Docsity
Docsity
Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity

Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan

Guidelines and tips
Guidelines and tips
Sell on Docsity
Sell on Docsity
Docsity AI
Docsity AI
Log inSign up

Prepare for your exams

Study with the several resources on Docsity

Find documents

Prepare for your exams with the study notes shared by other students like you on Docsity

Search for your university

Find the specific documents for your university's exams

Docsity AINEW

Summarize your documents, ask them questions, convert them into quizzes and concept maps

Explore questions

Clear up your doubts by reading the answers to questions asked by your fellow students

Earn points to download

Earn points by helping other students or get them with a premium plan

Share documents
download point icon20 Points

For each uploaded document

Answer questions
download point icon5 Points

For each given answer (max 1 per day)

All the ways to get free points
Get points immediately

Choose a premium plan with all the points you need

Study Opportunities

Choose your next study program

Get in touch with the best universities in the world. Search through thousands of universities and official partners

Community

Ask the community

Ask the community for help and clear up your study doubts

Free resources

Our save-the-student-ebooks!

Download our free guides on studying techniques, anxiety management strategies, and thesis advice from Docsity tutors

Population Genetics Exam 1 with Solved Problems | ANSC 446, Exams of Human Genetics

University of Illinois - Urbana-ChampaignHuman Genetics

Past Exam for ANSC 446 - Population Genetics with Roca at Illinois (UIUC)

Typology: Exams

2019/2020
On special offer
30 Points
Discount

Limited-time offer

Uploaded on 06/10/2020

brittani
brittani 🇺🇸

4.7

(30)

287 documents

1 / 5

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
Page 1 of 5
ANSC 446 / IB 416
Population Genetics
Exam 1, Sept. 19, 2008
Name _____________________
(5 pages) Please show decimals rounded to 4 significant digits. Show your
work or describe your logic to earn partial credit for incomplete answers.
(9) 1. A survey in one state finds that the proportion of newborn infants
who have the recessive autosomal disease cystic fibrosis is 1 in 2000.
(3) a) What is the estimated frequency of this disease allele?
q = P(a) = square root of 1/2000 = 0.02236
(3) b) What proportion of individuals would be carriers of the disease?
P(carriers) = 2pq = 2 X P(A) X P(a) = 2 X (1-0.02236) X (0.02236) = 0.04372
(3) c) Assuming a random-mating population, what proportion of
matings would be between two carriers?
(P(carriers))^2 = 0.04372^2 = 0.001911
(15) 2. Assume that the following mtDNA sequences were found in four
different individuals sampled from a population.
AATCGAGACTTTAGT
ATTCCAGATTTAAGC
ATTCCAGATTTAAGC
AATTGAGACTTTAGT
(3) a) How many sites are segregating? 6
(3) b) What proportion of nucleotide sites differ between the first and
second sequences? 5/15 = 0.3333
(3) c) How many transitions are present between the first and second
sequences? 2
(6) d) Estimate the population nucleotide diversity from this sample.
Three unique sequences: A (sequence 1), B (sequences 2, 3), C (sequence 4)
Frequencies P(A) = 0.25, P(B) = 0.5, P(C) = 0.25
πij for comparisons: A-B 5/15 = 0.3333, A-C 1/15 = 0.06666, B-C 6/15 = 0.4
pf3
pf4
pf5
Discount

On special offer

Partial preview of the text

Download Population Genetics Exam 1 with Solved Problems | ANSC 446 and more Exams Human Genetics in PDF only on Docsity!

ANSC 446 / IB 416

Population Genetics Exam 1, Sept. 19, 2008 Name _____________________ (5 pages) Please show decimals rounded to 4 significant digits. Show your work or describe your logic to earn partial credit for incomplete answers. (9) 1. A survey in one state finds that the proportion of newborn infants who have the recessive autosomal disease cystic fibrosis is 1 in 2000. (3) a) What is the estimated frequency of this disease allele? q = P(a) = square root of 1/2000 = 0. (3) b) What proportion of individuals would be carriers of the disease? P(carriers) = 2pq = 2 X P(A) X P(a) = 2 X (1-0.02236) X (0.02236) = 0. (3) c) Assuming a random-mating population, what proportion of matings would be between two carriers? (P(carriers))^2 = 0.04372^2 = 0. (15) 2. Assume that the following mtDNA sequences were found in four different individuals sampled from a population.

AATCGAGACTTTAGT

ATTCCAGATTTAAGC

ATTCCAGATTTAAGC

AATTGAGACTTTAGT

(3) a) How many sites are segregating? 6 (3) b) What proportion of nucleotide sites differ between the first and second sequences? 5/15 = 0. (3) c) How many transitions are present between the first and second sequences? 2 (6) d) Estimate the population nucleotide diversity from this sample. Three unique sequences: A (sequence 1), B (sequences 2, 3), C (sequence 4) Frequencies P(A) = 0.25, P(B) = 0.5, P(C) = 0. πij for comparisons: A-B 5/15 = 0.3333, A-C 1/15 = 0.06666, B-C 6/15 = 0.

π-hat (estimate) = (N/(N-1)) Σ (pi X pj) X πij = 4/3 X [(2 X 0.25 X 0.5 X 0.3333) + (2 X 0.25 X 0.25 X 0.06666) + (2 X 0.5 X 0.25 X 0.4)] = 4/3 X (.08333 + 0.008333 + 0.1)] = 0. (6) 3. A population of caracals was sampled to determine the weight of adult males. A normal distribution was present in which the arithmetic mean weight of the sampled caracals was 20 kg, with standard deviation of 5 kg. Hint: draw a bell shaped curve (3) a) What proportion of adult male caracals would be expected to weigh 29.8 kg or more? 29.8 is 9.8 kg above the mean. 9.8/5 = 1.96 standard deviations; 95% of values fall within 1.96 sd’s of the mean, only 5% are more extreme than this, with half above and half below. So 0.025 of adult male caracals would be > 29.8 kg. (3) b) Among 200 adult male caracals, how many individuals would be expected to weigh between 15 and 25 kg? This range of one standard deviation above and below the mean in a normal distribution includes 68% of samples; 0.68 X 200 = 136 adult male caracals. (16) 4. Red-green color blindness in humans is an X-linked recessive trait present in 7% of males in the United States. Assuming Hardy Weinberg Equilibrium: (3) a) What would be the expected frequency of the trait in females? P(Xa) = 0.07; in females P(trait) = P(Xa)^2 = 0. (3) b) What would be the expected frequency of carriers among females? Female heterozygotes are carriers, with P(het) = 2 X P(Xa) X P(XA) = 2 X(0.07) X (1-0.07) = 2 X 0.07 X 0.93 = 0. (3) c) Alice is a daughter born to a father with normal vision and a mother who is a carrier. Using just this information, what is the probability that Alice is also a carrier? Hint: draw Punnett square. P(carrier) = 0. (7) d) As an adult, Alice has four sons who all have normal vision.

P(C3) = square root of 490/1000 = 0. P(C2) = (square root of 640/1000) – P(C3) = 0.8-0.7 = 0. P(C1) = 1 – P(C2) – P(C3) = 1 – 0.7 – 0.1 = 0. (5) 8. An AFLP marker in xantusid lizards was found to be heterozygous in 80 lizards and homozygous in 10 lizards. Estimate the effective number of alleles at this marker. AE = 1 / (1-(80/90)) = 1/(1-0.8888) = 1/.1111 = 9 alleles (16) 9. A survey of MN blood type frequencies was conducted using samples from 400 Navaho in New Mexico. The phenotypic results were: 336 M 48 MN 16 N (3) a. What is the frequency of the M allele? P(M allele) = (336 + (48/2)) / 400 = 0. (3) b. What is the frequency of the N allele? P(N allele) = (16 + (48/2)) / 400 = 0. (6) c. What are the expected genotypic frequencies under Hardy- Weinberg equilibrium? MM = 0.9^2 = 0. MN = 2 X 0.9 X 0.1 = 0. NN = 0.1^2 = 0. (4) d. Using a chi-square (χ2) test, are the observed genotypes in the sampling consistent with Hardy-Weinberg equilibrium? Expected: MM 0.81 X 400 = 324, MN = 0.18 X 400 = 72, NN = 0.01 X 400 = 4 Σ ((O-E)^2/E) = [(336-324)^2/324] + [(48-72)^2/72] + [(16-4)^2/4] = 0.4444 + 8 + 36 = 44.

Since 44.44 > 3.84, we conclude that the population significantly differs from Hardy Weinberg expectations

    1. Give the best definition for the following terms: (2) a) Synonymous mutation A nucleotide substitution in a codon that does not alter the amino acid coded (2) b) Transversion (mutation) A substitution mutation between a purine and a pyrimidine, or vice versa (15) 11. Two populations of cattle were sampled and found to have the following allele frequencies for two SNP sites: Site 1 Site 2 G C G A Ten Asian cattle .20 .80 .30. Forty African cattle .60 .40 .35. (3) a) Estimate the mean allele frequency of G at Site 1. Mean P(G) = (0.2 X 10/50) + (0.6 X 40/50) = 0.04 + 0.48 = 0. (8) b) Calculate genetic identity and its three components for Site 1. Jxi = 0.2^2 + 0.8^2 = 0.04 + 0.64 = 0. Jyi = 0.6^2 + 0.4^2 = 0.36 + 0.16 = 0. Jxiyi = (0.2 X 0.6) + (0.8 X 0.4) = 0.12 + 0.32 = 0. I = 0.44/((0.68 X 0.52)^0.5)) = 0.44/0.5946 = 0. (4) c) Estimate Nei’s standard genetic distance between the two cattle populations at Site 1. D = - ln (I) = - ln (0.7399) = 0. Potentially useful chi square critical values. Degrees of freedom P value =. 1 3. 2 5. 3 7. 4 9.