Have the student use the infomration obtain regarding the USA population to determine the “Expected” frequency and counts as well as having them  to analyze the data and provide phenotype, genotype and allele frequencies plus the perform chi-square analyses. Have the student to complete every single empty square. For frequencies, (O-E)2 /E and x^2 round to two decimal places. Use whole numbers for counts

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Have the student use the infomration obtain regarding the USA population to determine the “Expected” frequency and counts as well as having them  to analyze the data and provide phenotype, genotype and allele frequencies plus the perform chi-square analyses. Have the student to complete every single empty square. For frequencies, (O-E)2
/E and x^2 round to two decimal places. Use whole numbers for counts… people are counted in whole numbers! 

I. Test S = Sodium benzoate:
BIO 340
USA
II. Test T = Phenylthiocarbamide (PTC)
TASTERS (p² + 2pq)
#
280
N.A.
III. Test U = Thiourea
1.
2.
TASTERS (p²+2pq)
#
225
N.A.
BIO 340
USA
3.
Observed BIO 340 Counts
Observed BIO 340 Frequency
Expected Frequency
Expected Counts
(O-E)²/E
4.
FREQUENCY
5.
0.25
BIO 340
USA
FREQUENCY
Observed BIO 340 Counts
Observed BIO 340 Frequency
Expected Frequency
Expected Counts
(O-E)²/E
0.70
TASTERS (p² + 2pq)
#
230
N.A.
Did you expect these results? Explain why.
FREQUENCY
Observed BIO 340 Counts
Observed BIO 340 Frequency
Expected Frequency
Expected Counts
(O-E)²/E
NON-TASTERS (9²)
#
75
N.A.
SS (p²)
#
20
N.A.
NON-TASTERS (q²)
TT (p²)
FREQUENCY
#
70
N.A.
Ss (2pq)
UU (p²)
FREQUENCY
NON-TASTERS (9²)
Tt (2pq)
FREQUENCY
Uu (2pq)
ALLELE FREQUENCIES
p
р
ss (q²)
75
tt (q²)
20
Mention 3 reasons why most populations are NOT in Hardy-Weinberg equilibrium:
ALLELE FREQUENCIES
Р
0.45
uu (q²)
70
TOTAL
300
1
1
x² =
q
=
TOTAL
300
ALLELE FREQUENCIES
Р
9
Use the Chi-square stable, which can be found in the Chi-Square Analysis of Mendel's Data page of your
online textbook as well as in the Module IV Chi-square analysis lecture, to determine whether you
should reject the null hypotheses (Ho) or not. Provide your answers next to each of the compounds.
Sodium benzoate Ho:
Phenylthiocarbamide (PTC) Ho:
Thiourea Ho:
x² =
How do the BIO 340 student results compare to those of previously published data from the USA
population? Note: Compare the genotype/allele frequencies of each of the test papers.
1
1
9
0.55
TOTAL
300
1
1
Transcribed Image Text:I. Test S = Sodium benzoate: BIO 340 USA II. Test T = Phenylthiocarbamide (PTC) TASTERS (p² + 2pq) # 280 N.A. III. Test U = Thiourea 1. 2. TASTERS (p²+2pq) # 225 N.A. BIO 340 USA 3. Observed BIO 340 Counts Observed BIO 340 Frequency Expected Frequency Expected Counts (O-E)²/E 4. FREQUENCY 5. 0.25 BIO 340 USA FREQUENCY Observed BIO 340 Counts Observed BIO 340 Frequency Expected Frequency Expected Counts (O-E)²/E 0.70 TASTERS (p² + 2pq) # 230 N.A. Did you expect these results? Explain why. FREQUENCY Observed BIO 340 Counts Observed BIO 340 Frequency Expected Frequency Expected Counts (O-E)²/E NON-TASTERS (9²) # 75 N.A. SS (p²) # 20 N.A. NON-TASTERS (q²) TT (p²) FREQUENCY # 70 N.A. Ss (2pq) UU (p²) FREQUENCY NON-TASTERS (9²) Tt (2pq) FREQUENCY Uu (2pq) ALLELE FREQUENCIES p р ss (q²) 75 tt (q²) 20 Mention 3 reasons why most populations are NOT in Hardy-Weinberg equilibrium: ALLELE FREQUENCIES Р 0.45 uu (q²) 70 TOTAL 300 1 1 x² = q = TOTAL 300 ALLELE FREQUENCIES Р 9 Use the Chi-square stable, which can be found in the Chi-Square Analysis of Mendel's Data page of your online textbook as well as in the Module IV Chi-square analysis lecture, to determine whether you should reject the null hypotheses (Ho) or not. Provide your answers next to each of the compounds. Sodium benzoate Ho: Phenylthiocarbamide (PTC) Ho: Thiourea Ho: x² = How do the BIO 340 student results compare to those of previously published data from the USA population? Note: Compare the genotype/allele frequencies of each of the test papers. 1 1 9 0.55 TOTAL 300 1 1
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