Q49. Which of the following is NOT true? If a genetic disease reduces fertility
ID: 56445 • Letter: Q
Question
Q49. Which of the following is NOT true?
If a genetic disease reduces fertility and the allele that causes the disease offers no other advantage, the allele will likely eventually disappear, due to natural selection.
Natural selection does not favor individuals who are homozygous for the sickle-cell allele, because these individuals typically die before they are old enough to reproduce.
Individuals who are heterozygous HbA/HbS are protected from malaria, and this is why sickle-cell disease persists in wetter, mosquito-prone regions in Africa.
In regions where malaria does not occur, individuals who are heterozygous HbA/HbS have a fitness advantage over those who are homozygous for the normal hemoglobin allele (HbA).
Q50. AFTER malaria is cured, the frequency of the HbS allele should decrease in regions with lots of mosquitoes because:
People will no longer die from sickle-cell disease in these regions.
Having one copy of the HbS allele will no longer be advantageous in these regions.
Natural selection will no longer act on the HbS allele at all in these regions.
All alleles associated with genetic diseases eventually disappear.
Q52. If the frequency of the HbS allele is 0.4 in a population, what is the frequency of the HbA allele (assuming this is a two-allele system)?
Q53. Which of the following would be sufficient for the Hardy-Weinberg equation to accurately predict genotype frequencies from allele frequencies?
p + q = 1
The population is not evolving due to natural selection.
The population is not evolving due to any of the conditions that disrupt Hardy-Weinberg equilibrium.
The population is infinitely large.
Q54. In a village, if the proportion of individuals who have sickle-cell disease is 0.25, and the population is assumed to be at Hardy-Weinberg equilibrium, what is the expected frequency of the HbS allele? (Hint: What is the genotype of people with sickle-cell disease, and how is that genotype represented in the Hardy-Weinberg equation?)
0.1250.250.50.75
Q55. In the same village, what proportion of the population should be heterozygous HbA/HbS, according to the Hardy-Weinberg equation?
Use the following passage to answer the next three questions.
It has been hypothesized that people who are heterozygous for the allele that causes the deadly genetic condition cystic fibrosis (which, among other symptoms, reduces fertility) are more resistant to the deadly disease tuberculosis.
Q56. If the cystic fibrosis allele protects against tuberculosis the same way the sickle-cell allele protects against malaria, then which of the following should be true of a comparison between regions with and without tuberculosis?
Cystic fibrosis deaths should be more common in regions with tuberculosis.
Cystic fibrosis deaths should be less common in regions with tuberculosis.
Cystic fibrosis deaths should be equally common in both types of regions.
Regional differences in the cystic fibrosis death rate should be purely random and unpredictable.
Q57. A person who is heterozygous for the cystic fibrosis allele moves to a small, isolated community where no one previously carried the allele. If the cystic fibrosis allele protects against tuberculosis the same way the sickle-cell allele protects against malaria, what should happen to the frequency of the cystic fibrosis allele in the community over time, and why?
The cystic fibrosis allele should disappear from the population, because a single individual with the allele is not enough for it to proliferate.
The cystic fibrosis allele should increase to a relatively high frequency, because heterozygotes with the allele will be more likely to survive than others.
The cystic fibrosis allele should become fixed in the population, due to genetic drift.
The cystic fibrosis allele should either disappear or increase in frequency, depending on chance as well as on tuberculosis prevalence and death rate.
Use the following additional passage to answer the next question.
Suppose you travel to the future, to a time when neither cystic fibrosis nor tuberculosis have caused any deaths for many generations. In all of these future populations, the cystic fibrosis allele still exists at a low frequency.
Q58. You visit a huge city with millions of people. If you were to start sampling the cystic fibrosis allele from one generation to the next, what should happen to its frequency over the next few generations, and why?
The allele frequency should change a lot from one generation to the next due to random genetic drift.
The allele frequency should not change much from one generation to the next because the population is large.
The allele frequency should steadily increase due to natural selection.
The allele frequency should steadily decrease due to natural selection.
Explanation / Answer
49.
The correct answer is
In regions where malaria does not occur, individuals who are heterozygous HbA/HbS have a fitness advantage over those who are homozygous for the normal hemoglobin allele (HbA).
Sickle cell anaemia is a disease which is beneficial to heterozygous people who are in the malaria prone areas. This heterozygous condition prevents them from the malaria and provides resistance. If there is no malaria in that region, there is no benefit of heterozygous HbA/HbS alleles.
50.
The correct answer is
Having one copy of the HbS allele will no longer be advantageous in these regions.
Sickle cell disease is exhibited in homozygous recessive condition. In heterozygous condition it offers resistance against malaria. The natural selection does not remove the sickle cell anaemia. So, if malaria is cured there is no use of HbS allele and its frequency decreases.
52.
According to the Hardy-Weinberg equillibrium
p + q = 1
Here the frequency of one allele (p) is 0.4
Hence,
p = 0.4
0.4 + q = 1
q = 0.6
Frequency of HbA allele (q) = 0.6
53.
The population is infinitely large would be sufficient for the Hardy-Weinberg equation to accurately predict genotype frequencies from allele frequencies
54.
Frequency of individuals having sickle cell anemia ( q^2) = 0.25
Frequency of HbS allele (q) = sqrt ( q^2) = sqrt (0.25) = 0.5
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