5. In human population A, hitchhiker\'s thumb is a simple Mendelian trait. For t
ID: 271730 • Letter: 5
Question
5. In human population A, hitchhiker's thumb is a simple Mendelian trait. For this trait, N is the dominant allele and n is the recessive allele. Individuals who are homozygous dominant (NN) or heterozygous [Nn) express non-hitchhiker's thumbs. Individuals who are homozygous recessive [nn) express hitchhiker's thumbs. In this population, 90% (0.9) of the alleles are dominant [N] and 10%(0.1) are recessive [n. Use the Hardy-Weinberg equation to determine the genotype frequencies we should expect in the next genera- tion. Be sure to show your work. You have collected data on the observed genotype frequencies of the next generation. They are 70% NN, 20% Nn. and 10% nn. Based on these observations and your expectations, is this trait currently evolving in this popula tion? Why or why not? fu vaccine (administered as a shot or a nasal spray) is made of a combination of different types of influenza viruses that research indicates will be most there are numerous specific viruses that can be selected for use in the ing small amounts of these influenza viruses to the body. This stimulates ies against these specific viruses that help p Each year, the specific viruses used in the vaccine will vary. Based on w be the case? Why can't the sam 6. The common during the upcoming season. For each type of influenza, vaccine. The vaccine works by introduc- the immune system to create antibod- rotect the individual from becoming infected with these viruses. hat you know of evolution, why might this t we reuse the same vaccine land combination of viruses) from year toExplanation / Answer
5.
According to HW law = (p+q)^2 =p^2+2pq+q^2 = 1.
p represents dominant allele i.e. N and q represents the recessive allele i.e. n.
The proportion of p allele i.e.N = 0.9 and the proportion of q allele i.e. n = 0.1
So, the proportions in the next generation is
p^2 = 0.9*0.9 = 0.81 *100 = 81%
2pq = 2*0.9*0.1 = 0.18 *100 = 18%
q^2 = 0.1*0.1 = 0.01 *100 = 1%
If the population is evolving there is change in the frequency of genotypes. To study wherther the change is significant or not, the best method is chi square test. If the hypothesis is acceptable which indicates that the population is not evolving, if the hypothesis is rejected which indicates that the population is evolving.
Null hypothesis: The observed frequencies are not deviating from the expected frequencies.
Test Static – Chisquare test:
Category
NN
Nn
nn
Observed values
70
20
10
100
Exprected Values
81
18
1
100
Deviation
-11
2
9
0
D^2
121
4
81
206
D^2/E
1.493827
0.222222
81
82.71605
X^2
82.71605
Degrees of freedom
1
Inference: The calculated chisquare value i.e. 82.71 is greater than the table value i.e. 3.84 at 1 DF and 0.05 probability. Hence the null hypothesis is rejected. Which means the population is not in HW equilibrium in another way, the population is evolving…
ACCORDING TO CHEGG GUIDELINES WE HAVE TO ANSWER ONE QUESTION AT A TIME. POST THE REST AS SEPERATE QUESTIONS THEN I CAN HELP YOU.
Category
NN
Nn
nn
Observed values
70
20
10
100
Exprected Values
81
18
1
100
Deviation
-11
2
9
0
D^2
121
4
81
206
D^2/E
1.493827
0.222222
81
82.71605
X^2
82.71605
Degrees of freedom
1
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