.5). Therefore, any LD observed between them cannot be due to physical linkage. Such LD is a classic signature of population structure or recent admixture between populations with different allele and haplotype frequencies.

Question 5

1. 0.18
2. -0.07
3. 0.07 (correct answer)
4. 0.43

Explanation: The coefficient of linkage disequilibrium (D) is calculated as the difference between the observed frequency of a haplotype and its expected frequency if the loci were independent. The expected frequency of the AB haplotype is the product of the individual allele frequencies: E[P(AB)] = p(A) × p(B) = 0.6 × 0.3 = 0.18. The observed frequency is given as P(AB) = 0.25. Therefore, D = P(AB) - E[P(AB)] = 0.25 - 0.18 = 0.07.

Question 6

1. A narrow region of very low linkage disequilibrium centered on locus X, because selection eliminates linked deleterious alleles.
2. A region of high linkage disequilibrium and reduced haplotype diversity extending a significant distance from locus X. (correct answer)
3. Strong LD only between locus X and its immediately adjacent markers, decaying to zero within a few kilobases.
4. The generation of linkage equilibrium across the region as the single optimal haplotype becomes fixed.

Explanation: A selective sweep occurs when a beneficial mutation increases in frequency so rapidly that there is insufficient time for recombination to break down the association between the new allele and the alleles at nearby loci on the original chromosome. This phenomenon, known as genetic hitchhiking, results in a large region surrounding the selected locus where one haplotype predominates. This manifests as a distinct signature of high linkage disequilibrium and very low genetic/haplotype diversity.

Question 7

1. The loci will be in linkage equilibrium because the original populations were at equilibrium.
2. Linkage disequilibrium will be generated with D > 0, due to the non-random association of alleles from the source populations. (correct answer)
3. Linkage disequilibrium will be generated with D < 0, because only two of the four possible haplotypes exist.
4. No significant linkage disequilibrium will be generated without strong selection or physical linkage.

Explanation: In the admixed population, P(AB) = 0.5 and P(ab) = 0.5. The allele frequencies are p(A) = 0.5 and p(B) = 0.5. The expected frequency of the AB haplotype is p(A) × p(B) = 0.25. D = P(AB) - p(A)p(B) = 0.5 - 0.25 = +0.25. Population admixture is a major cause of linkage disequilibrium. The positive value of D indicates an excess of coupling haplotypes, which is expected since the original populations contributed only AB and ab haplotypes.

Question 8

1. 0.50
2. 0.25
3. 0.00 (correct answer)
4. Cannot be determined from the information given.

Explanation: We can use the definitions of D and allele frequency to solve this. We are given D = P(AB) - p(A)p(B). Plugging in the values: 0.25 = P(AB) - (0.5)(0.5), which gives 0.25 = P(AB) - 0.25, so P(AB) = 0.50. The frequency of an allele is the sum of the frequencies of the haplotypes that contain it. Thus, p(A) = P(AB) + P(Ab). We know p(A)=0.5 and we just calculated P(AB)=0.5. So, 0.5 = 0.5 + P(Ab). This implies that P(Ab) must be 0.

Question 9

1. D = +0.15, indicating an excess of coupling haplotypes (AB, ab) compared to linkage equilibrium. (correct answer)
2. D = -0.15, indicating an excess of repulsion haplotypes (Ab, aB) compared to linkage equilibrium.
3. D = 0.00, indicating the loci are in linkage equilibrium and assort independently.
4. D = 0.25, which is the expected frequency of the AB haplotype under independent assortment.

Explanation: First, calculate the allele frequencies: p(A) = P(AB) + P(Ab) = 0.40 + 0.10 = 0.50. p(B) = P(AB) + P(aB) = 0.40 + 0.10 = 0.50. The expected frequency of the AB haplotype under linkage equilibrium is p(A) × p(B) = 0.50 × 0.50 = 0.25. The coefficient of linkage disequilibrium, D, is the observed frequency minus the expected frequency: D = P(AB) - p(A)p(B) = 0.40 - 0.25 = +0.15. A positive D value indicates that the coupling haplotypes (AB and ab) are more common than expected by chance.

Question 10

1. The two SNPs are perfectly correlated; knowing the allele at one SNP predicts the other with 100% certainty.
2. At least one of the four possible two-SNP haplotypes is absent, but the alleles are not perfectly predictive of each other. (correct answer)
3. The low r² value relative to D' indicates that the population has experienced a recent bottleneck.
4. The value of r² is likely suppressed due to genotyping error; with perfect data, r² would also equal 1.0.

Explanation: D' = 1.0 indicates that at least one of the four possible haplotypes is not observed in the population. However, r² measures the correlation between the alleles at the two loci. An r² value less than 1.0 indicates that knowing the allele at one SNP does not perfectly predict the allele at the other. This situation arises when three of the four possible haplotypes are present, which satisfies the D' = 1.0 condition but results in an imperfect correlation (r² < 1.0). If r² were 1.0, only two haplotypes would exist.

Question 11

1. LD will be lower on the island due to the purging of rare haplotypes during the founding event.
2. LD will be similar to the mainland, as the founders were a random genetic sample.
3. Extensive long-range LD will exist on the island due to the founder effect and subsequent genetic drift. (correct answer)
4. LD will be high only for loci under strong selection, but otherwise low due to random mating.

Explanation: The colonization by a small number of individuals constitutes a founder effect, a form of genetic drift. By chance, the founders will carry only a subset of the haplotypes from the source population, and their frequencies will likely differ. This non-random association of alleles, coupled with subsequent generations of drift in the small island population, will create and maintain extensive linkage disequilibrium, often over much longer chromosomal distances than in the large source population.

Question 12

1. LD between loci on different chromosomes never decays because the recombination fraction 'c' is effectively zero.
2. LD between unlinked loci will decay by 5% each generation, as c is a small constant.
3. The model is not applicable to loci on different chromosomes, as it only accounts for physical linkage.
4. LD between loci on different chromosomes will decay to half its value in a single generation of random mating. (correct answer)

Explanation: When analyzing linkage disequilibrium (LD) decay, you need to understand what the recombination fraction 'c' represents for different types of loci. The equation $D_t = D_0(1-c)^t$ describes how LD decreases over generations based on recombination frequency. For loci on different chromosomes (unlinked loci), they assort independently during meiosis, meaning they recombine 50% of the time. This gives us $c = 0.5$. Substituting into the equation: $D_t = D_0(1-0.5)^t = D_0(0.5)^t$. After just one generation (t=1), $D_1 = D_0(0.5) = 0.5D_0$, meaning LD drops to exactly half its original value in a single generation of random mating. Option A is wrong because unlinked loci have maximum recombination (c = 0.5), not zero recombination. Option B incorrectly states that c is small for unlinked loci and misrepresents the decay rate—with c = 0.5, LD decays by 50% per generation, not 5%. Option C is incorrect because the LD decay model absolutely applies to loci on different chromosomes; the equation works for any recombination frequency, whether loci are linked (c < 0.5) or unlinked (c = 0.5). Remember this key distinction: linked loci (same chromosome) have c < 0.5 and LD decays slowly, while unlinked loci (different chromosomes) have c = 0.5 and LD decays rapidly—halving each generation. This makes biological sense because independent assortment quickly breaks up associations between unlinked genes.

Question 13

1. Frequencies: P(AG) = 0.5, P(TC) = 0.5. The AC and TG haplotypes are absent.
2. Frequencies: P(AG) = 0.25, P(AC) = 0.25, P(TG) = 0.25, P(TC) = 0.25.
3. Frequencies: P(AG) = 0.5, P(AC) = 0.2, P(TG) = 0.3. The TC haplotype is absent. (correct answer)
4. The observed frequency of the AG haplotype is exactly equal to the product of the frequencies of the A and G alleles.

Explanation: D' = 1 when at least one of the four possible haplotypes is absent. r² = 1 only when just two of the four haplotypes exist (perfect correlation). Scenario C describes a population where three haplotypes are present and one is absent (TC). The absence of the TC haplotype ensures that D' = 1. However, because three haplotypes exist, the alleles are not perfectly predictive of each other (e.g., an A allele can be paired with either a G or a C), meaning r² must be less than 1.

Question 14

1. D = +0.15, indicating an excess of coupling haplotypes (AB, ab) compared to linkage equilibrium. (correct answer)
2. D = -0.15, indicating an excess of repulsion haplotypes (Ab, aB) compared to linkage equilibrium.
3. D = 0.00, indicating the loci are in linkage equilibrium and assort independently.
4. D = 0.25, which is the expected frequency of the AB haplotype under independent assortment.

Explanation: First, calculate the allele frequencies: p(A) = P(AB) + P(Ab) = 0.40 + 0.10 = 0.50. p(B) = P(AB) + P(aB) = 0.40 + 0.10 = 0.50. The expected frequency of the AB haplotype under linkage equilibrium is p(A) × p(B) = 0.50 × 0.50 = 0.25. The coefficient of linkage disequilibrium, D, is the observed frequency minus the expected frequency: D = P(AB) - p(A)p(B) = 0.40 - 0.25 = +0.15. A positive D value indicates that the coupling haplotypes (AB and ab) are more common than expected by chance.

Question 15

1. The association is a Type I error (false positive) because intronic SNPs are non-functional and cannot cause disease.
2. The intronic SNP is in linkage disequilibrium with a nearby ungenotyped variant that is the true causal factor for the disease. (correct answer)
3. The intronic SNP must have a novel, undiscovered regulatory function that directly influences disease risk.
4. The population is in complete linkage equilibrium, which increases the statistical power to detect associations with non-causal variants.

Explanation: The principle of GWAS relies on linkage disequilibrium. A genotyped marker SNP does not need to be causal itself. If it is in high LD with a true, ungenotyped causal variant, it will serve as a proxy and show a statistical association with the trait or disease. This is the most common and likely explanation for signals in non-coding regions.

Question 16

1. The hotspot would increase LD by creating novel, advantageous allele combinations.
2. The hotspot would have little effect on LD, as physical distance is the primary determinant.
3. The hotspot would cause LD to be significantly lower than expected based on physical distance alone. (correct answer)
4. The hotspot would generate long-range LD by suppressing crossing over in the flanking regions.

Explanation: Recombination is the primary force that breaks down linkage disequilibrium. A recombination hotspot is a small region of the genome with a much higher rate of recombination than average. Therefore, even if two markers are physically close, the presence of a hotspot between them will lead to more frequent crossing over, which will break down associations between their alleles more rapidly. This results in lower LD than would be expected for that physical distance.

Question 17

1. 0.100
2. 0.180
3. 0.020
4. 0.118 (correct answer)

Explanation: The decay of linkage disequilibrium due to recombination is modeled by the equation D_t = D_0(1-c)^t, where D_t is the disequilibrium at generation t, D_0 is the initial disequilibrium, c is the recombination fraction, and t is the number of generations. Plugging in the values: D_5 = 0.20 × (1 - 0.10)^5 = 0.20 × (0.9)^5 = 0.20 × 0.59049 ≈ 0.118.

Question 18

1. The two SNPs are perfectly correlated; knowing the allele at one SNP predicts the other with 100% certainty.
2. At least one of the four possible two-SNP haplotypes is absent, but the alleles are not perfectly predictive of each other. (correct answer)
3. The low r² value relative to D' indicates that the population has experienced a recent bottleneck.
4. The value of r² is likely suppressed due to genotyping error; with perfect data, r² would also equal 1.0.

Explanation: D' = 1.0 indicates that at least one of the four possible haplotypes is not observed in the population. However, r² measures the correlation between the alleles at the two loci. An r² value less than 1.0 indicates that knowing the allele at one SNP does not perfectly predict the allele at the other. This situation arises when three of the four possible haplotypes are present, which satisfies the D' = 1.0 condition but results in an imperfect correlation (r² < 1.0). If r² were 1.0, only two haplotypes would exist.

Question 19

1. Two SNPs located 2 kb apart on chromosome 3 show an r² of 0.9 in a large, randomly mating population.
2. In an isolated island population, almost all pairs of loci on chromosome 10 show moderate LD.
3. A SNP on chromosome 1 and a SNP on chromosome 5 show a strong statistical association in an admixed population. (correct answer)
4. Two alleles at adjacent loci within the MHC region show strong LD across many human populations.

Explanation: Loci on different chromosomes (chromosome 1 and 5) are, by definition, not physically linked and assort independently during meiosis (recombination fraction = 0.5). Therefore, any LD observed between them cannot be due to physical linkage. Such LD is a classic signature of population structure or recent admixture between populations with different allele and haplotype frequencies.

Question 20

1. 0.14 (correct answer)
2. 0.24
3. 0.25
4. 0.56

Explanation: The value of D is constrained by the allele frequencies. For a positive D (excess of coupling haplotypes AB and ab), the maximum value is given by D_max = min[p(A)p(b), p(a)p(B)]. First, we find the frequencies of the other alleles: p(a) = 1 - p(A) = 1 - 0.8 = 0.2, and p(b) = 1 - p(B) = 1 - 0.7 = 0.3. Then, we calculate D_max = min[0.8 × 0.3, 0.2 × 0.7] = min[0.24, 0.14] = 0.14.