To tal # of rerpondents = 215 + 85 300A Phenoty pic FrequencyCan Roll fongue =Cannat Roll tongue =B. Allele FrequenuitsUse the frequency of the hono zygous recessive tvont (cannat nll tingue)C. Expected Genaty pe Frequencier (arcuming HWE)RR p2 =2 pgRr =D. Expected Numb er of Individuals of each GenotypePR =Rr =E. Expected Number of tndividuals of coch PhonotypeCan Roll tongue (p² and 2pg).Cannot Rall tongue (q²).-rr =F. Chi - square (x)Since the obscrved data (215 and 85) arepheno ty pie humber , then use theexpected nunber of individuals of cach phunotype. The dere of fredom is. 2. A survey was conducted for a certain trait (the ability to roll tongue or inability to roll thetongue) in a population. The result shows that 215 of the respondents can roll theirtongue and 85 cannot roll their tongue. The picture below shows a person rolling histongue. Rolling the tongue into a tube shape is often described as a dominant trait.If the dominant allele is represented by R (can roll tongue) and rrepresents the recessive allele, calculate the phenotypic frequencies,allele frequencies, the expected numbers of the three genotypicclasses and the expected numbers of the two phenotypic classes(assuming random mating). Using chi-square test, determine whetheror not this population is in Hardy-Weinberg equilibrium.REMEMBER:PhenotypeRR and Rr Can roll tongueCannot roll tongueGenotypeHWE termp2 and 2pqq?rrAgain, follow the format below for your answers. Regarding letter "F" on the format, you willshow the computations on how you arrive to the chi-square value and also include theinterpretation like what we did in the examples.

A chi-square value is shows the deviation of the populations from the expected values.

Answered: 2. A survey was conducted for a certain…

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Publisher:Matthew Douglas, Jung Choi, Mary Ann Clark

Chapter19: The Evolution Of Populations

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To tal # of rerpondents = 215 + 85 300
A Phenoty pic Frequency
Can Roll fongue =
Cannat Roll tongue =
B. Allele Frequenuits
Use the frequency of the hono zygous recessive tvont (cannat nll tingue)
C. Expected Genaty pe Frequencier (arcuming HWE)
RR p2 =
2 pg
Rr =
D. Expected Numb er of Individuals of each Genotype
PR =
Rr =
E. Expected Number of tndividuals of coch Phonotype
Can Roll tongue (p² and 2pg).
Cannot Rall tongue (q²).
-rr =
F. Chi - square (x)
Since the obscrved data (215 and 85) are
pheno ty pie humber , then use the
expected nunber of individuals of cach phunotype. The dere of fredom is.

2. A survey was conducted for a certain trait (the ability to roll tongue or inability to roll the
tongue) in a population. The result shows that 215 of the respondents can roll their
tongue and 85 cannot roll their tongue. The picture below shows a person rolling his
tongue. Rolling the tongue into a tube shape is often described as a dominant trait.
If the dominant allele is represented by R (can roll tongue) and r
represents the recessive allele, calculate the phenotypic frequencies,
allele frequencies, the expected numbers of the three genotypic
classes and the expected numbers of the two phenotypic classes
(assuming random mating). Using chi-square test, determine whether
or not this population is in Hardy-Weinberg equilibrium.
REMEMBER:
Phenotype
RR and Rr Can roll tongue
Cannot roll tongue
Genotype
HWE term
p2 and 2pq
q?
rr
Again, follow the format below for your answers. Regarding letter "F" on the format, you will
show the computations on how you arrive to the chi-square value and also include the
interpretation like what we did in the examples.

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1. The ability to taste the compound PTC is controlled by a dominant allele T, while individuals homozygous for the recessive allele (t) cannot taste PTC. In a population consisting of 500 individuals, 347 are tasters and 153 are non-PTC tasters. Calculate the frequency of the T and t alleles in this population, and frequency of the genotypes.
(Please train yourself to use the Hardy-Weinberg equation.)
To present your answers, follow the format in the picture below.

p' 2 pg +: 1
a. Phenetypic Freguenty
PTC tasters :
Non-PTC tackae =
b. Frequenty of the T and + aleler
Manipulating
p+: 1...
C. Freguenty of the geroty pes
TT p :
Tt 2p9
H: 9° =
* Note : If we add p?t tpg+ q*, if chould be equal to 1.

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