Download Pedigree analysis

Probability
Probability = the # of times an event happens ⁄
the # of opportunities for an event to happen
Pedigree analysis
Probability: Product rule
• The probability of two independent events
occurring simultaneously is the product of
each of their respective probabilities
– Ie. RrYy x RrYy
Jan 11
• What is the probability of getting rryy genotype ?
• ¼ * ¼ = 1/16
• The probability of rolling a 3 on a die in one trial is:
– P (of a 3) = 1/6
– P (of two 3’s) = (1/6) x (1/6) = 1/36
• The second example show the product rule:
– The probability of two independent events occurring
simultaneously is the product of each of their respective
probabilities
Text p.39
Probability: Product rule
• The probability of two independent events
occurring simultaneously is the product of
each of their respective probabilities
– Ie. RrYy x RrYy
•
•
•
•
Probability: sum rule
• The probability of either of two mutually
exclusive events occurring is the sum of
their individual probabilities
– Probability of a 3 or a 5 when rolling a die is?
• The probability of either of two mutually
exclusive events occurring is the sum of
their individual probabilities
– Probability of a 3 or a 5 when rolling a die is
What is the probability of getting rryy genotype ?
¼ * ¼ = 1/16
What is the probability of getting RRYY genotype?
¼ * ¼ = 1/16
• P (of a 3)= 1/6 and p (of a 5) = 1/6 so
• P (of a 3 or a 5) = 1/6 + 1/6 =2/6= ⅓ .
Text p.40
Text p.39
Probability: sum rule
• The probability of either of two mutually
exclusive events occurring is the sum of
their individual probabilities
– Probability of a 3 or a 5 when rolling a die is
• P (of a 3)= 1/6 and p (of a 5) is 1/6 so
• P (of a 3 or a 5) = 1/6 + 1/6 =2/6= ⅓ .
– In the cross RrYy x RrYy what is the
probability of getting either the rryy or the
RRYY genotype?
Text p.40
Probability: sum rule
Text p.40
Probability: sum rule
• The probability of either of two mutually
exclusive events occurring is the sum of
their individual probabilities
– Probability of a 3 or a 5 when rolling a die is
• P (of a 3)= 1/6 and p (of a 5) is 1/6 so
• P (of a 3 or a 5) = 1/6 + 1/6 =2/6= ⅓ .
– In the cross RrYy x RrYy what is the
probability of getting either the rryy or the
RRYY genotype?
• 1/16 + 1/16 = 2/16 = ⅛
Probability and pedigrees eg:
albinism
• IV-3 came to the
genetic consult visit
he is planning to have
a child with a normally
pigmented woman
who also has an
albino brother. What
is the probability of
their first son being
albino?
Text p.40
1
Probability and pedigrees eg:
albinism
• IV-3 came to the genetic
consult visit he is
planning to have a child
with a normally
pigmented woman who
also has an albino
brother. What is the
probability of their first
son being albino?
• P (both parents being
heterozygous) = P (Aa) *
P (Aa) = (2/3)*(2/3)
• And P (having an albino
child) = P (aa)= ¼
• And P (having a son)= ½
• Thus P (of first son being
aa)=
(2/3)*(2/3)*(1/4)*(1/2)
=4/72=1/8
Probability and pedigrees eg:
albinism
• IV-3 came to the genetic
consult visit he is
planning to have a child
with a normally
pigmented woman who
also has an albino
brother. What is the
probability of their first
son being albino?
• P (both parents being
heterozygous) = P (Aa) *
P (Aa) = (2/3)*(2/3)
• And P (having an albino
child) = P (aa)= ¼
• And P (having a son)= ½
• Thus P (of first son being
aa)=
(2/3)*(2/3)*(1/4)*(1/2)
=4/72=1/8
Albinism in mice
• In a test cross what is the probability of
getting a male albino mouse?
• C/c X c/c
• P (of being male)= ½ * p (of being cc)= ½
=¼
Probability and pedigrees eg:
albinism
• IV-3 came to the genetic
consult visit he is
planning to have a child
with a normally
pigmented woman who
also has an albino
brother. What is the
probability of their first
son being albino?
• P (both parents being
heterozygous) = P (Aa) *
P (Aa) = (2/3)*(2/3)
• And P (having an albino
child) = P (aa)= ¼
• And P (having a son)= ½
• Thus P (of first son being
aa)=
(2/3)*(2/3)*(1/4)*(1/2)
=4/72=1/8
Albinism in mice
• The allele c causes albinism in mice (C causes
mice to be black). In a test cross what is the
probability of getting a male albino mouse?
Do you recognize the pattern?
• The seed colour Yellow is
dominant to green
• The F1 heterozygotes
crossed to plants
homozygous recessive
for seed colour
• From a monohybrid test
cross a 1:1 progeny ratio
is expected
Probability and pedigrees eg:
albinism
• IV-3 came to the genetic
consult visit he is
planning to have a child
with a normally
pigmented woman who
also has an albino
brother. What is the
probability of their first
son being albino?
• P (both parents being
heterozygous) = P (Aa) *
P (Aa) = (2/3)*(2/3)
• And P (having an albino
child) = P (aa)= ¼
• And P (having a son)= ½
• Thus P (of first son being
aa)=
(2/3)*(2/3)*(1/4)*(1/2)
=4/72=1/8
Albinism in mice
• In a test cross what is the probability of
getting a male albino mouse?
• C/c X c/c
Albino mice
• The allele c causes albinism in mice (C
causes mice to be black). A student tested
the hypothesis that a black mouse is a
heterozygote.
2
Albino mice
Chi-Square (Χ2 ) Test
Chi-Square (Χ2 ) Test
• The allele c causes albinism in mice (C
causes mice to be black). A student tested
the hypothesis that a black mouse is a
heterozygote.
• C/c X c/c
• What is the expected ratio of black to
albino progeny
• The allele c causes albinism in mice (C causes
mice to be black). A student tested the
hypothesis that a black mouse is a heterozygote.
• The allele c causes albinism in mice (C causes
mice to be black). A student tested the
hypothesis that a black mouse is a heterozygote.
– H0: based on Mendel’s first law of equal segregation
the observed progeny is 50% C/c and 50% c/c.
– H0: based on Mendel’s first law of equal segregation
the observed progeny is 50% C/c and 50% c/c.
Class
O
E
Class
O
E
Albino
55
60
Albino
55
60
Black
65
60
Black
65
60
120
Chi-Square (Χ2 ) Test
Chi-Square (Χ2 ) Test
• The chi-square test is useful for comparing
observed results against those predicted
by a hypothesis
• The chi-square test is useful for comparing
observed results against those predicted
by a hypothesis
• Various deviations from the expected
occur by chance even if the hypothesis is
true
Chi-Square (Χ2 ) Test
Chi-Square (Χ2 ) Test
• The chi-square test is useful for comparing
observed results against those predicted by a
hypothesis
• Various deviations from the expected occur by
chance even if the hypothesis is true
• How likely is it that the deviations from the
expected occur by chance alone?
• The chi-square test is a way of quantifying the
various deviations expected by chance if the
hypothesis is true.
•
Procedure:
1. State the simple hypothesis that gives the precise
expectation: the null hypothesis
2. Calculate the chi-square, (using observed values
not percentages): Χ2 = Σ (O-E)2 /E
3. Estimate the probability of obtaining a deviation
from the expected at least this large by chance
alone
4. Report the chi-square value, the degrees of
freedom, and the probability and the rejection level.
120
Chi-Square (Χ2 ) Test
• The chi-square test is useful for comparing
observed results against those predicted by a
hypothesis
• Various deviations from the expected occur by
chance even if the hypothesis is true
• How likely is it that the deviations from the
expected occur by chance alone?
• The chi-square test is a way of quantifying the
various deviations expected by chance if the
hypothesis is true.
Chi-Square (Χ2 ) Test
•
Procedure:
1. State the simple hypothesis that gives the precise
expectation: the null hypothesis
2. Calculate the chi-square, (using observed values
not percentages): Χ2 = Σ (O-E)2 /E
3. Estimate the probability of obtaining a deviation
from the expected at least this large by chance
alone
4. Report the chi-square value, the degrees of
freedom, and the probability and the rejection level.
3
Chi-Square (Χ2 ) Test
•
Procedure:
Chi-Square (Χ2 ) Test
•
1. State the simple hypothesis that gives the precise
expectation: the null hypothesis
2. Calculate the chi-square, (using observed values
not percentages): Χ2 = Σ (O-E)2 /E
3. Estimate the probability of obtaining a deviation
from the expected at least this large by chance
alone
4. Report the chi-square value, the degrees of
freedom, and the probability and the rejection level.
Procedure:
1. State the simple hypothesis that gives the precise
expectation: the null hypothesis
2. Calculate the chi-square, (using observed values
not percentages): Χ2 = Σ (O-E)2 /E
3. Estimate the probability of obtaining a deviation
from the expected at least this large by chance
alone
4. Report the chi-square value, the degrees of
freedom, and the probability and the rejection level.
Chi-Square (Χ2 ) Test eg. 1
• The allele c causes albinism in mice (C
causes mice to be black). A student tested
the hypothesis that a black mouse is a
heterozygote.
• Procedure:
1.State the simple hypothesis that gives the
precise expectation: the null hypothesis
1.H0: based on Mendel’s first law of equal
segregation the observed progeny is 50% C/c and
50% c/c.
–
in our example df = 2-1=1
2. Calculate the chi-square, (using observed values not
percentages)
O
E
(O-E)2
(O-E)2/E
Class
O
E
(O-E)2
(O-E)2/E
55
60
25
25/60 = 0.42
Albino
55
60
25
25/60 = 0.42
Black
65
60
25
25/60 = 0.42
Black
65
60
25
Total= Χ2 = 0.84
•
in our example Χ2 =0.84
Degrees of freedom (df) = number of independent variables minus
1 (in this case phenotypic classes -1)
Chi-Square (Χ2 ) Test
Albino
Chi-Square (Χ2 ) Test
•
1.H0: based on Mendel’s first law of equal
segregation the observed progeny is 50% C/c and
50% c/c.
Class
3. Estimate the probability of obtaining a deviation from the expected at
least this large by chance alone
Χ2 = Σ (O-E)2 /E
1.State the simple hypothesis that gives the
precise expectation: the null hypothesis
2. Calculate the chi-square, (using observed values not
percentages)
Chi-Square (Χ2 ) Test
–
• The allele c causes albinism in mice (C
causes mice to be black). A student tested
the hypothesis that a black mouse is a
heterozygote.
• Procedure:
Chi-Square (Χ2 ) Test
3. Estimate the probability of obtaining a deviation from the expected at
least this large by chance alone
•
Chi-Square (Χ2 ) Test eg. 1
Χ2 = Σ (O-E)2 /E
–
•
in our example Χ2 =0.84
Degrees of freedom (df) = number of independent variables minus
1 (in this case phenotypic classes -1)
–
in our example df = 2-1=1
25/60 = 0.42
Total= Χ2 = 0.84
Chi-Square (Χ2 ) Test
3. Estimate the probability of obtaining a deviation from
the expected at least this large by chance alone
•
•
•
Χ2 = 0.84
Degrees of freedom (df) = 2-1=1
0.5>p>0.1
4
4. Report the chi-square value, the degrees
of freedom, and the probability and the
rejection level.
4. Report the chi-square value, the degrees
of freedom, and the probability and the
rejection level.
4. Report the chi-square value, the degrees
of freedom, and the probability and the
rejection level.
• Χ2 =0.84
• Degrees of freedom (df) = 2-1 = 1
• Probability of observing a deviation from the expected
results at least this large on the basis of chance alone, in
our example 0.5>p>0.1
• Rejection level is p=0.05
• Conclusion: At the 5% rejection level we fail to reject the
null hypothesis that the observed progeny is 50% C/c
and 50% c/c, a 1:1 ratio of pigmented to albino mice. We
conclude that the black mouse used in the test cross
was a heterozygote.
• Χ2 =0.84
• Degrees of freedom (df) = 2-1 = 1
• Probability of observing a deviation from the expected
results at least this large on the basis of chance alone, in
our example 0.5>p>0.1
• Rejection level is p=0.05
• Conclusion: At the 5% rejection level we fail to reject the
null hypothesis that the observed progeny is 50% C/c
and 50% c/c, a 1:1 ratio of pigmented to albino mice. We
conclude that the black mouse used in the test cross
was a heterozygote.
• Χ2 =0.84
• Degrees of freedom (df) = 2-1 = 1
• Probability of observing a deviation from the expected
results at least this large on the basis of chance alone, in
our example 0.5>p>0.1
• Rejection level is p=0.05
• Conclusion: At the 5% rejection level we fail to reject the
null hypothesis that the observed progeny is 50% C/c
and 50% c/c, a 1:1 ratio of pigmented to albino mice. We
conclude that the black mouse used in the test cross
was a heterozygote.
The dihybrid situation
The dihybrid situation
The dihybrid situation
• In dogs dark coat colour (C) is dominant to
albino and short hair (S) is dominant over long
hair. The ratio of progeny from a backcross to
the homozygous recessive is
–
–
–
–
140 CS
135 cs
110 Cs
115 cS
• In dogs dark coat colour (C) is dominant
to albino and short hair (S) is dominant
over long hair. ….a backcross to the
homozygous recessive …
• C/c;S/s X c/c;s/s
–
–
–
–
• Is this ratio different from the expected 1:1:1:1
ratio?
Chi-Square (Χ2 ) Test
Χ2
(O-E)2
•
= Σ
/E
• Degrees of freedom (df) = number of independent
variables minus 1 (in this case phenotypic classes -1)
• Probability of observing a deviation from the expected
results at least this large on the basis of chance alone.
• In dogs dark coat colour (C) is dominant to
albino and short hair (S) is dominant over long
hair. ….a backcross to the homozygous
recessive …
• C/c;S/s X c/c;s/s
• The ratio of progeny from a backcross to the
homozygous recessive is
140 CS
135 cs
110 Cs
115 cS
• Parental vs recombinant
Chi-Square (Χ2 ) Test eg. 2
1. H0: based on Mendel’s first law of equal segregation the
observed progeny fits a 1:1:1:1 ratio of phenotypes
CS:cs:Cs:cS (coat colour C= dark and hair length S= short).
2. Calculate chi-square
Class
CS
cs
CS
cS
df= 4-1
O
135
135
110
115
500
E
125
125
125
125
(O-E)2
(O-E)2/E
225
1.8
100
0.8
225
1.8
100
0.8
Chi-square = 5.2
Chi-Square (Χ2 ) Test eg. 2
1. H0: based on Mendel’s first law of equal segregation the
observed progeny fits a 1:1:1:1 ratio of phenotypes
CS:cs:Cs:cS (coat colour C= dark and hair length S= short).
2. Calculate chi-square
Class
CS
cs
CS
cS
df= 4-1
O
135
135
110
115
500
E
125
125
125
125
(O-E)2
(O-E)2/E
225
1.8
100
0.8
225
1.8
100
0.8
Chi-square = 5.2
5
Chi-Square (Χ2 ) Test eg. 2
Chi-Square (Χ2 ) Test eg. 2
Other 1:1 ratios?
• A mouse cross:
• A/a;B/b x a/a;b/b gave:
–
–
–
–
• Χ2 =5.2 , df = 3, p>10%, rejection level =5%
• Conclusion: given the chi-square =5.2 and df=3 a
deviation from the expected 1:1:1:1 ratio at least this
large would occur by chance alone 10% of the time so
we fail to reject the null hypothesis that the observed
ratio of CS:cs:Cs:cS is 1:1:1:1.
• Χ2 =5.2 , df = 3, p>10%, rejection level =5%
• Conclusion: given the chi-square =5.2 and df=3 a
deviation from the expected 1:1:1:1 ratio at least this
large would occur by chance alone 10% of the time so
we fail to reject the null hypothesis that the observed
ratio of CS:cs:Cs:cS is 1:1:1:1.
Other 1:1 ratios?
• A mouse cross:
• A/a;B/b x a/a;b/b gave:
–
–
–
–
25% A/a;B/b
25% A/a;b/b
25% a/a;B/b
25% a/a;b/b
• What is the probability of
getting a female mouse
that is a/a;b/b?
• Another mouse cross:
• C/c;D/d X c/c;d/d gave:
–
–
–
–
38% C/c;D/d
12% C/c;d/d
38% c/c;D/d
12% c/c;d/d
• The student could not find
a female cc;d/d mouse. In
fact all the _;d/d mice
were male!
• Why?
Other 1:1 ratios?
• Another mouse cross:
• C/c;D/d X c/c;d/d gave:
–
–
–
–
38% C/c; D/d
12% C/c; d/d
38% c/c; D/d
12% c/c; d/d
• C/c;D/d X c/c;d/d gave:
–
–
–
–
38% C/c; D/d
12% C/c; d/Y
38% c/c; D/d
12% c/c; d/Y
• The student could not find
a female cc;d/d mouse. In
fact all the _;d/d mice
were male!
Other 1:1 ratios?
Other 1:1 ratios?
• For autosomal traits males and females
are affected in equal proportions
• From the mouse cross that gave no _;d/d
female progeny we would test whether the
d is X-linked
• The c allele is likely autosomal
• In the P generation the homozygous
recessive mice were female
• For autosomal traits males and females
are affected in equal proportions
• From the mouse cross that gave no _;d/d
female progeny we would test whether the
d is X-linked
• The c allele is likely autosomal
• In the P generation the homozygous
recessive mice were female
• Another mouse cross:
• C/c;D/d X c/c;d/d gave:
25% A/a;B/b
25% A/a;b/b
25% a/a;B/b
25% a/a;b/b
• What is the probability of
getting a female mouse
that is a/a;b/b?
–
–
–
–
38% C/c;D/d
12% C/c;d/d
38% c/c;D/d
12% c/c;d/d
• The student could not find
a female cc;d/d mouse. In
fact all the _;d/d mice
were male!
• Why?
Other 1:1 ratios?
• For autosomal traits males and females
are affected in equal proportions
• From the mouse cross that gave no _;d/d
female progeny we would test whether the
d is X-linked
• The c allele is likely autosomal
• In the P generation the homozygous
recessive mice were female
Other 1:1 ratios
• What is the probability of 6 children being girls?
• P (girl)= ½
– P (all 6 being girls)= (½)6 = 1/64.
• Probability that of 6 children all will be the same sex?
– P (all 6 being girls) + P (all 6 being boys) = (½ )6 + (½)6 = 1/32
The distribution of boys and girls in 240 families
(O-E)2/E
Sex O
E
(O-E)2
Girls 129
120
81
0.67
Boys 111
120
81
0.67
6
Other 1:1 ratios
• What is the probability of 6 children being girls?
• P (girl)= ½
– P (all 6 being girls)= (½)6 = 1/64.
• Probability that of 6 children all will be the same sex?
– P (all 6 being girls) + P (all 6 being boys) = (½ )6 + (½)6 = 1/32
Other 1:1 ratios
•
•
•
H0: based on Mendel’s first law of equal segregation the observed
progeny fits a 1:1 ratio of girls : boys.
Chi-square = 1.34, df= 1, p>0.1, rejection level is p=0.05
Therefore we conclude that for Χ2 =1.34, df=1 we would expect a
deviation from the 1:1 ratio at least this large would occur by chance
alone more than 1% of the time so we fail to reject the null
hypothesis that the observed ratio of progeny fits a 1:1 ratio of girls
to boys.
Other 1:1 ratios
•
•
•
H0: based on Mendel’s first law of equal segregation the observed
progeny fits a 1:1 ratio of girls : boys.
Chi-square = 1.34, df= 1, p>0.1, rejection level is p=0.05
Therefore we conclude that for Χ2 =1.34, df=1 we would expect a
deviation from the 1:1 ratio at least this large would occur by chance
alone more than 1% of the time so we fail to reject the null
hypothesis that the observed ratio of progeny fits a 1:1 ratio of girls
to boys.
The distribution of boys and girls in 240 families
Sex O
E
(O-E)2
(O-E)2/E
Girls 129
120
81
0.67
Boys 111
120
81
0.67
Other 1:1 ratios
•
•
•
Pedigree of an autosomal dominant
condition
H0: based on Mendel’s first law of equal segregation the observed
progeny fits a 1:1 ratio of girls : boys.
Chi-square = 1.34, df= 1, p>0.1, rejection level is p=0.05
Therefore we conclude that for Χ2 =1.34, df=1 we would expect a
deviation from the 1:1 ratio at least this large would occur by chance
alone more than 1% of the time so we fail to reject the null
hypothesis that the observed ratio of progeny fits a 1:1 ratio of girls
to boys.
Polydactyly a rare dominant
phenotype
• The unaffected couple in
generation IV have12
offspring. What is the
probability of all of 12
being boys?
• His brother is affected, if
before having children the
probability of having an
affected son was
P(son)*P(affected) = ½ *
½ = ¼ Is it strange that
they have 2 affected
sons?
Polydactyly a rare dominant
phenotype
• The unaffected couple in
generation IV have12
offspring. What is the
probability of all of 12
being boys?
• His brother is affected, if
before having children the
probability of having two
affected sons was
P(son)*P(affected) = ½ *
½ = ¼ Is it strange that
they have 2 affected
sons?
Pedigree: X-linked recessive
• For I_1and I-2 the
probability of grandson
inheriting red-green
colour blindness is:
Pedigree: X-linked recessive
• For I_1and I-2 the
probability of grandson
inheriting red-green
colour blindness is:
– Their daughter is a
heterozygote so
• P(grandson)*P(grandson
is afflicted)= (½ ) * (½)= ¼
7
Sex linkage
Sex linkage
• Recessive genes showing X-linked
inheritance can be detected in pedigree
charts by
• Recessive genes showing X-linked
inheritance can be detected in pedigree
charts by
– More males than females are affected
– None of the offspring of an affected male
show the phenotype
– More males than females are affected
– None of the offspring of an affected male
show the phenotype
• but his daughters are carriers and half of their sons
show the recessive phenotype
• And none of his sons show the recessive
phenotype nor will they pass it on to their offspring
Pedigree: X-linked dominant
• Hypophosphatemia
• but his daughters are carriers and half of their sons
show the recessive phenotype
• And none of his sons show the recessive
phenotype nor will they pass it on to their offspring
Pedigree: X-linked recessive
• Haemophilia a clotting disorder most
commonly attributed to the absence or
malfunction of factor VIII .
Sex linkage
• Dominant genes showing X-linked
inheritance can be detected as follows:
– Affected males pass on the gene to all of their
daughters and to none of their sons
– Affected females pass on the gene to half of
their offspring
Course Overview
Outline
Week
1
2
3
4
5
6
7
8
9
10
11
12
Topic
Course objectives and Introduction to genetics
Human Pedigrees
Patterns of Inheritance: sex-linkage
Chromosomal basis of inheritance
Changes in chromosome number
Gene Mapping
Gene to Phenotype
Modified Mendelian ratios
Model organisms and mutants
Genetics of Plant Development (Arabidopsis)
Genetics of Animal Development (Drosophila)
Behaviour Genetics/Quantitative genetics
Chapter
Ch. 1 & Ch. 2
Ch. 2
Ch. 2
Ch. 3
Ch. 15
Ch. 4 (Ch. 16)
Ch. 6
Ch. 6
Ch. 6 (Ch. 16)
Ch. 18
Ch. 18
Ch. 16 + papers
8