Statistics
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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
xx
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.
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
xx
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.
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