Question

Will Be Given on Exam: p + q = 1 p = AA + ½Aa q...

Will Be Given on Exam: p + q = 1 p = AA + ½Aa q = aa + ½Aa p2 + 2pq + q2 = 1 (p + q)2 = 1 X2 =  [(observed - expected)2 / expected] X2 2, 0.05 = 6.00

1) Blue eyes are a homozygous recessive genotype within humans. Within the human population, you have 225 individuals with blue eyes, 480 individuals that are homozygous dominant for brown eyes, and 125 people that are heterozygous for brown eyes.

a. Calculate the observed allelic frequency of the gene encoding blue eyes (b) and the gene encoding brown eyes (B).

b. Calculate the observed genotypic frequencies.

c. Calculate the expected genotypic frequencies.

d. Without conducting the Chi-Square yet, do you think this population is in Hardy-Weinberg equilibrium? Why or why not?

e. Confirm or refute your prediction using the Chi-Square Test

Math   Statistics-And-Probability   
0 0
Add a commentTranscribed image text
Next >

Homework Answers

Answer #1

blue eyes homozygous recessive genotype(q^2),aa=225

homozygous dominant for brown eyes(p^2),AA=480

heterozygous for brown eyes(2pq),Aa=125

total=225+480+125=830

p = AA + ½Aa

p=(480+(1/2)*125)/830=0.6536

q = aa + ½Aa

q=(225+(1/2)*125)/830=0.3464

p^2=0.4272

q^2=0.1200

2pq=0.4528

chi square test-

Categories Observed Expected (fo-fe)2/fe
homozygous dominant 480 830*0.4272=354.576 (480-354.576)2/354.576 = 44.366
homozygous recessive genotype 225 830*0.12=99.6 (225-99.6)2/99.6 = 157.883
heterozygous 125 830*0.4528=375.824 (125-375.824)2/375.824 = 167.399
Sum = 830 830 369.649

null hyp:there is no significant differences between expected and observed gene frequencies

alternate hyp:Some of the population proportions differ from the values stated in the null hypothesis

there is no significant differences between expected and observed gene frequencies.

0 0
Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Log In

Sign Up

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
On a separate sheet of paper, plug in the following values for p and q into...
On a separate sheet of paper, plug in the following values for p and q into the Hardy-Weinberg equation (p2+2pq+q2=1): p=0.5, q=0.5 p=0.6, q=0.4 p=0.7, q=0.3 p=0.8, q=0.2 p=0.9, q=0.2 p=1.0, q=0.0 You have just modeled genetic drift acting on a population. After you have plugged these values in, what trend do you see with regards to the frequencies of the homozygous q (q2) and heterozygous (2pq) genotypes? What does this tell you about genetic drift?
A moth census on Wooded Island counted 100 black (BB) moths, 40 gray (Bb) and 60...
A moth census on Wooded Island counted 100 black (BB) moths, 40 gray (Bb) and 60 white (bb) moths. What are the genotypic frequencies? What are the allelic frequencies? Is the population in Hardy-Weinberg equilibrium?    Phenylketonuria is a severe form of mental retardation due to a rare autosomal recessive allele. About 1 in 10,000 newborn Caucasians are affected with the disease. Calculate the frequency of carriers (i.e., heterozygotes).   Note what assumption you are making. A study on blood types in...
3) In general, p = allele frequency of the dominant allele and q = allele frequency...
3) In general, p = allele frequency of the dominant allele and q = allele frequency of the recessive allele. Using the product and/or sum rule(s), use these variables (p and q) to determine the probabilities of an individual in the next generation being: a) Homozygous dominant (A dominant allele (A) AND another dominant allele (A))? b) Homozygous recessive (A recessive allele (a) AND another recessive allele (a))? c) Heterozygote (Aa OR aA)? 4a) In a population in Hardy-Weinberg equilibrium,...
p+q =1 All of the allele frequencies p2 + 2pq + q2 =1 All of the...
p+q =1 All of the allele frequencies p2 + 2pq + q2 =1 All of the genotype frequencies together equals 1 p2 and q2 Frequencies for each homozygotes 2pq Frequencies for Heterozygotes. p2 + 2pq= + q2 Homozygous Dominant Heterozygous Homozygous recessive If the frequencies of the color in Homozygous recessive frequencies pea plants are 30% in the population what are the genotype and phenotype frequencies in the populations. A homozygous recessive plant would be (pp) and a Homozygous dominant...
) Imagine that you are a grad student in Biological Anthropology, and you travel to the...
) Imagine that you are a grad student in Biological Anthropology, and you travel to the land of “Jenesaiquoivia” to study Jenesaiquoivian genetics. [Footnote 1] You notice that 9 out of every 100 people in your study group have blue eyes. The rest of the population have brown eyes. You know that the brown eyes allele is dominant, and blue eyes is recessive. [2] Construct the Hardy-Weinberg equation for that population. You know that q2=.09 (9%) so p2 plus 2pq...
In 1783, a British sailing ship crashes on an island in the South Pacific. 100 individuals...
In 1783, a British sailing ship crashes on an island in the South Pacific. 100 individuals survive, only to discover that the island is already inhabited by a native population of 300. With no hope of returning home, they join the local population. A) In Great Britain, the allele frequency for blue eyes (b) is 0.8. If the native population is true-breeding for brown eyes (B, which has complete dominance over b), what is the new allele frequency for this...
1. You are studying a population in Hardy-Weinberg equilibrium, where eye colour is controlled by a...
1. You are studying a population in Hardy-Weinberg equilibrium, where eye colour is controlled by a single gene. This gene has 3 alleles: A1 is dominant to A2 and A3, and the phenotype is brown eyes A2 is dominant to A3, and the phenotype is blue eyes A3 phenotype is green eyes. a) What eye colour would be expected for someone whose genotype is A2A3? a) Green b) Blue c) Brown b) b) If the frequency of the A1 allele...
Use the equation for Hardy-Weinberg equilibrium for the following questions. p + q = 1 p2...
Use the equation for Hardy-Weinberg equilibrium for the following questions. p + q = 1 p2 + 2pq + q2 = 1 In domestic cats, the allele for short hair (L) is dominant and the allele for long hair (l) is recessive. You are studying a population of cats found on Iwato Island and have determined that the frequency of the recessive long hair allele is q = 0.3. A) What is the frequency of the dominant allele for short...
1) Rabbit’s ears can be either short (dominant) or floppy (recessive). If a population of rabbits...
1) Rabbit’s ears can be either short (dominant) or floppy (recessive). If a population of rabbits in Hardy-Weinberg equilibrium has 250 short eared rabbits and 100 floppy eared rabbits, what is the allele frequency of the “short” allele? A.0.714 B.0.465 C.0.845 D.0.535 2)You are observing the suit (clubs, diamonds, hearts, or spades) of a randomly drawn card from a deck. What is the degree(s) of freedom if you were to perform the Chi-square test on the expected and observed outcome...
Cystic fibrosis is an autosomal recessive disorder caused by a loss-of-function mutation in a chlorine channel,...
Cystic fibrosis is an autosomal recessive disorder caused by a loss-of-function mutation in a chlorine channel, which causes a thickening of the mucus, leading to respiratory failure. This disease affects approximately 1 in 1353 newborns in the Republic of Ireland every year. Fill in the following values: p = q = Current population of Ireland is approximately 4.8 million. Based on the allele frequencies that you calculated, what is the expected number of cystic fibrosis carriers (heterozygotes) in Ireland? Show...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT