1. In a population of lizards, if the S gene has a frequency of p=0.7 then answer the following using the HardyWeinberg equations and the Punnett square. In this case the S gene shows incomplete dominance over the s gene, and the trait is visible to the naked eye ( S is for smooth skin and s is for rough skin). A) if the S gene is incompletely dominant over the s gene will you be able to distinguish the 5S and S5 genotypes? BRIEFLY explain your answer. If p=0.7, then q= B) Calculate the following, and show your work Frequency of ss genotype, p2= Frequency for the S s genotype, 2pq= Frequency for the ss genotype, q2= What do your frequencies add up to? = 2. In a different population of the same lizards from question 1 you observe the following numbers: 50 smooth skinned (SS): 300 intermediate roughness (Ss): 450 rough skinned (ss). Count the number of alleles to determine the gene frequency. BE CAREFULI it is easy to forget that the number of alleles is twice the number of individuals since these are diploid organisms. p= Use the Hardy-Weinberg equation to calculate the following: For SS p2= , for 552pq= for 55q2= 3. The allele L ( which codes for longer bill) shows complete dominance over the allele I (which codes for short bili). In at population of 400 birds, you see 360 with longer bills and 40 with short bills. Calculate the allele, genotype and phenotype frequencles from these data, and show your work! HINT: Since you do not know the number of LI individuals, think about what you can use to calculate either p or q. Once you know either p or q youshould be able to calculate the other value. A. The frequency of L,P= The frequency of 1,9 in B. Genotypic frequencies are: p2 (for LL) =2,2pq(for(U)=, and q2( for (1)= C. The phenotyple frequencies are: Longer bill = whorter bill =q2 D. If the population is in Hardy-Weinbere cquilibrium and the next generation contains 1,200 individuats, what would you the expected numbers of different senotypes to be? 4L: 4. II E. What would your expected number of phenotypes be? Longer billed: Shorter billed 4. K is the aliele for a normal antiporter protein in a cell membrane. The allele frequency for k is p= 0.8 . The normal allele is codominant with a mutated aflele k (allele frequency q=0.2 ), which means both are expressed and both antiporter proteins are found on the surface of heterozygous celis. For a population of 1,000 individuals, the expected numbers for the different genotypes are: 640kk:320kk:40kk. Your observed genotypes were 665KK:315kk:20kk. Perform a Chi-squared lest to see if this difference is significant or not. The number of degrees of freedom is df = From the table of Chi-squared values, the critical value, meaning Chi Square that will provide p=0.05 (the highlighted column) for the number of degrees of freedom (di) is The sample Chi-squared value calculated from the data = Is the nul hypothesis accepted or rejected?.

1. 1. In a population of lizards, if the S gene has a frequency of p=0.7 then answer the following
using the HardyWeinberg equations and the Punnett square. In this case the S gene shows
incomplete dominance over the s gene, and the trait is visible to the naked eye ( S is for smooth
skin and s is for rough skin). A) if the S gene is incompletely dominant over the s gene will you
be able to distinguish the 5S and S5 genotypes? BRIEFLY explain your answer. If p=0.7, then
q= B) Calculate the following, and show your work Frequency of ss genotype, p2= Frequency
for the S s genotype, 2pq= Frequency for the ss genotype, q2= What do your frequencies add up
to? = 2. In a different population of the same lizards from question 1 you observe the following
numbers: 50 smooth skinned (SS): 300 intermediate roughness (Ss): 450 rough skinned (ss).
Count the number of alleles to determine the gene frequency. BE CAREFULI it is easy to forget
that the number of alleles is twice the number of individuals since these are diploid organisms.
p= Use the Hardy-Weinberg equation to calculate the following: For SS p2= , for 552pq= for
55q2= 3. The allele L ( which codes for longer bill) shows complete dominance over the allele I
(which codes for short bili). In at population of 400 birds, you see 360 with longer bills and 40
with short bills. Calculate the allele, genotype and phenotype frequencles from these data, and
show your work! HINT: Since you do not know the number of LI individuals, think about what
you can use to calculate either p or q. Once you know either p or q youshould be able to calculate
the other value.
A. The frequency of L,P= The frequency of 1,9 in B. Genotypic frequencies are: p2 (for LL)
=2,2pq(for(U)=, and q2( for (1)= C. The phenotyple frequencies are: Longer bill = whorter bill
=q2 D. If the population is in Hardy-Weinbere cquilibrium and the next generation contains
1,200 individuats, what would you the expected numbers of different senotypes to be? 4L: 4. II
E. What would your expected number of phenotypes be? Longer billed: Shorter billed 4. K is the
aliele for a normal antiporter protein in a cell membrane. The allele frequency for k is p= 0.8 .
The normal allele is codominant with a mutated aflele k (allele frequency q=0.2 ), which means
both are expressed and both antiporter proteins are found on the surface of heterozygous celis.
For a population of 1,000 individuals, the expected numbers for the different genotypes are:
640kk:320kk:40kk. Your observed genotypes were 665KK:315kk:20kk. Perform a Chi-squared
lest to see if this difference is significant or not. The number of degrees of freedom is df = From
the table of Chi-squared values, the critical value, meaning Chi Square that will provide p=0.05
(the highlighted column) for the number of degrees of freedom (di) is The sample Chi-squared
value calculated from the data = Is the nul hypothesis accepted or rejected?