Hardy Weinberg lab - Pop Gen Lab PDF

Preview

Summary

Pop Gen Lab...

Description

ANSWER SHEET – Lab 2: Hardy-Weinberg Name: Jazmyne Garcia Section: 2014 478 Date: 9/9/2020 1. Using the formulas listed in the lab sheet, calculate the observed genotype and allele frequencies. Record them below. ƒ RR = 250/1000=0.25 p = 0.5

ƒ Rr = 500/1000= 0.5

ƒ rr = 250/1000=0.25

q = 0.5

2. Calculate the expected genotype frequencies and record them below. p2 = 0.25

q2 = 0.25

2pq = 0.5

3. Use the expected genotype frequencies above to establish the expected number of individuals of each phenotype in a sample of 200 under Hardy-Weinberg equilibrium RR = 50

Rr = 100

rr = 50

4. Fill out the table below to calculate the χ2 value and transcribe Ho and Ha below it. Phenotypes Observed (O) Expected (E) (O - E) (O - E)2 (O - E)2 / E 26 50 24 576 11.52 Red Pink 70 100 30 900 9 White 104 50 54 2916 58.32 200 200 Σ χ2 value = 78.88 H0: Non random mating does not disrupt Hardy-Weinberg equilibrium HA: Non random mating does disrupt Hardy-Weinberg equilibrium. 5. a. Is the greenhouse population in Hardy-Weinberg equilibrium? The green house population is not in Hardy-Weinberg equilibrium because non random mating does disrupt the Hardy Weinberg equilibrium.

b. What would be a possible biological explanation for the results of your experiment? (Hint: think about the pollinators as mediators of plant mating) The pollinators didn’t randomly pollinate since the white flowers were preffered according to our data.

EXERCISES Use the information below for questions 6 and 7. Space Station Alpha (SSA) was opened in the year 2500. One hundred fifty young people were selected (reproductive age, 1:1 sex ratio) and sent out into the outer space colony, and SSA was completely autonomous regarding resources needed for population growth. Individuals of the founding population varied in having free (dominant) or attached (recessive) earlobes, and scientists calculated the frequency of the recessive allele in the founding population as 0.4. 6. Assuming Hardy-Weinberg equilibrium and complete dominance, a) how many individuals of the founding population are expected to have free earlobes and b) how many are expected to have attached earlobes? Show your work. q = 0.4 p = 0.6 1-0.4=0.6

7. 100 years later, SSA population census reported 519 individuals, 83 of whom with attached earlobes. a) Demonstrate that this population is likely to be in Hardy-Weinberg equilibrium by calculating the expected number of individuals with attached earlobes out of 519. Show your work. 519(0.4) = 207.6 0.4 is attatched earlobes

b) Propose an explanation for what happened. The expected number is not 83, that means that the one of the conditions for Hardy-Weinberg is not met.

Use the information below for questions 8-10.

The snow goose (Chen caerulescens) has blue and white morphs. Inheritance of color is Mendelian: BB and Bb individuals are blue, and bb individuals are white. In a population of 140 geese, 105 were blue and 35 were white. Researchers estimated that the frequency of the dominant allele was p = 0.5 in this population, and that it was in Hardy-Weinberg equilibrium. An introduced mammal preyed upon the geese during a particularly snowy winter, reducing its population to 46 individuals, 19 of which were white. The blue geese were genotyped and 12 were found to be homozygous (BB) and 15 heterozygous (Bb). 8. Before predation, what would be the expected genotype frequencies? Show your work. p = 0.5 q = 1-0.5 = 0.5 Expected genotype frequencies p^2 = 0.25 2pq= 2 x 0.5 x 0.5 = 0.5 q^2 = 0.25 expected white population size 140 x 25 /100 = 35

9. Calculate observed genotype frequencies after predation (selection). Calculate allele frequencies after predation. Make sure to round your genotype frequencies to two decimal places and the closest even number (i.e. if a genotype frequency comes out to 0.17, round to 0.18). Genotype frequencies must equal 1. Show your work. BB= 12 Bb=15

bb= 19 Observed genotype 12/46=0.3 15/46= 0.3 19/46= 0.4 Allele frequencies p - 0.3+0.15= 0.45 q- 0.4+0.15= 0.55

10. a) Was Hardy-Weinberg equilibrium disrupted by predation? Provide support for your answer by using a chi-square test. Genotypes Observed (O) Expected (E) (O - E) (O - E)2 (O - E)2 / E BB 12 9.2 2.8 7.84 0.85 Bb 15 23 8 64 2.78 bb 19 13.8 5.2 27 1.95 Total 46 46 Σ χ2 value = 5.58

  

Σχ2 = __5.58_ Number of degrees of freedom: 2 P > or < 0.05 (circle one to indicate if p-value is above or below 0.05)

b) Explain how predator behavior led to the observed genotype and allele frequencies. The predator ate blue geese with heterozygous genotype. The population would have have maximum number of those with the Bb genotype compared to the BB or bb genotypes....