Statistics

Chi square Χ2

Χ2 = ∑(O – E)2/E Class

Red

Observed 705 • •

White 224

Null hypothesis : Ratio of red : white is 3 : 1

Class

Red

Observed 705 Expected

White

Total

224

929

696.75 232.25 929

∀ Χ2 = (705-696.75)2/696.75 + (224-232.25)2/232.25 = 0.0977 + 0.2931 = 0.03908 •

From the Χ2 table, Χ2 at probability 0.05 (5%) and 1 degree of freedom is 3.84

•

The null hypothesis is accepted. The ratio between red and white is 3 : 1

MN Blood Group - Codominance Parents

Children

Number Frequency Number Frequency

M MN N

948 1568 644

1024 1627 722

Total

3160

3373

Allele LM Allele LN

3464 2856

Total

6320

0.548 0.452

3675 3071 6746

0.304 0.482 0.214

0.545 0.455

Χ2 = Σ(Ο – E)2/Ε •

Class

Allele LM

Allele LN

Observed

3675

3071

Null hypothesis: No difference between the allele frequencies of the children and the parents i.e. p = 0.548 and q = 0.452.

2. Class

Allele LM

Allele LN

Total

Observed

3675

3071

6746

Expected

3697

3049

6746

∀

Χ2 = (3675-3697)2/3697 + (3071-3049)2/3049 = 0.14 + 0.16 = 0.30.

•

From the Χ2 table, Χ2 at the 5% level with 1 degree of freedom is 3.84.

•

The null hypothesis is accepted, i.e. the is no difference in the LM and LN frequencies between the children and the parental populations.

Class

L ML M

Observed 948

•

2.

∀

L ML N

LNLN

1568

644

LM

LN

0.548 0.452

Null hypothesis: Genotypic frequency LMLM, LMLN, LNLN is in accordance with the formula p2, 2pq, q2 where p is the frequency of allele LM and q is the frequency of LN. Class

L ML M

L ML N

LNLN

Total

Observed 948

1568

644

3160

Expected 949

1565

646

3160

Χ2 = (948-949)2/949 + (1568-1565)2/1565 + (644-646)2/646 = 0.001 + 0.005 + 0.006 = 0.01.

•

From the Χ2 table, Χ2 at the 5% level with 2 degrees of freedom is 5.99.

•

The null hypothesis is accepted, i.e. genotypic frequency LMLM, LMLN, LNLN is in accordance with the formula p2, 2pq, q2 .

2 x 2 Table Generatio n Parents

Χ² =

Χ² =

Allele LM

Allele LN Total

Children

3464 3675

2856 3071

6320 6746

Total

7139

5927

13066

n(|ad-bc| -½n)²

B1

(a+b)(c+d)(a+c)(b+d)

B2 Total

A1

A2

Total

a c

b d

(a + b) (c + d)

(a + c)

(b + d) n = (a+b+c+d)

13066(|3464x3071 – 2856 x 3675| - 6533)² 6320 x 6746 x 7139 x 5927

= 0.13 with 1 degree of freedom

Probability Theory Carrier

Carrier

Aa

x

Aa

AA

Aa

aa

Normal

Normal

Albino

¼

½

¼

Addition rule: If an event can occur in more than one way, the probability of the event in the sum of the probability of each event. Example: the probability of getting a normal baby = … +  =  Multiplication rule: The probability of several independent events occurring together is the product of the probability of each event. Example: the probability of getting the 1st child normal, 2nd child albino and the 3rd child normal =  x … x  =9/64

The Probability Method (phenotypic or genotypic ratio) RrYyCc x RrYyCc Rr x Rr

Yy x Yy

Cc x Cc

RR Rr rr

YY Yy yy

CC Cc cc

¼

½

¼

¼

½

¼

¼

½

¼

Example 1: Frequency of genotype RRYy cc Answer: ¼ x ½ x ¼ = 1/32 or 2/64 Example 2: Phenotype Wrinkled–Yellow–Red Answer: ¼ x (¼ + ½) x (¼ + ½ ) = 9/64

Forked-Line Method (phenotypic or genotypic ratio)

1st child

2nd child

3rd child

Proportion

¾A

xx

¼a

¾x¾x…

¾A

A 

A

¾x…x¾ …a

…

a



A

¾x¼x …

A …

a …a



Forked-Line Method

A …a

…

a

Binomial Theorem (p + q)4 = p4 + 4p3q + 6p2q2 + 4pq3 + q4 A couple, both carriers of albino (Aa) intends to have 4 children. Aa

x

Aa

AA

Aa

aa

¼

½

¼

1. What is the probability of having all normal children? p4 =(¾)4 •

Probability of having 3 normal and 1 albino children? 4p3q = 4(¾)3(¼)

•

Probability of having at least 1 normal child? p4 + 4p3q + 6p2q2 + 4pq3

(p + q)4 = p4 + 4p3q + 6p2q2 + 4pq3 + q4

Pascal triangle 1 1 1 1

1 2

3 4

1 3

6

1 4

1

4

6

4

1

1p4

4p3

6p2

4p1

1p0

1p4q0

4p3q1

6p2q2

4p1q3

1p0q4

p4

4p3q

6p2q2

4pq3

q4

1

Key Points • Most organisms have sex chromosomes but sex genes may also be found on autosomes. • Evidence for sex determination in Drosophila comes from study on Drosophila genetics and gynandromorph. • Evidence for sex determination in human comes from the study on human sex anomalies, sex mosaics and Sry genes.