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BSC 315 Lecture 28: 19

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silvercicada472
27 Apr 2017
School
University of Alabama
Department
Biological Sciences
Course
BSC 315
Professor
Edwin Stephenson

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Document Summary

Population genetic terms: population: a group of breeding individuals, population genetics: allele frequencies in populations and mechanisms of change over time, evolution: change in allele frequency within a population over time. Calculating allele frequencies: a static view of allele frequency: the hardy-weinberg equation, imagine gene with alleles a and a, 36% have genotype a/a (recessive phenotype). Hardy-weinberg equilibrium: hardy-weinberg equilibrium: the population is stable and allele frequencies do not change over time. If so, allele frequencies can be calculated using the h-w equation. Assumptions: no migration, no mutation, mating is random, populations infinitely large, no selection, since all are never true, h-w is an approximation. Hw can be expanded to 3 or more alleles: example: ia, ib and i alleles for blood type. Many genes in human populations are in hw equilibrium: most loci tested are in hw equilibrium, practical importance: applicability of codis database.

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Related Questions

Consider a wildflower population with the following allele and genotype frequencies.

Frequency of the CRallele: p = 0.6

Frequency of the CWallele: q = 0.4

Frequency of CRCR: 50%

Frequency of CRCW: 20%

Frequency of CWCW: 30%

Is this population in Hardy-Weinberg equilibrium?

No, the frequency of genotype CRCWis too low.
No, the frequency of genotype CRCRis too low.
Yes, the genotype frequencies are what we would expect for a population in Hardy-Weinberg equilibrium.
No, the frequency of genotype CWCWis too low.

Suppose that red flower color (A/A or A/a) is dominant to white flower color (a/a) in petunias. The common allele (A) encodes the red pigment for petal colour. The rare allele (a) encodes a nonsense mutation in the pigment gene, so no pigment is produced (such that petals are not coloured and appear white). Among the petunias in a greenhouse population, 480 are A/A and 90 are A/a. There are no white-flowered petunias observed. Is this population of petunias in genetic equilibrium? Show all of your steps and calculations clearly (including use of a ƏĀ‡ 2 test) and write a brief (2-3 sentence) conclusion about whether this population of petunias is in HardyĆ¢Ā€Ā“Weinberg equilibrium. Make sure the expected genotype frequencies equal to 1

You have selected a petunia plant from the greenhouse with red flowers and want to determine whether the plant is A/A or A/a.

a) What kind of experiment (what kind of cross) could you perform to determine the genotype of the selected petunia plant? Explain briefly.

b) How will the cross in (a) help you determine the genotype of your red-flowered petunia? That is, explain how will the results from this cross differ if the red-flowered petunia is A/A versus A/a?

c) You end up choosing two red plants that are both of the genotype A/a and you cross them. If the cross produces six plants, what is the probability that four of them will be white and two will be red?

MY ANSWERS SO FAR

Allele frequency of p= 1/2 (2pq)+ p2

P= 1/2 (0.1578) + 0.8421

=0.921

Allele frequency of q = 1/2 (2pq) + q2

= 1/2 (0.1578) + 0.0

= 0.0789

Expected genotype of the population:

AA(p2)= 0.848 aa (q2)= 0.006

(2pq)Aa =0.0105

I'm confused as to why this doesn't total upto 1

1. Data from the caffeine metabolism experiment was pooled together using information from internal and external students across a number of years. The data collected is summarized in the table below. The genotypes are shown in the first column and the other two show the number of genotypes in the two categories of caffeine tolerance (tolerant/intolerant Ć¢Ā€Ā“ the classes are derived from the questionnaire data). Use the data in this table to answer this question and others throughout the quiz.

Genotype

Tolerant

Intolerant

AA

148

37

AC

106

99

CC

83

137

What are the genotype frequencies using data from all the individuals measured in this experiment?

Select one:

a. AA=0.40 AC=0.44 CC=0.16

b. AA=0.165 AC=0.439 CC=0.396

c. AA=0.25 AC=0.50 CC=0.25

d. AA=0.303 AC=0.336 CC=0.361

2. What are the allele frequencies using all individuals in the experiment?

Select one:

a. A=0.5 C=0.5

b. A=0.471 C=0.529

c. A=0.371 C=0.629

d. A=0.25 C=0.75

3. What are the expected genotype frequencies under Hardy-Weinberg equilibrium?

Select one:

a. AA=0.25 AC=0.5 CC=0.25

b. AA=0.333 AC=0.333 CC=0.333

c. AA=0.222 AC=0.498 CC=0.280

d. AA=0.166 AC=0.668 CC=0.166

4. Test if the population is in equilibrium using a chi-square test. What is the chi-square value for this population?

5. What is the correct number of degrees of freedom for the chi-square test from the previous question?

6. Is this population in Hardy-Weinberg equilibrium?

Select one:

True

False

7. What is the genomic location of CYP1A2?

Select one:

a. 21p24.1

b. 7q24.1

c. 15p24.1

d. 15q24.1

8. What is the reported world-wide frequency for the C allele in human populations?

9. Test if the allele frequencies in our population are significantly different from the world-wide frequencies using a chi-square test.

Select one:

a. chi-square = 131.44; 1 degree of freedom, our population's allele frequencies are significantly different from the world-wide estimates

b. chi-square = 1.314; 1 degree of freedom, our population's allele frequencies are not significantly different from the world-wide estimates

c. chi-square = 1.314; 2 degrees of freedom, our population's allele frequencies are not significantly different from the world-wide estimates

d. chi-square = 131.44; 2 degrees of freedom, our population's allele frequencies are significantly different from the world-wide estimates

10. Use a chi-square independence test to determine if the genotypic frequencies differ between the tolerant and intolerant individuals. The chi-square value is: