Culling Plan for a Recessive Trait in a Goat Herd
A goat breeder in Mafraq keeps a closed herd of 400 adult animals. Sixteen of them show a recessive coat condition that lowers their market value but does not affect survival or fertility. The breeder has read that the condition is caused by a single recessive allele and assumes the herd is in Hardy–Weinberg equilibrium for this locus. He plans to cull the 16 affected animals before the next breeding season and expects the problem to disappear quickly. A veterinary extension officer points out that unaffected carriers can pass the allele on, and offers DNA testing at 15 Jordanian dinars per animal. The breeder has only 1,000 dinars available for testing this year and does not want to cull animals that are not carriers. The officer must estimate the allele frequency, predict what culling affected animals alone will achieve in one generation, and decide whether the testing budget can support a better strategy. The herd is closed, mating is random, and no other evolutionary forces are acting.
1. Estimate q, p, and the number of carriers. State one assumption that could make this estimate wrong.
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The recessive genotype frequency is [q^2=\frac{16}{400}=0.04.] Therefore [q=\sqrt{0.04}=0.20,] and [p=1-q=0.80.] The carrier frequency is [2pq=2(0.80)(0.20)=0.32,] so the expected number of carriers is [0.32\times 400=128.] Thus about 128 unaffected animals carry the allele.
A key assumption is Hardy–Weinberg equilibrium: random mating, no selection, no migration, and a large population. If carriers or affected animals mate non-randomly, or if the condition slightly reduces fertility even though it does not affect survival, the estimate of 128 carriers could be biased.
2. Predict the proportion of affected animals in the next generation if only the 16 affected animals are culled and all other adults breed randomly.
Read the worked answer
After culling the 16 affected animals, 384 adults remain. From the Hardy–Weinberg frequencies before culling: [\text{AA}=p^2(400)=0.64(400)=256,] [\text{Aa}=2pq(400)=0.32(400)=128,] [\text{aa}=16.] Removing the 16 aa leaves 256 AA and 128 Aa. The new allele frequency of the recessive allele among breeders is [q'=\frac{128}{2(384)}=\frac{128}{768}=0.1667.] If these adults mate randomly, the next generation's affected frequency is [q'^2=(0.1667)^2=0.0278.] So about 2.8% of the next generation, or about 11 of 400 animals, would be affected. This is a fall from 4.0% to 2.8%, but the condition does not disappear; most of the recessive alleles remain hidden in carriers.
3. The testing budget is 1,000 dinars at 15 dinars per animal. Evaluate the breeder's plan and propose a feasible alternative. What is the main limitation?
Read the worked answer
Testing all 400 animals would cost [400\times 15=6000\text{ dinars},] which is far above the 1,000-dinar budget. With 1,000 dinars, the breeder can test at most [\frac{1000}{15}\approx 66\text{ animals}.] Culling only the 16 affected animals costs nothing in testing but leaves about 128 carriers, so it reduces the affected frequency only slowly.
A feasible alternative is to cull the affected animals and use the limited testing on the highest-risk unaffected animals, such as their offspring, full siblings, or animals born to known carriers. Up to 66 such animals can be tested and any carriers culled. This targets the animals most likely to carry the allele instead of testing the whole herd at random.
The main limitation is that partial testing cannot remove all carriers. The recessive allele will remain in untested animals, so the breeder needs several generations of selection, or a larger testing budget, to reduce the condition substantially. A complementary measure is to use only tested non-carrier males for breeding.