题目
AGRI10051_2025_SM2 Supplementary or Special Exam: Genetics for Agriculture (AGRI10051_2025_SM2)- Requires Respondus LockDown Browser
单项选择题
You are researching a population of 100 cows, where 91 of them are grey and 9 are black. You know that black colour is a recessive trait for this type of cow. Using the Hardy-Weinberg Equilibrium equation (p2 + 2pq + q2 = 1), what number would be heterozygous?
选项
A.42
B.80
C.70
D.32
查看解析
标准答案
Please login to view
思路分析
In this population genetics problem, we start from the phenotype data: 9% black cows, which are recessive, meaning aa. Therefore q^2 = 0.09, so q = sqrt(0.09) = 0.3. The proportion of the other allele, p, is 1 - q = 0.7. The heterozygous genotype frequency is 2pq = 2 × 0.7 × 0.3 = 0.42.
Option by option an......Login to view full explanation登录即可查看完整答案
我们收录了全球超50000道考试原题与详细解析,现在登录,立即获得答案。
类似问题
Assuming the population is in Hardy-Weinberg equilibrium, if 16% of the individuals in a population show a recessive trait (q2=16%), what is the allelic frequency for the dominant allele?
What occurs to allele frequencies when a population is at Hardy-Weinberg equilibrium?
What would disrupt the conditions required for Hardy-Weinberg equilibrium?
The allele for brown eyes (B) is found within a population at a frequency of 0.7; the allele for blue eyes (b) at a frequency of 0.3. If the conditions for Hardy-Weinberg equilibrium are satisfied, what will be the expected frequency of heterozygous (Bb) individuals?
更多留学生实用工具
希望你的学习变得更简单
加入我们,立即解锁 海量真题 与 独家解析,让复习快人一步!