o AT&T; LTE 11:01 AM Assignment 4.docx 4. Sickle cell disease is a recessive dis
ID: 198203 • Letter: O
Question
o AT&T; LTE 11:01 AM Assignment 4.docx 4. Sickle cell disease is a recessive disorder. Suppose the disease allele is s while the normal allele is S. Suppose the frequency of individuals with sickle cell disease in a population is 0.09. What is the frequency of the normal allele? (2 pts.) 5. Suppose you observe a population of black mice and brown mice. Suppose black mice is recessive to brown mice. There are 1000 mice in this population. You count 100 black mice in the population. What would the frequency of heterozygotes in the population be? (2 pts. for showing work; 2 bonus points for correct answer) Open with Print
Explanation / Answer
Hardy Weinberg law states that genetic variation in a population will remain constant from one generation to another in the absence of any evolutionary influence.
Using Hardy Weinberg principle frequency of the allele can be calculated
p2 + 2pq + q2 = 1 and p + q = 1
p = frequency of the dominant allele in a population
q = frequency of the recessive allele in a population
p2 = percentage of homozygous dominant individuals
q2 = percentage of homozygous recessive individuals
2pq = percentage of heterozygous individuals
Question: 4
In a sampled population of sickle cell anemia given we have homozygous recessive genotype s frequency is 0.09
Population with sickle cell anaemia –s is : 0.09= q2 then q=0.09=0.3
The frequency of q = 0.3
Since p + q = 1, then p = 1 - q, p = 1 - 0.3 = 0.7
Frequency of S allele is: 0.7
2pq = 2 ( 0.7 x 0.3 ) = 0.42
The percentage of homozygous normal allele is p2= 0.7x0.7=0.49 = 49%
The percentage of homozygous sickle cell condition is q2=0.3x0.3=0.09 is 9%
Heterozygous carrier Ss is: 2pq = 0.42 = 42%
Answer: the frequency of normal allele SS is: 0.49
Question: 5
Black mice: bb recessive
Brown mice: BB dominant
Total population is:1000
Total black mice is: 100, 10% is black
From that calculating frequency of b is: q2 = 10% = 0.1 square root of q is = 0.32
Percentage of b is: 32%
If q is 0.31 than using p+q = 1 from the formula can be used to calculate p
p + 0.32 = 1
p = 1 - 0.32 = 0.68
Percentage of B is: 68%
2pq= 2 (0.68 x 0.32) = 0.44
The percentage of homozygous BB - brown is: p2 = 0.68 x 0.68 = 0.46 = 46%
The percentage of homozygous bb – black is:q2 = 0.32 x 0.32 = 0.1 = 10%
The percentage of heterozygous Bb - brown is: 2pq = 0.44 = 44%
Answer: The frequency of individuals to be heterozygous is = 0.44 = 44%
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.