1. Which of the following is FALSE?
A. 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.
B. 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.
C. Individuals who are heterozygous HbA/HbS are
protected from malaria, and this is why sickle-ce disease persists
in wetter, mosquito-prone regions in Africa.
D. In regions where malaria does not occur,
individuals who are heterozygous HbA/HbS have a fitness advantage
over those who homozygous for the hemoglobin allele (HbA).
2. AFTER malaria is cured, the
frequency of the HbS allele should decrease in regions with lots of
mosquitoes because:
A. People will no longer die from sickle-cell
disease in these regions.
B. Having one copy of the HbS
allele will no longer be advantageous in these regions.
C. Natural selection will no longer act on the
HbS allele at all in these regions.
D. All alleles associated with genetic diseases
eventually disappear.
3. If the frequency of the HbS
allele is 0.1 in a population, what is the frequency of the HbA
allele (assuming this is a two-allele system)?
4. Which of the following would be sufficient for the
Hardy-Weinberg equation to accurately predict genotype frequencies
from allele frequencies?
p+ q = 1
A. The population is not evolving due to natural
selection.
B. The population is not evolving due to any of
the conditions that disrupt Hardy-Weinberg equilibrium.
C. The population is infinitely large.
.
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1. Which of the following is FALSE? A. If a genetic disease reduces fertility and the...
Q5.1. Which of the following is FALSE? 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...
Use the following information to answer the next two questions. Sickle cell anemia is a disease that is caused by a mutation in the gene that produces haemoglobin. Hemoglobin carries oxygen in red blood cells. The HbA allele produces normal hemoglobin and the HbS allele produces haemoglobin that sticks together and causes red blood cells to sickle. Heterozygous individuals (HbAHbS) produce both normal and "sickle" hemoglobin so the HbA and HbS alleles are codominant. Heterozygotes do not develop sickle cell...
5. Sickle-cell disease is an interesting genetic disease. Normal homozygous individuals (SS) have normal blood cells that are easily infected with the malarial parasite. Thus, many of these individuals become very ill from the parasite and many die. Individuals homozygous for the sickle-cell trait (ss) have red blood cells that readily collapse when deoxygenated. Although malaria cannot grow in these red blood cells, individuals often die because of the genetic defect. However, individuals with the heterozygous condition (Ss) have some...
In a hypothetical population which is in Hardy Weinberg equilibrium, the frequency for a recessive allele is 30%. What percentage of the population would be expected to show the dominant trait in the next generation? Humans who are born homozygous for the recessive sickle cell allele die of sickle cell anemia, while those who are heterozygous are resistant to malaria (see chapter 4 for more information on this balanced polymorphism). 4% of the population of the Congo are homozygous recessive for...
Hardy-Weinberg Practice Problems: You need to list equations used and provide steps of problem solving. Providing answer itself is not enough for full grade. 1. You have sampled a population in which you know that the percentage of the homozygous recessive genotype (aa) is 36%. Calculate the frequency of the heterozygous genotype, homozygous dominant genotype and homozygous recessive genotype. 2. You have sampled a population in which you know that the percentage of the homozygous recessive genotype (aa) is 49%....
JUST PART B PLEASE AND THANK YOU
TABLE OF CHI SQUARED VALUES. D.F = DEGREES OF FREEDOM
df
<0.10
<0.05
<0.01
<0.001
1
2.706
3.841
6.635
10.8
2
4.605
5.991
9.210
13.8
3
6.251
7.815
11.345
16.3
4
7.779
9.488
13.277
18.5
5
9.236
11.070
15.086
20.5
6
10.645
12.592
16.812
22.4
8
13.362
15.507
20.090
26.1
10
15.987
18.307
23.209
29.6
12
18.549
21.026
26.217
32.9
20
28.412
31.410
37.566
45.3
25
34.382
37.652
44.314
52.6
Rules for Assigning...
It is exceedingly rare for all the Hardy-Weinberg assumptions to be met in nature. Evolution is a change in allele frequencies in a population over time, so a population in Hardy-Weinberg equilibrium is not evolving. Match the following terms with the most correct statement. Each of these relates to evolutionary forces or conditions that violate the Hardy- Weinberg assumptions. (Each term only matches to one statement). Genetic drift - Migration - Inbreeding - 4 Mutation - Natural selection - Nonrandom mating a. Does...
2.3 Problem 3 The Hardy-Weinberg equation is useful for predicting the percent of a hu- man population that may be heterozygous carriers of recessive alleles for certain genetic diseases. Phenylketonuria (PKU) is a human metabolic dis- order that results in mental retardation if it is untreated in infancy. In the United States, one out of approximately 10.000 babies is born with the disor- der. Approximately what percent of the population are heterozygous carriers of the recessive PKU allele? If you...
Type 2: In a population of 2000 gaboon vipers, a genetic difference with respect to venom exists at a single locus. The alleles are incompletely dominant. The population shows 100 individuals homozygous for the ‘t’ allele (genotype “tt” = poisonous) 800 heterozygous (genotype Tt = mildly poisonous, and 1100 homozygous for eth T allele (genotype TT = lethally poisonous). (a) What is the frequency of the ‘t’ allele in the population? (b) Are the genotypes in Hardy-Weinberg equlibrium?
Two people who are “carriers” of (heterozygous) for Tay Sachs disease marry and plan a family. What is the probability that a child from this union will suffer from Tay Sachs disease. (Recall that this is an autosomal recessive disorder, that is, homozygous recessives have the disease.) a. Zero b. 0.25 c. 0.5 d. 0.75 e. 1.0 6. At Hardy-Weinberg equilibrium, heterozygotes are the most common genotype in the population when- a. b. c. d . p> 0.67 q>0.67 and...