round all answers to 2 decimal places 6. In a population of 500 individuals, of
ID: 150191 • Letter: R
Question
round all answers to 2 decimal places
6. In a population of 500 individuals, of which 280 were Rh+ and 220 were Rh-. 1. Calculate the allele frequencies of D and d (DD individuals have the Rh+ phenotype. dd individuals have the Rh- phenotype. The phenotype of Dd is Rh+). 2. How many of the Rh+ individuals would be expected to be heterozygous? 7. The IA "allele" for the ABO blood groups actually consists of two subtypes, Al and A2, either being considered "A". In a subpopulation of humans, 3/4 of the IA alelles are Al and 1/4 are 12. What would be the expected proportions of A14A1, Al/A2, and IA24A2 among IA individuals? 8. 1 in 1700 US Caucasian newborns have cystic fibrosis. A is the normal allele, dominant over the recess ive a. Individuals must be homozygous for the recessive allele to have the disease. 1. 2. What percent of the above population have cystic fibrosis? Assuming a Hardy-Weinberg Equilibrium, how many newborns would have cystic fibrosis in a population of 15,000 people?Explanation / Answer
Answer:
6).1.
The frequency of Rh- = 220/500 = 0.044
The frequency of allele, d = 0.21
As p+q=1
The frequency of allele, D = 1-0.21 = 0.79
6).2.
The frequency of heterozygoud Rh+ genotype = 2Dd = 2 * 0.79 * 0.21 = 0.33
Number of heterozygous Rh+ individuals = 0.33 * 500 =165
7).
The frequency of IA1 IA1 = ¾ * ¾ = 9/16
The frequency of IA2 IA2 = ¼ * ¼ = 1/16
The frequency of IA1 IA2 =2 * ¾ * ¼ = 2*3/16 = 3/8
8).1.
The frequency of cystic fibrosis = 1/1700 = qq =0.00059
The percentage of population have cystic fibrosis = 0.00059 * 100 = 0.059%
8).2.
Number of cystic fibrosis newborns in 15000 people = 0.00059 *15000 = 8.85 = 9
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.