PG Calculations (W4b)

Question 1

Chi² test formula?

Answer 1

Chi² = Sum of (O–E)² / E

Question 2

What do symbols of chi² formula mean?

Answer 2

• O = Observed value.
• E = Expected value (calculated).

Question 3

Chi² use?

Answer 3

To evaluate how well your data fits the model (goodness of fit test).

Question 4

Interpreting chi² steps? (2)

Answer 4

• Determine degrees of freedom (df).
• Determine probability value (P) from a standard table/graph.

Question 5

If significance level < 0.05 attributes? (3)

Answer 5

• Reject Ho.
• Significant deviation from HWE.
• Population is evolving.

Question 6

If significance level >= 0.05 attributes? (3)

Answer 6

• Fail to reject Ho.
• No significance deviation from HWE.
• Population not evolving.

Question 7

If calculated chi² >= table chi², what does it mean? (2)

Answer 7

• Reject Ho at significance level chosen.
• Population is evolving.

Question 8

Things to note in PG PG calculations? (2)

Answer 8

• Significance level will always be given but if not stick with 0.05.
• Include 3 significant figures for decimal places.

Question 9

Other formats for the genotypes? (3)

Answer 9

• Homozygous dominant = P1P1.
• Heterozygous = P1P2.
• Homozygous recessive = P2P2.

Question 10

Given the following information:

• Total individuals = 800.
• BB + Bb = 490 individuals.
• bb = 310 individuals.

Find B & b.

Answer 10

1) Find q (b)

bb = 310/800 = 0.387 = q²
q² = 0.387.

Therefore,
q = /q² = /0.387 = 0.622
q = 0.622
b = 0.622.

2) Find p (B)

From p + q = 1
p = 1 – q = 1 – 0.622 = 0.377.
B = 0.377

Therefore,
B = 0.377
b = 0.622

Question 11

In our population of 250 red stars:
• 84 dark pink tepals (P1P1).
• 124 light pink tepals (P1P2).
• 42 white tepals (P2P2).

Questions:
1) Calculate allele frequency.
2) Calculate expected genotype frequency & no. of individuals based on HWE.
3) Do the observed genotypes differ from the HWE expectations?
4) Do the observed genotypes differ from HWE?

Answer Q1? (4)

Answer 11

1) Calculate allele frequency (p & q)

(i) Get total no. of alleles
N = 250
Total alleles = 250×2 = 500.

(ii) Get no. of alleles for each genotype (P1P1 & P2P2)

• P1P1 = 84
P1P1 alleles
= (P1P1×2)+P1P2 = (84×2)+124 = 292

• P2P2 = 42
P2P2 alleles
= (P2P2×2)+P1P2 = (42×2)+124 = 208

(iii) Get allele frequency for each genotype (P1P1 & P2P2 ~ p & q)

• p
= P1P1
= P1P1 alleles ÷ Total alleles
= 292/500
= 0.584

• q
= P2P2
= P2P2 alleles ÷ Total alleles
= 208/500
= 0.416

(iv) Therefore,

p = 0.585
q = 0.416

Question 12

In our population of 250 red stars:
• 84 dark pink tepals (P1P1).
• 124 light pink tepals (P1P2).
• 42 white tepals (P2P2).

Questions:
1) Calculate allele frequency.
2) Calculate expected genotype frequency & no. of individuals based on HWE.
3) Do the observed genotypes differ from the HWE expectations?
4) Do the observed genotypes differ from HWE?

Answer Q2? (4)

Answer 12

2) Calculate expected genotype frequency & no. of individuals based on HWE.

(i) Use ratio of p²:2pq:q², allele frequencies & total no. of individuals

• p²
= P1P1
= (0.584)² × 250
= 85.26

• 2pq
= P1P2
= 2(0.584)(0.416) × 250
= 121.5

• q²
= P2P2
= (0.416)² × 250
= 43.3

Question 13

In our population of 250 red stars:
• 84 dark pink tepals (P1P1).
• 124 light pink tepals (P1P2).
• 42 white tepals (P2P2).

Questions:
1) Calculate allele frequency.
2) Calculate expected genotype frequency & no. of individuals based on HWE.
3) Do the observed genotypes differ from the HWE expectations?
4) Do the observed genotypes differ from HWE?

Answer Q3? (3)

Answer 13

3) Do the observed genotypes differ from HWE expectations/expected genotypes?

(i) Observed genotypes (What we saw)

P1P1 = 84.
P1P2 = 124.
P2P2 = 42.

(ii) Expected genotypes (What is predicted-HWE)

P1P1 = x = 85.26
P1P2 = y = 121.5
P2P2 = z = 43.3

(iii) Compare both & conclude

The observed genotypes do not differ significantly from the HWE expectations.

Question 14

In our population of 250 red stars:
• 84 dark pink tepals (P1P1).
• 124 light pink tepals (P1P2).
• 42 white tepals (P2P2).

Questions:
1) Calculate allele frequency.
2) Calculate expected genotype frequency & no. of individuals based on HWE.
3) Do the observed genotypes differ from the HWE expectations?
4) Do the observed genotypes differ from HWE?

Answer Q4? (4)

Answer 14

4) Do the observed genotypes differ from HWE?

(i) Calculate chi²

Chi² = Sum of (O–E)² / E
= (84–85.26)² / 85.26 + (124–121.5)² / 121.5 + (42–43.3)² / 43.3

= 0.01862 + 0.05144 + 0.03903
= 0.108

Therefore, chi² = 0.108

(ii) Check table chi²

Significance level (alpha)= 0.05
Degrees of freedom = 3-2 = 1

Table chi² = 3.84

(iii) Compare calculated chi² & table chi²

Calculated chi² = 0.108 < 3.84

(iv) Conclusion

Since calculated chi² = 0.108 < 3.84, we fail to reject Ho at a significance level of 0.05, therefore the population is not evolving.

Question 15

How do we calculate the expected genotypes frequencies?

Answer 15

We use the allele frequency & then multiply by the total no. of individuals.

Question 16

Explain symbols of p² + 2pq + q² = 1? (3)

Answer 16

• p² = P1P1.
• 2pq = P1P2.
• q² = P2P2.

Question 17

Explain symbols of p + q = 1? (2)

Answer 17

• p = B.
• q = b.