Thank you Let\'s imagine an amazing trait in this beetle that is controlled by a
ID: 206384 • Letter: T
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Thank you Let's imagine an amazing trait in this beetle that is controlled by a single locus. Beetles who are B2B2 have the trait and are much more fit than beetles with BiBi or BiB2. (a) Suppose you have data from year 2017 that shows the following genotype counts for 1000 beetles: B2B2 = 50, Bi B2-600, Bi Bi-350. What are the allele frequencies of Bi and Bi? (b) Is this population in Hardy-Weinberg equilibrium at this locus? Why or why not? (Hint: this will take a few steps of math!) (c) Do you think the frequency of the B2 allele will be the same, lower, or larger in year 2200? Why? 21. In the human populations in Angola, Africa, the hemoglobin allele has two alleles: HbS and HbA. (a) The HbS allele has a frequency of 0.18. What is the frequency of HbA? (b) Did you need to use the Hardy-Weinberg equations to calculate the previous answer? Yes OR no? (c) What are the possible genotypes? (d) Assuming Hardy-Weinberg, what are these genotype frequencies? (e) People who are homozygous for the HbS allele have sickle-cell anemia, a disease that can block blood flow and often leads to early death. If you tested 1000 people in Angola, how many would you predict to have sickle-cell anemia? (?) The actual number is typically much much lower than the prediction. Why do you think that is? (g) Do you think the hemoglobin gene is in Hardy-Weinberg equilibrium? Why or why not?Explanation / Answer
20. a) frequency of B1= no. of indivuals 350/1000=0.35 B2= 1-0.35=0.65
21. a)frequency of HbA= p+q=1 1-0.18=0.82 b) yes; c)HbS-HbS,,HbS-HbA,,,HbA-HbA d) p2= HbS= 0.18*0.18=0.0324 HbA=q2=0.82*0.82=0.6782 HbS.HbA=2pq=2*0.18*0.82=0.2952 e) no. of persons having sickle cell anaemia= 0.18*1000=180 f) yes it is lower than prediction but depends upon the genetic history g) yes it is. we used this for screening of new borns.
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