Download 4. F Phenotypes Class Observed

4. F2 Phenotypes
Class
Normal
Hydrocephalic
Observed
136
42
5. F2 Phenotypes
Class
Smooth, purple
Wrinkled, purple
Smooth, white
Wrinkled, white
Observed
2418
869
740
330
6. F2 Phenotypes (Epistasis)
Class
Observed
Purple
870
Yellow
730
7. F2 Phenotypes (Epistasis)
Class
Observed
White
725
Yellow
194
Green
81
Answers to additional problems # 1-5
1. df = 1
X2 = 6.00
2. df = l
X2 = 1.33
3. df = 1
X2 = 0.0029
4. df = 1
X2 = 0.187
5. df = 3
X2 = 23.38
6. df = 1
X2 = 2.29
7. df = 2
X2 = 6.95
95
Chi-Square (χ2) Test:
1. Degrees of Freedom (df) = 1
Hypothetical Ratio = 1:1
O
E
O-E = D
D2
D2/E
Σ(D2/E) = χ2
1950 : 2106
2028 : 2028
-78 : 78
6084 : 6084
3:3
6
χ2 (6) @ 1 df
is between 3.841 and 6.635
.01 < P < .05
Because P ≤ .05, take H1, and
Reject 1:1 hypothetical ratio.
Chi-Square (χ2) Test:
6. Degrees of Freedom (df) = 1
Hypothetical Ratio = 9:7. Compare to 13:3 and
15:1 by fitting total number of observed into
phenotypic classes according to ratios.
O
870 : 730
E
900 : 700
O-E = D
-30 : 30
D2
900 : 900
D2/E
1.0 : 1.29
2.29
Σ(D2/E) = χ2
χ2 (2.29) @ 1 df
is between 1.642 and 3.841
.05 < P < .20
Because P > .05, take H0, and Accept 9:7
hypothetical ratio.
PEDIGREE ANALYSIS
Introduction
1.
Practical applications of the Mendelian principles and the
laws of probability are made by human geneticists and by
animal breeders in analyzing pedigrees.
2.
In human genetics, pedigrees are important not only
because man has a long generation time and few offspring,
but also because, unlike experimental animals, his matings
cannot be controlled by a geneticist.
3.
Traits with a simple pattern of inheritance may sometimes
be traced accurately enough to justify predictions concerning
the likelihood of their expression in future children.
4.
Examinations of family histories are an important approach in
the field of genetic counseling.
5.
The first step in such an analysis is to determine whether the
trait in question is behaving as a dominant or a recessive.
6.
Recessive genes are difficult to keep track of because they may
remain hidden by their dominant alleles generation after
generation.
7.
Carriers in the population usually cannot be identified until an
expression occurs.
8.
Recessives are expressed more frequently in families in which
the father and mother are more closely related than they are in
the general population.
9.
Some phenotypes may behave as dominant in some families
and recessive in others.
10.
In absence of data to indicate which individuals are carriers,
probability becomes the best tool for determining likelihood of
expression of a certain trait in a family.
11.
It may be necessary to use estimates of the frequency of a gene
in the general population.
12.
Probability can be based on the family history, which may be
recorded in a pedigree chart. (see figure in the next page.)
98
Pedigree analysis:
The study of an inherited trait in a group of
related individuals to determine the pattern
and characteristics of the trait, including its
mode of inheritance, age of onset, and
phenotypic variability.
Normal female
Normal male
Unknown
Affected
Affected
Marriage line
Offspring line
Dizygotic
twins
Monozygotic
twins
99
Offspring, in
order of birth
(left to right)
Monozygotic twins: Twin offspring that
have developed from one oocyte
Diozygotic twins: Twin offspring that have
developed from two separate oocytes
fertilized at the same time
Parents Same phenotype x same phenotype
Progeny Contrasting phenotype
Based on the above information, you know . . .
1. The phenotype of the parents is dependent upon the dominant gene,
and they are heterozygous.
2. The phenotype of the progeny is dependent upon the recessive
gene, and the progeny is homozygous recessive.
Aa
Aa
A - light
a - dark
Aa x Aa
AA Aa
Aa aa
If the parents are heterozygous and the phenotype of the progeny is
dependent upon the dominant gene, the probability that the progeny is
homozygous is 1/3 and the probability that the progeny is
heterozygous is 2/3.
1/1 Aa
1/1 Aa
1/3 AA
2/3 Aa
1/1 aa
100
Pedigree problem on page 101 of workbook:
1) Parents: Aa × Aa
sister: aa
Therefore, you are :
2)
A a
A AA Aa
a Aa aa
1/3 AA
×
2/3 Aa
A gamete for AA: 1/3 × 1/1 = 1/3
A gamete for Aa: 2/3 × 1/2 = 2/6 = 1/3
A gamete: 1/3 + 1/3 = 2/3
a gamete for AA: 1/3 × 0 = 0
a gamete for Aa: 2/3 × 1/2 = 1/3
a gamete: 0 + 1/3 = 1/3
3) In progeny:
aa genotype:
AA genotype:
Aa genotype:
1/3 AA
2/3 Aa
1/1 Aa
A gamete = 1/2
a gamete = 1/2
1/3 × 1/2 = 1/6
2/3 × 1/2 = 2/6
2/3 × 1/2 = 2/6
1/3 × 1/2 = 1/6
2/6 + 1/6 = 3/6
Total of aa, AA, and Aa genotypes = 1/6 + 2/6 + 3/6 = 6/6 = 1
I
1
2
II
1
2
3
III
1
2
3
4
5
103
6
7
8
I
1
2
1/1Aa 1/1Aa
1/2 A 1/2 A
1/2 a
1/2 a
II
1
2
3
1/1aa 1/1AA
1/1 a 1/1 A
1/3AA 1/1AA
2/3Aa 1/1 A
2/3 A
1/1AA 1/3AA
1/1 A 2/3Aa
2/3 A
1/3 a
1/3 a
III
1
2
1/3Aa
2/3AA
5/6 A
3
4
5
1/1Aa
1/2 A
1/2 a
6
7
1/3Aa
2/3AA
5/6 A
1/6 a
1/6 a
103
8
Forming gametes from parent
genotypes (addition rule):
• II-1: A; (1/3 x 1/1) + (2/3 x 1/2) = 1/3 + 2/6
= 2/3
• a; (1/3 x 0) + (2/3 x 1/2) = 0 + 2/6 = 1/3
• 2/3 A + 1/3 a = 3/3 = 1
• III-1,2,3: A; (1/3 x 1/2) + (2/3 x 1/1) = 1/6 +
2/3 = 5/6
• a; (1/3 x 1/2) + (2/3 x 0) = 1/6 + 0 =1/6
• 5/6 A + 1/6 a = 6/6 = 1
Forming progeny genotypes from
parent gametes (product rule):
• II: 1/3AA x 1/1 AA:
• 2/3 A x 1/1 A = 2/3 AA
• 1/3 a x 1/1 A = 1/3 Aa
• III-2 x III-8 :
• AA: 5/6 A x 5/6 A = 25/36 AA
• Aa: (5/6 A x 1/6 a) + (1/6 a x 5/6 A) = 10/36 Aa
• aa: 1/6 a x 1/6 a = 1/36 aa
• 25/36 AA + 10/36 Aa + 1/36 aa = 36/36 = 1
Chart 2
1
1
2
4
3
2
1
2
Chart 2
1
2
1/1 aa
1/1 a
1/1 Aa
1/2 A
1/2 a
1
1/1 Aa 1/1 AA
1/2 A 1/1 A
2
1/1 Aa
1/2 A
1/1 Aa
1/2 A
1/2 a
1/2 a
1/2 a
1/2 AA
1/2 Aa
3/4 A
1/4 a
1/1 aa
1/1 a
1/3 AA
2/3 Aa
2/3 A
3
1/1 aa
1/1 a
1/1 AA
1/1 A
4
1/1 Aa
1/2 A
1/1 AA
1/1 A
1/2 a
1/1 AA 1/1 AA 1/1 Aa
1/1 A 1/1 A 1/2 A
1/2 a
1/3 a
1
2/3 AA
1/3 Aa
5/6 A
2
1/2 AA
1/2 Aa
3/4 A
1/6 a
1/4 a
1/2 AA
1/2 Aa
3/4 A
1/4 a
Chart #3
1
1
2
2
3
1
4
2
1
Chart #3
2
1/1 aa
1/1 a
1/1 Aa
1/2 A
1/2 a
1
2
1/1 Aa 1/1 AA 1/1 Aa
1/2 A 1/1 A 1/2 A
1/1 Aa
1/2 A
1/2 a
1/2 a
1/2 a
1/2 AA
1/2 Aa
3/4 A
1/4 a
1/1 aa
1/1 a
3
1/1 aa
1/1 a
1/3 AA
2/3 Aa
2/3 A
1/1 AA 1/1 aa 1/1 Aa
1/1 A 1/1 a 1/2 A
1/2 a
1/3 a
1/1 AA
1/1 A
4
1/1 Aa
1/2 A
1/1 Aa
1/2 A
1/2 a
1/2 a
1/3 AA 1/1 aa
2/3 Aa
2/3 A 1/1 a
1/3 a
1
2
2/3 AA
1/3 Aa
5/6 A
1/6 a
1/1 Aa
1/2 A
1/2 a