- In a large genome-wide association study (GWAS), a SNP in an intron is found to have a highly significant statistical association with risk for a specific disease. This intronic SNP has no known biological function. What is the most plausible explanation for this finding?
Answer: The intronic SNP is in linkage disequilibrium with a nearby ungenotyped variant that is the true causal factor for the disease.
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.
- Consider two loci where the combination of alleles determines fitness. The AB/AB and ab/ab genotypes have high fitness, while all genotypes containing repulsion haplotypes (Ab or aB) have significantly reduced fitness. If a population starts in linkage equilibrium, what is the expected long-term effect of this form of epistatic selection?
Answer: Linkage disequilibrium with D > 0 will be generated, as selection favors the coupling haplotypes.
This scenario describes epistatic selection where fitness depends on the combination of alleles at two different loci. By favoring the AB and ab combinations and selecting against the Ab and aB combinations, this selection pressure will cause the frequencies of the AB and ab (coupling) haplotypes to increase beyond what is expected by chance. This leads to the generation of positive linkage disequilibrium (D > 0).
- A new, highly advantageous allele arises at locus X and rapidly sweeps to fixation in a population. What is the expected signature of this selective sweep on the pattern of linkage disequilibrium in the genomic region surrounding locus X?
Answer: A region of high linkage disequilibrium and reduced haplotype diversity extending a significant distance from locus X.
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.
- Two large, isolated populations exist in linkage equilibrium. Population 1 is fixed for the 'A' and 'B' alleles (AB haplotype). Population 2 is fixed for the 'a' and 'b' alleles (ab haplotype). If these two populations merge in equal proportions, what is the immediate state of linkage disequilibrium in the newly formed admixed population?
Answer: Linkage disequilibrium will be generated with D > 0, due to the non-random association of alleles from the source populations.
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.
- The decay of linkage disequilibrium is modeled by Dt = D0(1−c)t. Based on this equation, which statement accurately describes the relationship between recombination and LD for loci on different chromosomes?
Answer: LD between loci on different chromosomes will decay to half its value in a single generation of random mating.
When analyzing linkage disequilibrium (LD) decay, you need to understand what the recombination fraction 'c' represents for different types of loci. The equation Dt=D0(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: Dt=D0(1−0.5)t=D0(0.5)t. After just one generation (t=1), D1=D0(0.5)=0.5D0, 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. - In a study population, the frequencies of haplotypes for two biallelic loci (A/a and B/b) are determined to be: P(AB) = 0.40, P(Ab) = 0.10, P(aB) = 0.10, and P(ab) = 0.40. Which of the following correctly states the coefficient of linkage disequilibrium (D) and its interpretation?
Answer: D = +0.15, indicating an excess of coupling haplotypes (AB, ab) compared to linkage equilibrium.
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.
- Consider two large, randomly mating populations, P1 (effective size Nₑ = 20,000) and P2 (effective size Nₑ = 200). Both start with identical allele frequencies and are in linkage equilibrium. Assuming no selection or migration, how will the average level of background linkage disequilibrium likely differ between P1 and P2 after 500 generations?
Answer: P2 will have higher average LD because genetic drift is a more potent force, creating random associations between loci.
When you encounter questions about linkage disequilibrium (LD) and population genetics, focus on how genetic drift interacts with recombination to shape patterns of association between loci.
Genetic drift creates random sampling effects that are inversely proportional to effective population size. In smaller populations, drift is much stronger and can overwhelm the homogenizing effects of recombination. When drift randomly changes allele frequencies at different loci, it creates chance associations between alleles at those loci - this is background linkage disequilibrium. The smaller the population, the stronger these random associations become because there are fewer individuals to "average out" the sampling variance.
Choice A correctly identifies that P2 (Ne=200) will have higher average LD than P1 (Ne=20,000) because genetic drift is roughly 100 times stronger in the smaller population, creating more random allelic associations between loci.
Choice B incorrectly suggests larger populations maintain higher LD through more haplotypes. Actually, larger populations have weaker drift, allowing recombination to more effectively break down chance associations. Choice C misses the key point that while recombination rate affects LD, population size determines how effectively recombination can counteract drift-induced associations. Choice D contains a partial truth about fixation reducing polymorphism, but this doesn't eliminate LD - fixed alleles at different loci can still show complete association.
Remember: smaller effective population size equals stronger genetic drift, which creates more background linkage disequilibrium by generating random allelic associations that recombination cannot fully eliminate. - In a given population, the frequencies of alleles A and B are p(A)=0.5 and p(B)=0.5. The two loci are in maximum possible linkage disequilibrium for these allele frequencies, with D = +0.25. Based on this information, what must be the frequency of the Ab haplotype?
Answer: 0.00
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.
- Two linked loci initially exhibit linkage disequilibrium with a coefficient D = 0.20. The recombination fraction (c) between them is 0.10. Assuming random mating and no other evolutionary forces, what will be the approximate value of D after 5 generations?
Answer: 0.118
The decay of linkage disequilibrium due to recombination is modeled by the equation Dt = D0(1−c)t, where Dt is the disequilibrium at generation t, D0 is the initial disequilibrium, c is the recombination fraction, and t is the number of generations. Plugging in the values: D5 = 0.20 × (1−0.10)5 = 0.20 × (0.9)5 = 0.20 × 0.59049 ≈ 0.118. - A large, ancient mainland population shows linkage equilibrium for loci more than 50 kb apart. A small group of 100 individuals from this population colonizes a remote island. After several hundred generations of isolation, what is the most likely pattern of linkage disequilibrium in the island population compared to the mainland source?
Answer: Extensive long-range LD will exist on the island due to the founder effect and subsequent genetic drift.
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.
- A population genetics study finds that two SNPs are in strong linkage disequilibrium, with a normalized value of D' = 1.0 but a squared correlation coefficient of r² = 0.6. What is the most accurate interpretation of these values?
Answer: At least one of the four possible two-SNP haplotypes is absent, but the alleles are not perfectly predictive of each other.
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.
- A new, neutral mutation occurs, creating allele 'a' from 'A' on a single chromosome that carries the 'B' allele at a linked locus. Immediately following this event (in the first generation), what is the state of linkage disequilibrium between the new 'a' allele and the 'B' allele?
Answer: The 'a' allele is in complete linkage disequilibrium (D' = 1) with the 'B' allele.
When analyzing linkage disequilibrium (LD), you need to understand that it measures the non-random association between alleles at different loci. LD occurs when certain allele combinations appear together more or less frequently than expected by chance.
In this scenario, a mutation creates allele 'a' from 'A' on a chromosome that already carries the 'B' allele. This means that initially, every single copy of the new 'a' allele exists exclusively with the 'B' allele on the same chromosome. This creates perfect, complete linkage disequilibrium with D' = 1, making answer A correct.
Answer B is wrong because while mutation is random, the physical location where it occurs is not. The 'a' allele didn't appear randomly throughout the population—it appeared specifically on a chromosome carrying 'B', creating immediate non-random association.
Answer C misses the key point about timing. While recombination rate will eventually determine how quickly LD breaks down over generations, in the very first generation after mutation, no recombination has occurred yet. The strength of initial LD is always maximal regardless of map distance.
Answer D incorrectly suggests that allele frequency determines LD existence. LD can exist at any allele frequency, even when an allele is extremely rare. Frequency affects the statistical power to detect LD, but not its actual presence.
Remember: New mutations always create maximum LD with nearby alleles because they occur on specific chromosomal backgrounds. LD then decays over generations through recombination, but starts at its highest possible value.
- In which of the following scenarios describing haplotype frequencies for two SNPs (Alleles A/T and G/C) would the normalized disequilibrium coefficient D' be equal to 1, while the squared correlation coefficient r² is definitively less than 1?
Answer: Frequencies: P(AG) = 0.5, P(AC) = 0.2, P(TG) = 0.3. The TC haplotype is absent.
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.
- For two loci, the frequency of allele A is 0.6 and the frequency of allele B is 0.3. The observed frequency of the AB haplotype is 0.25. What is the coefficient of linkage disequilibrium, D?
Answer: 0.07
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.
- Which of the following scenarios provides the strongest evidence that two genetic loci are in linkage disequilibrium due to a mechanism other than tight physical linkage?
Answer: A SNP on chromosome 1 and a SNP on chromosome 5 show a strong statistical association in an admixed population.
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.
- How would the presence of a recombination hotspot between two physically close genetic markers affect the linkage disequilibrium between them, compared to two markers the same distance apart in a region with an average recombination rate?
Answer: The hotspot would cause LD to be significantly lower than expected based on physical distance alone.
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.
- Given allele frequencies p(A) = 0.8 and p(B) = 0.7 for two loci, what is the theoretical maximum positive value for the linkage disequilibrium coefficient, D?
Answer: 0.14
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.
- A study in a large, randomly mating human population finds two SNPs on the same chromosome, separated by a physical distance of 2 Mb, that exhibit very strong linkage disequilibrium (r² = 0.85). Which of the following is the LEAST likely explanation for this observation?
Answer: The LD is due solely to the fact that the two mutations occurred recently and have not had time to be separated by recombination.
While recent mutation on a shared background is the ultimate origin of all LD, it is extremely unlikely to be the sole explanation for maintaining high LD (r² = 0.85) over a vast 2 Mb distance in a large population. Recombination would break down this association very quickly over such a distance. The other options provide valid mechanisms that can create or maintain long-range LD: inversions (A) suppress recombination, recombination deserts (B) have inherently low recombination, and selective sweeps (C) drive a haplotype to high frequency too fast for recombination to act. Option D lacks such a mechanism, making it the least plausible stand-alone explanation.
