Genetic Drift Practice Problems With Answers

This is a complete set of genetic drift practice problems with fully worked answers, carefully arranged from easy to hard. Each problem gives the question, a detailed step-by-step solution, and the key formula, so you can check your method as well as your final number. The set covers the calculations a population genetics course is most likely to ask: the probability of allele fixation, the rate of heterozygosity loss, effective population size, the founder effect, and telling genetic drift apart from natural selection, all worked through with real numbers.

Work each problem yourself before reading the solution, because the goal is to master the method, not just to see the final answer. The core formulas are few in number, and most problems are simply a rearrangement of the same handful of relationships. If you need to review the underlying ideas first, our guide on what genetic drift is covers the concepts these problems test. Watching allele frequencies wander in a simulator is another good way to build intuition as you practise.

The Key Formulas You Will Need

Before the problems, here is a quick reference of the formulas used throughout this set. Memorizing these few relationships lets you solve the large majority of genetic drift problems. Each one connects population size to the strength or outcome of drift.

The probability that a neutral allele eventually drifts to fixation equals its current frequency. The rate of heterozygosity lost per generation is 1 divided by twice the population size, written 1/(2N). The heterozygosity remaining after t generations is the starting heterozygosity times the quantity one minus 1/(2N), raised to the power t. The frequency of a brand-new mutation in a diploid population of size N is 1/(2N), because it is a single copy among 2N total alleles. For effective size with an unequal sex ratio, Ne equals 4 times Nm times Nf divided by the sum of Nm and Nf. For a population that varies in size across generations, the effective size is the harmonic mean of the generational sizes.

The table below collects these for easy reference as you work.

Quantity	Formula	Notes
Probability of fixation	equals allele frequency	For a neutral allele
Probability of loss	1 minus allele frequency	The complement of fixation
Heterozygosity lost per generation	1 / (2N)	Higher in small populations
Heterozygosity after t generations	H0 × (1 − 1/2N)^t	Compounds over time
New mutation frequency	1 / (2N)	One copy among 2N alleles
Ne with unequal sex ratio	4·Nm·Nf / (Nm + Nf)	Rarer sex limits Ne
Ne with varying size	harmonic mean of sizes	Dominated by smallest values

Keep this table beside you while you solve. Notice how often population size, N or Ne, appears, because population size is the master variable controlling drift. With these formulas in hand, the problems become a matter of identifying which one applies.

A useful way to approach any genetic drift problem is to ask three quick questions. First, does the problem involve a probability, such as fixation or loss? If so, the answer often comes straight from the allele frequency. Second, does it involve the loss of diversity over time? If so, reach for the heterozygosity formulas built around 1/(2N). Third, does it involve finding the genetically relevant population size? If so, the sex-ratio or harmonic-mean formula gives the effective size. Almost every problem in this set falls into one of these three categories, and recognizing which category a problem belongs to is the single most useful skill for solving it quickly. Once you have identified the category, the right formula is usually obvious, and the calculation itself is short.

Easy Problems: Fixation and Loss

These first problems use the simplest and most elegant result in all of drift theory: the probability that a neutral allele drifts to fixation equals its current frequency. Make sure you are comfortable with this before moving on, because it appears in several later problems too.

Problem 1. A neutral allele has a frequency of 0.3 in a population. If genetic drift is the only force acting, what is the probability it will eventually become fixed?

Solution. The probability of fixation of a neutral allele equals its current frequency. So the probability is simply 0.3, or 30 percent. There is no calculation beyond recognizing the principle, which is what makes this the most elegant result in drift theory. The probability it is instead lost is 1 minus 0.3, which is 0.7, or 70 percent.

Problem 2. In a diploid population of 50 individuals, a new neutral mutation appears in one individual. What is its initial frequency, and what is the probability it eventually fixes?

Solution. A diploid population of 50 has 2 times 50, or 100, total alleles. The new mutation is a single copy, so its frequency is 1 divided by 100, which is 0.01. Because the probability of fixation equals the frequency, the chance this new mutation drifts to fixation is also 0.01, or 1 percent. This shows how unlikely any single new mutation is to take over a population by drift alone.

Problem 3. Two neutral alleles, A and a, have frequencies of 0.6 and 0.4 in a population evolving only by drift. Which is more likely to fix, and with what probability?

Solution. Each allele's fixation probability equals its frequency, so A has a 0.6 chance and a has a 0.4 chance of eventually fixing. Allele A is more likely to fix, with a 60 percent probability, simply because it is more common to begin with. The two probabilities sum to 1, as they must, since one allele or the other will eventually fix.

Medium Problems: Heterozygosity Loss

These problems use the rate at which drift erodes genetic diversity, measured as the loss of heterozygosity. The key formula is that heterozygosity falls by 1/(2N) each generation, where smaller populations lose diversity faster.

Problem 4. An endangered species has an effective population size of 50. What is the expected loss of heterozygosity each generation?

Solution. The heterozygosity lost per generation is 1 divided by twice the effective size, so 1 divided by 2 times 50, which is 1 divided by 100, equal to 0.01. The population loses about 1 percent of its heterozygosity per generation. This is a meaningful rate, illustrating why a population this small is genetically at risk.

Problem 5. Compare the per-generation heterozygosity loss in a population with an effective size of 100 versus one with an effective size of 1,000.

Solution. For the population of 100, the loss is 1 divided by 200, which is 0.005, or 0.5 percent per generation. For the population of 1,000, the loss is 1 divided by 2,000, which is 0.0005, or 0.05 percent per generation. The smaller population loses heterozygosity ten times faster, demonstrating the inverse relationship between population size and the strength of drift.

Problem 6. A population has an effective size of 25 and starts with a heterozygosity of 0.5. Approximately what heterozygosity remains after one generation?

Solution. The fraction retained each generation is one minus 1/(2N), which is one minus 1 divided by 50, or 0.98. Multiply the starting heterozygosity by this: 0.5 times 0.98 equals 0.49. So after one generation, heterozygosity falls from 0.5 to about 0.49. Over many generations this compounds, with the heterozygosity after t generations equal to 0.5 times 0.98 raised to the power t. The measurement of this diversity is explored in our guide on genetic drift in conservation.

Harder Problems: Effective Population Size

These problems calculate effective population size, the genetically relevant size that governs drift. They use the sex-ratio and fluctuating-size formulas, which often reveal an effective size far below the headcount.

Problem 7. A population has 100 breeding males and 400 breeding females. What is its effective population size?

Solution. Use the unequal-sex-ratio formula: Ne equals 4 times Nm times Nf divided by the sum of Nm and Nf. That is 4 times 100 times 400, divided by 500, which is 160,000 divided by 500, equal to 320. So although 500 adults breed, the effective size is only 320, because the rarer sex, the males, limits the genetic contribution.

Problem 8. A population has only 10 breeding males but 1,000 breeding females. Calculate the effective size and comment on the result.

Solution. Applying the formula, Ne equals 4 times 10 times 1,000, divided by 1,010, which is 40,000 divided by 1,010, approximately 40. Despite 1,010 breeding adults, the effective size is only about 40. The tiny number of males creates a severe genetic bottleneck, dragging the effective size down near the danger threshold of the 50/500 rule. This is why skewed sex ratios are a serious conservation concern. The full concept is covered in our guide on effective population size.

Problem 9. Over five generations, a population has sizes of 1,000, 1,000, 10, 1,000, and 1,000. What is its effective size across this period?

Solution. The effective size across generations is the harmonic mean, not the arithmetic mean. The harmonic mean is the number of values divided by the sum of their reciprocals. The reciprocals are 1/1,000, 1/1,000, 1/10, 1/1,000, and 1/1,000, which sum to 0.001 plus 0.001 plus 0.1 plus 0.001 plus 0.001, equal to 0.104. The harmonic mean is 5 divided by 0.104, approximately 48. So the single bottleneck generation of 10 drags the effective size down to about 48, far below the arithmetic average of 802. This is why a single crash has such lasting genetic impact.

Challenge Problems: Founder Effect and Drift Versus Selection

These final problems apply drift to the founder effect and ask you to distinguish drift from natural selection, integrating the concepts from across the topic.

Problem 10. In a source population, a recessive allele has a frequency of 1 percent. A new colony is founded by 25 individuals, one of whom carries a single copy of the allele. What is the allele frequency in the new colony?

Solution. The 25 founders carry 2 times 25, or 50, total alleles. The single copy gives a frequency of 1 divided by 50, which is 0.02, or 2 percent. The allele frequency has doubled compared to the source population's 1 percent, purely because the small founding sample happened to include a carrier. This is the founder effect, explored in our guide on the founder effect.

Problem 11. Two isolated island populations of the same beetle species start identical. After many generations with no migration, they have markedly different frequencies of a color allele that has no effect on survival. What process most likely caused the divergence?

Solution. Genetic drift is the most likely cause. The allele is described as having no effect on survival, so it is neutral, which rules out natural selection as the driver. Two isolated populations evolving by drift will diverge randomly, in different directions, exactly as described. The absence of any fitness difference, combined with random divergence in isolation, is the signature of drift rather than selection.

Problem 12. A beneficial allele appears in a small population but is lost within a few generations. Is this possible, and what does it illustrate?

Solution. Yes, this is entirely possible, and it illustrates that genetic drift can override natural selection in small populations. Even though the allele is beneficial and selection favors it, in a small population random sampling can eliminate it by chance before selection has time to increase it. This shows that drift, not just selection, decides the fate of alleles when populations are small, a key reason small populations behave differently from large ones. The contrast between the two forces is detailed in our guide on genetic drift versus natural selection.

Bonus Problems: Putting It Together

These final problems combine multiple concepts, reflecting the integrated questions that exams and courses often ask. They reward a solid grasp of how the pieces fit together.

Problem 13. A population with an effective size of 25 starts with a heterozygosity of 0.5. Approximately what heterozygosity remains after 10 generations?

Solution. Use the compounding formula: heterozygosity after t generations equals the starting value times one minus 1/(2N), raised to the power t. Here 1/(2N) is 1 divided by 50, which is 0.02, so the retained fraction per generation is 0.98. Over 10 generations, that is 0.98 raised to the tenth power, which is about 0.817. Multiply by the starting heterozygosity: 0.5 times 0.817 equals about 0.41. So heterozygosity falls from 0.5 to roughly 0.41 over ten generations, a noticeable erosion of diversity in this small population.

Problem 14. A conservation biologist finds a population whose census count is 2,000 animals, but estimates its effective size at only 200. What is the Ne to N ratio, and is this typical?

Solution. The ratio is the effective size divided by the census size, so 200 divided by 2,000, which is 0.1, or 10 percent. This is entirely typical, because real populations average an Ne to N ratio of about 0.1 to 0.14. The result means this population drifts as though it had only 200 individuals, losing heterozygosity at a rate of 1 divided by 400, about 0.25 percent per generation, despite the larger headcount.

Problem 15. In a population evolving only by drift, allele B has a frequency of 0.25. Over thousands of generations across many identical replicate populations, in what fraction of those populations would you expect allele B to eventually fix?

Solution. Since the probability of fixation of a neutral allele equals its frequency, allele B should fix in about 0.25, or 25 percent, of the replicate populations. In the other 75 percent it will be lost. This replicate-population view captures what fixation probability really means: across many independent populations starting at the same frequency, the proportion that end up fixed for the allele equals that starting frequency. It is a powerful way to see drift as a probabilistic outcome rather than a single fixed result.

Common Mistakes to Avoid

A few recurring errors trip students up on genetic drift problems, and watching for them will protect your marks. Each is easy to avoid once you are aware of it.

The most common mistake is forgetting that the probability of fixation equals the allele frequency directly, with no further calculation needed. Students sometimes overthink this elegant result and look for a complicated formula that does not exist. A second frequent error is using the arithmetic mean instead of the harmonic mean for effective size across fluctuating generations. The harmonic mean is always smaller and is the correct choice, because drift depends on it; using the arithmetic mean badly overestimates the genetic health of a fluctuating population.

A third error involves the factor of 2 in the formulas. Remember that a diploid population of N individuals has 2N alleles, so heterozygosity loss is 1/(2N), not 1/N, and a new mutation's frequency is 1/(2N). Dropping the 2 doubles your answer incorrectly. A fourth mistake is confusing census size with effective size; problems often give a headcount, but the genetically relevant number is the effective size, which the sex-ratio or harmonic-mean formula produces. Finally, when asked to identify drift versus selection, watch for the keyword neutral or the phrase no effect on fitness, which signals drift, and for any mention of a survival or reproductive advantage, which signals selection.

Frequently Asked Questions

How do you calculate the probability that an allele fixes by drift?

For a neutral allele, the probability of fixation equals its current frequency. An allele at a frequency of 0.4 has a 40 percent chance of eventually drifting to fixation. The probability it is lost instead is 1 minus the frequency, so 60 percent in that example.

What is the formula for heterozygosity loss from drift?

A population loses heterozygosity at a rate of 1 divided by twice the effective population size, or 1/(2N), per generation. The heterozygosity remaining after t generations equals the starting heterozygosity times one minus 1/(2N), raised to the power t.

How do you find effective population size with unequal sex ratios?

Use Ne equals 4 times the number of breeding males times the number of breeding females, divided by their sum. The result is strongly limited by whichever sex is rarer, so a skewed sex ratio produces an effective size much smaller than the total number of breeding adults.

How can you tell if a change was caused by drift or selection?

Look for clues about fitness and randomness. If the allele is neutral or the change shows no link to survival or reproduction, and especially if isolated populations diverge randomly, drift is the likely cause. If a change tracks a survival or reproductive advantage, selection is indicated.

Where Chance Meets the Math

You now have twelve worked genetic drift problems spanning every common type: fixation and loss probabilities, heterozygosity decay, effective population size with skewed sex ratios and fluctuating numbers, the founder effect, and distinguishing drift from selection. The same few formulas solve nearly all of them, with population size, N or Ne, appearing again and again as the master variable that controls how strongly drift acts.

Return to these problems until the method is automatic, then test yourself with fresh numbers. The biggest habits to build are recognizing that fixation probability equals frequency, using the harmonic mean for fluctuating size, and never dropping the factor of 2 that comes from diploidy. You can generate endless practice by watching allele frequencies drift and fix with the genetic drift simulator, which turns these formulas into a visible process. For more worked treatment of the underlying equations, this Nature Education article on genetic drift and effective population size is a thorough reference, and this Study.com lesson on heterozygosity and allele fixation walks through the diversity-loss mathematics with clear worked examples you can study alongside this set.