## Introduction the cystic fibrosis allele by disposing

Introduction
For this third lab, each group was supposed to figure out the allele and genotypic frequencies in a population utilizing varying types of colored beads. In class we discussed what the Hardy-Weinberg equilibrium means in relation to populations, and also listed the necessary conditions for it to exist. Next each group was then given a case to explore whether or not they meet Hardy-Weinberg equilibrium by testing it with Chi Square. Our group, was assigned to case five which followed Hardy-Weinberg equilibrium and our chi square test demonstrated that. In conclusion, we examined the differing sorts of Natural Selection processes and how they influence allele frequencies in a populace after some time.

1. The genetic factors that are necessary for Hardy Weinberg to occur are mutation, random genetic drift, migration, assortative mating, and natural selection

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The expected frequencies for the three genotypes are

Shown work: 1 =.0004 from question. q= .02. 1 p =.98. =.9604. 2*.98*.02=.0392 homozygous recessive, heterozygous normal, homozygous normal
We cannot assume that the population is at Hardy-Weinberg equilibrium because cystic fibrosis usually affects adolescents because they die before reaching adulthood.

We can’t expect that the population is at Hardy-Weinberg harmony because cystic fibrosis normally affects youths since they die young. Because of this, the vast majority of the affected people never have kids, implying that natural selection is choosing against the cystic fibrosis allele by disposing of the individuals who have it.
Since we know y=0.8, we can find p because p + q=1 and q = 0.8 therefore p = 0.2
(1-0.8=0.20) also since p=0.2 and q=0.8 therefore 2pq = 2 x 0.2 x 0.8 = 0.32 YY=(0.2)(0.2)=0.04 Yy = 2(0.2)(0.8)=0.32 yy = (0.8)(0.8) = 0.64

ii) q =.2, and p+q =1, at that point p =.08 or 80%

iii) Heterozygous Frequency = 2pq ? (2*.8*.2) = .32 or 32%

Results

1) The most fit Allele A1A1 because it’s frequency is 100 and based on the graph we predicted and the actual graph it continuously increases even though it is not as steep, therefore I predict that the frequency of this allele will be the most prevalence in the population. Due to this information we can see that allele A1 will survive and reproduce. Likewise, because of the fact that frequency is based on phenotype, the frequency of A1A2 will diminish.

2) Inconclusive because even though A1 is the most fit we don’t know what specific allele is being selected for.

3) It is directional selection because only 1 of the allele’s (A1A1) is favored

4) The recessive allele will never die out because heterozygotes will always be carriers and will continue on unharmed in society because it does not express phenotypically this will cause the allele to be passed on continuously.

5) A1A1 is more dominant because it is less fit ad we are selecting against the dominant allele.

6) The heterozygotes will eventually die out because both extremes are needed to make a heterozygote

7) Selection will always move in the favor of the recessive allele, therefore the dominant allele will eventually be eliminated because the recessive allele has a greater and the A2 phenotype is 100%

8) Huntington’s disease gets passed on before people find out that they have the disease. The disease does not show symptoms until the affected individual turns 40 so there is no way to know whether someone is affected unless you test beforehand.

9) The allele was brought to the same frequency then stabilized

10) The people that are heterozygous for the attribute will have a superior shot of surviving on the grounds that they are impervious to jungle fever. For sickle cell paleness, being a heterozygous bearer is great over some other genotype in nature. There will be a higher recurrence of heterozygous transporters in light of the fact that their genotype great, which is a case of balancing out choice. The homozygous characteristics will vanish.

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