Get the answers you need at Westonci.ca, where our expert community is dedicated to providing you with accurate information. Explore thousands of questions and answers from a knowledgeable community of experts ready to help you find solutions. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.
Sagot :
Answer:
0.9342
Explanation:
The Hardy-Weinberg equation states that p² + 2pq + q² = 1,
where p is the frequency of the dominant 'normal' (n) allele and q is the frequency of the recessive 'albino' (a) allele in the population, while q² represents the frequency of the homo-zygous albino genotype (aa), p² represents the frequency of the homo-zygous normal genotype (nn) and 2pq represents the frequency of the heterozygous genotype (na).
In this case, the frequency of individuals in the population that have the genotype aa (q²) is equal to 26/6000 = 0.004333. In consequence, q is equal to √ 0.004333 = 0.0658.
Moreover, the allele frequency of the normal (n) allele p is equal to 1 - q = 1 - 0.0658 = 0.9342, so p² (nn) = (0.9342)² = 0.8727.
Finally, the frequency of the heterozygous genotype (na) is 2pq = 2 x 0.9342 x 0.0658 = 0.123.
Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.