Homework 2-Chapter 23 The Hardy-Weinberg Principle 1 of 1 Keep in mind that you
ID: 276862 • Letter: H
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Homework 2-Chapter 23 The Hardy-Weinberg Principle 1 of 1 Keep in mind that you just leamed in Part A that The allele frequency of T is 0.4 The allele frequency of T is 0.6 The equation for Hardy-Weinberg equilibrium states that at a lacus with two alleles, as in this cat population, the three genotypes will occur in specific propartions: p2+2p+2-1 Enter the values for the expected frequency of each genotype: T^, TTS, and TSTS. Enter your answers numericaly to two decimal places, not as percentages. For help applying the Hardy Weinberg equation to this cat population, see Hints 1 and 2. View Available Hint(s) ? help Phenotype (tail length) Number of individuals in population Genotype Expected frequency ong 60 medium 40 short 100 Submit Part C This question will be shown after you complete previous question(s).Explanation / Answer
Answer:
Gene frequency: The proportion of different alleles of a gene in a population is called as Gene frequency. If locus has two alleles A & a existing with frequencies p &q respectively.
A=p and a=q, p+q=1.
Genotypic frequencies: The proportion of different genotypes of a character in a population is called as Genotypic frequency. If locus has two alleles A & a existing with frequencies p &q respectively.The possible genotypes are
AA-p2 ; Aa: pq ; aa: q2
p+q=1
(p+q)2= 1
p2 +2pq+q2 =1.
Here it may due to incomplete / co-dominance. Heterozygous clearly differentiate from either of homozygous. Existing locus has two allelea (TL and TS). Here three phenotypes and three genotypes observed.
The frequency of TL is 0.4; p=0.4. The frequency of TS is 0.6; q=0.6.
p+q=1; 0.4+0.6=1.
The genotypic frequencies are: TLTL:TLTS: TSTS
Total population is 200.
Observed population:
Population with TLTL genotype: 60
Population with TLTS genotype: 40
Population with TSTS genotype: 100
Expected and observed Frequency:
TL allele frequency= [TLTL+ 1/2(TLTS)]/Total; 60+20/200= 0.4
Expected TLTL genotype frequency = p2; p=0.4 then p2=0.16
Observed TLTL genotype frequency= 60/200= 0.3
Expected TLTS genotype frequency= 2pq; 2*0.4*0.6 then 2pq= 0.48
Observed TLTS genotype frequency= 40/200= 0.2
TS allele frequency= [TSTS+ 1/2(TLTS)]/Total; 100+20/200= 0.6
Expected TSTS genotype frequency = q2; p=0.6 then p2=0.36
Observed TSTS genotype frequency= 100/200= 0.5
Total expected and observed frequency
1. TLTL frequency= expected: 0.16 and observed: 60/200= 0.3
2. TLTS frequency= expected: 0.48 and observed: 40/200= 0.2
3. TSTS frequency= expected: 0.36 and observed: 100/200= 0.5
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