Download Modified Mendelian ratios

Course Overview
Modified Mendelian ratios
http://www.erin.utoronto.ca/~w3bio/bio207/index.htm
March 1st 2007
Test and Tutorials
• You need to do the assigned questions at the
end of chapter 6
• As stated on the website you need to present a
doctor’s note to the instructor within 48h of
missing a test (or a quiz)
• If you have a legitimate reason for missing a
quiz email IN ADVANCE
• NB. Academic conflict is NOT a legitimate
reason for missing a quiz
• Late penalty is 20% per day for quizzes that are
to be handed in to drop boxes.
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. 6
Ch. 6
Ch. 6 (Ch. 16)
Ch. 18
Ch. 18
Ch. 16 + papers
Interaction between alleles
• Petal colour in Fouro’clock plants:
– P:
– F1:
Red
pink
X white
X pink
• How many loci are
involved?
1
Interaction between alleles
• Petal colour in Fouro’clock plants:
– P:
Red
pink
– F1:
gave
– ¼ red petals
– ½ pink petals
– ¼ white petals
X white
X pink
• How many loci are
involved?
• The monohybrid ratio 3:1
is altered by incomplete
dominance to 1:2:1
Incomplete dominance
• Petal colour in Fouro’clock plants: let + red
and c = different
–
–
–
–
So F1 X F1: + c X + c
¼++
red petals
½+c
pink petals
¼cc
white petals
• Implies a pigment
synthesis pathway: e.g.
2x amount of pigment
gives red, 1x amount
gives pink and none give
white petals.
Dihybrid cross
• Three of four F2 are
round R_ ie 3:1 ratio
for seed shape
• Three of four F2 are
yellow Y_ ie 3:1 ratio
for seed colour
• F2 progeny in a
9:3:3:1 ratio implies
• Two genes
independently
assorting
2
•
•
•
•
Epistasis
Epistasis
Epistasis: two or more genes interact to influence a trait
E.g. Labrador Retriever dogs
Black coat colour (B), brown coat colour (b)
Another gene (e) is epistatic to these alleles giving a
golden coat colour
– P: B/B; e/e X b/b ; E/E
black Lab X brown Lab
– F1:
B/b ; E/e
Black lab
– F1 X F1:
B/b ; E/e X B/b; E/e gave…..
Genotype
B/_;E/_
b/b;E/_
B/_;e/e
b/b;e/e
Epistasis
Genotype
Phenotype
Ratio
B/_;E/_
Black
9
b/b;E/_
Brown
3
B/_;e/e
Golden
4
b/b;e/e
Golden
Phenotype
Black
Brown
Golden
Golden
Ratio
9
3
4
Coat colour in mice
• The two alleles of the
pigment gene B or b
• And the separate
gene E that allows
colour deposition in
the coat, e/e prevents
colour deposition
•
•
•
•
•
The A gene: A= agouti, a= nonagouti
The B gene: B= black pigment, b= brown
The C gene: C= pigmented, c= albino
The D gene: D=nondilution, d= dilution
The S gene: S= unspotted, s= pigmented
spots on a white background
Text p.219
Text p.204
3
Coat colour in mice
•
•
•
•
•
The A gene: A= agouti, a= nonagouti
The B gene: B= black pigment, b= brown
The C gene: C= pigmented, c= albino
The D gene: D=nondilution, d= dilution
The S gene: S= unspotted, s= pigmented
spots on a white background
Agouti
• Agouti pattern: a
yellow band around
each of the hairs, a
wildtype pattern
• P: A / A X a / a
• F1: 3:1 ratio agouti to
non agouti:
– 1 A / A (agouti)
– 2 A / a (agouti)
– 1 a / a (nonagouti)
Text p.219
Coat colour in mice
•
•
•
•
•
The A gene: A= agouti, a= nonagouti
The B gene: B= black pigment, b= brown
The C gene: C= pigmented, c= albino
The D gene: D=nondilution, d= dilution
The S gene: S= unspotted, s= pigmented
spots on a white background
Black pigment
• Black pigmented hair
is the wildtype hair
colour, and B is
dominant to b which
confers brown hair
colour
• P: B / B X b / b
• F1: 3:1 ratio black hair
colour to brown:
– 1 B / B (black)
– 2 B / b (black)
– 1 b / b (brown)
Text p.219
4
Cinnamons
• Brown-agouti mice
are called cinnamons
• They are A/_; b/b
• An agouti (black)
pure-breeding mouse
is crossed with a
pure-breeding
cinnamon
Chocolate
• Chocolate mice have a
solid rich dark brown
colour and are a/a; b/b.
A/a ; B/b black agouti
X
a/a ; B/b chocolate
¼ A/a; B/b wildtype
¼ A/a ; b/b cinnamon
¼ a/a ; B/b black nonagouti
¼ a/a ; b/b chocolate
– A/A; b/b X a/a; B/B
• The F1 are?
–
–
–
–
9 A/ _ ; B/ _
3 a/a ; B/_
3 A/_ ; b/b
1 a/a ; b/b
Coat colour in mice
•
•
•
•
•
The A gene: A= agouti, a= nonagouti
The B gene: B= black pigment, b= brown
The C gene: C= pigmented, c= albino
The D gene: D=nondilution, d= dilution
The S gene: S= unspotted, s= pigmented
spots on a white background
• A wildtype mouse (A/a;
B/b) is testcrossed.
• Diagram the cross:
Albino
• Albino mice are
homozygous for the
recessive member of
the C/c allelic pair.
• The c/c constitution is
epistatic to the other
coat colour genes
• An albino mouse is
crossed with a truebreeding wildtype
(black agouti) mouse.
– c/c ; ?/? ; ?/? X C/C,
A/A; B/B
• The F1 were
intercrossed to get
the following F2:
– Ie. C/c ; A/?; B/? X
C/c; A/?; B/?
Text p.219
5
ABC of coat colour
• P: c/c ; ?/? ; ?/? X
C/C, A/A; B/B
• F1: C/c ; A/?; B/? X
C/c; A/?; B/?
• F2:
– 87 wildtype
– 32 cinnamon
– 39 albino
• F1 x F1 have
cinnamon offspring
therefore F1 parents
must be B/b
• The F1 x F1 do not
have chocolate
offspring therefore the
F1 parents must have
been A/A
ABC of coat colour
• P: c/c ; ?/? ; ?/? X
C/C, A/A; B/B
• F1: C/c ; A/A; B/b X
C/c; A/A; B/b
• F2:
– 87 wildtype
– 32 cinnamon
– 39 albino
ABC of coat colour
• P: c/c ; A/A ; b/b X C/C, A/A; B/B
• F1: C/c ; A/A; B/b X C/c; A/A; B/b
• F2:
– 87
– 32
– 39
C/_ ; A/A; B/_
C/_ ; A/A; b/b
c/c ; A/A ; ?/?
wildtype
cinnamon
albino
• From the F1 x F1 we
would expect ¼ of the
offspring to be albino
(c/c)
• 87+32+39=
• 39/158
• Approx 1/4
Coat colour in mice
•
•
•
•
•
The A gene: A= agouti, a= nonagouti
The B gene: B= black pigment, b= brown
The C gene: C= pigmented, c= albino
The D gene: D=nondilution, d= dilution
The S gene: S= unspotted, s= pigmented
spots on a white background
Text p.219
6
D gene
• The D gene controls the
intensity of pigment
specified by other coat
colour genes
– D/_; A/_; B/_ full
expression of coat colour
– Dilute agouti dd; A/_; ?
– Dilute black: dd; ?; B/_
– Dilute brown dd; ?; b/b
– Dilute cinnamon:
• dd; a/_ ; b/b
• D is a modifier gene,
detectable by two grades
of mutant phenotypes
Coat colour in mice
•
•
•
•
•
The A gene: A= agouti, a= nonagouti
The B gene: B= black pigment, b= brown
The C gene: C= pigmented, c= albino
The D gene: D=nondilution, d= dilution
The S gene: S= unspotted, s= pigmented
spots on a white background
The S gene
• The S gene controls
the presence or
absence of spots on
the mouse
• No spots: S/_
• Spotted or piebald:
s/s
• Write the genotype of
a dilute spotted
cinnamon mouse.
• The mouse must be:
A/_; b/b; C/_; d/d; s/s
• Because to be dilute=
dd, to be spotted= ss,
to be cinnamon=
A/_;b/b and C/_
Text p.219
7
Testing interactions
The S gene?
• A history student ask you what might be the genetic
basis of a story in an ancient text:
– Two people Jacob and Laban divided their flock of animals by
coat colour Jacob took the unspotted goats and Laban the
spotted goats but only Jacob’s flock increased in number giving
both spotted and unspotted goats.
• From the coat colour genetics in mice: S_ gives no
spots and ss gives spots. So perhaps in goats Ss x Ss
goats that would produce both spotted and unspotted
progeny while the spotted goats (ss) would only give
spotted progeny
• P: c/c ; A/A ; b/b X C/C, A/A; B/B
• F1: C/c ; A/A; B/b X C/c; A/A; B/b
• F2:
– 87
– 32
– 39
Phenotype
Wildtype
Cinnamon
Albino
total
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.
C/_ ; A/A; B/_
C/_ ; A/A; b/b
c/c ; A/A ; ?/?
Observed
87
32
39
158
wildtype
cinnamon
Albino
albino x wildtype
wildtype F1 x F1
= C/_ ; B/_
= C/_ ; b/b
= c/c ; b/b
Expected
9/16*158
3/16*158
4/16 *158
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.
8
Testing interactions
Chi-Square (Χ2 ) Test
F2:
– 87
– 32
– 39
C/_ ; A/A; B/_
C/_ ; A/A; b/b
c/c ; A/A ; ?/?
wildtype = C/_ ; B/_
cinnamon = C/_ ; b/b
Albino = c/c ; b/b
• H0 : The observed coat colour ratio is not different from
the 9:3:4 ratio expected for recessive epistasis.
Phenotype Observed Expected (O-E)2 (O-E)2
/E
Wildtype
87
88.9
3.61 0.041
Cinnamon 32
29.6
5.76 0.194
Albino
39
39.5
0.25 0.063
total 158
0.298
4. Report the chi-square value, the degrees
of freedom, and the probability and the
rejection level.
• Χ2 =0.30
• Degrees of freedom (df) = 3-1 = 2
• Probability of observing a deviation from the expected
results at least this large on the basis of chance alone, in
our example 0.9>p>0.5
• Rejection level is p=0.05
• Conclusion: At the 5% rejection level we fail to reject the
null hypothesis that the observed progeny is 9 C/_;B_ :
3 C/_; b/b :4 c/c;b/b, the recessive epistasis ratio of
wildtype to cinnamon to albino mice. We conclude that
observed data support the hypothesis of recessive
epistasis.
3. Estimate the probability of obtaining a deviation from
the expected at least this large by chance alone
•
•
•
Χ2 = 0.30
Degrees of freedom (df) = 3-1=2
0.9>p>0.5
Testing interactions
F2:
– 32
– 87
– 39
cinnamon = C&/C&
wildtype = C&/C$
Albino = C$/C$
1
2
1
• H0 : The observed coat colour ratio is not different from
the 1:2:1 ratio expected for incomplete dominance.
Phenotype O
Cinnamon
Wildtype
Albino
total
Expected (O-E)2
32 ⅓ * 158
87 ⅔ * 158
39 ⅓ * 158
158
(32-52.6)2
(87-105.3)2
(39-52.6)2
(O-E)2
/E
8.07
3.18
3.52
14.77
9
4. Report the chi-square value, the degrees
of freedom, and the probability and the
rejection level.
Chi-Square (Χ2 ) Test
3. Estimate the probability of obtaining a deviation from
the expected at least this large by chance alone
•
•
•
Χ2 = 14.77
Degrees of freedom (df) = 3-1=2
0.005>p i.e. p< 0.05
• Χ2 =14.77
• Degrees of freedom (df) = 3-1 = 2
• Probability of observing a deviation from the expected
results at least this large on the basis of chance alone, in
our example p<< 0.05
• Rejection level is p=0.05
• Conclusion: At the 5% rejection level we reject the null
hypothesis that the observed progeny is 1 C&/C& : 2
C&/C$ : 1 C$/C$, the incomplete dominance ratio of
cinnamon to wildtype to albino mice. We conclude that
observed data does NOT support the hypothesis of
incomplete dominance.
Testing interactions
• Pure strains of pineapples were crossed and the
following types of leaves were noted:
– S= spiny, ST= spiny tip P= non-spiny; piping
Cross
1
2
3
Parental
ST x S
P x ST
PxS
Phenotypes
F1 F2
ST 99 ST: 34 S
P 120 P: 39 ST
P 95 P: 25 ST:
8S
Testing interactions
Cross
1
2
3
•
•
•
•
Parental
ST x S
P x ST
PxS
Phenotypes
F1 F2
ST 99 ST: 34 S
P 120 P: 39 ST
P 95 P: 25 ST: 8 S
Cross 1 indicates a single gene, note 3:1 ratio
Cross 2 indicates a single gene, note 3:1 ratio
Cross 3 indicates the two genes are interacting
What is the modified ratio?
10
Testing interactions
Cross
1
2
3
Parental
ST x S
P x ST
PxS
Testing interactions
Phenotypes
F1 F2
ST 99 ST: 34 S
P
120 P: 39 ST
P
95 P: 25 ST: 8 S
• Using the assigned gene
symbols P for piping and
T for tip
• Cross 3: ?
• Hypothesis: independent
assortment
–
–
–
–
• Assign gene symbols:
– P for piping
– T for tip
• So for cross 1: ST x S could be written
– P: T/T x t/t
– F1: T/t
– F2: 99 T/_ : 34 t/t
–
–
–
–
9: P/_; T/_
3 P/_ ; t/t
3 p/p ; T/_
1 p/p ; t/t
– 95 P/_ ; T/_
– 25 p/p; T/_
– 8 p/p; t/t
=P
=ST
=S
• So it seems that only
whenever P is dominant
we can see” the piping
phenotype
Text p.213
Dominant epistasis
Testing interactions
• Using the assigned gene
symbols P for piping and
T for tip
• Cross 3: ?
• Hypothesis: independent
assortment
9: P/_; T/_
3 P/_ ; t/t
3 p/p ; T/_
1 p/p ; t/t
• The dihybrid ratio is
modified
• The dihybrid ratio is
modified
– 95 P/_ ; T/_
– 25 p/p; T/_
– 8 p/p; t/t
=P
=ST
=S
• So it seems that only
whenever P is dominant
we can “see” the piping
phenotype
• Dominant epistasis:
–
–
–
–
9: P/_; T/_
3 P/_ ; t/t
3 p/p ; T/_
1 p/p ; t/t
Phenotype O
P
ST
S
12
3
1
Expected
(O-E)2
95 12/16 * 128 (95-96)2
25 3/16 * 128 (25-24)2
8
1/16 * 128 (8-8)2
total 128
(O-E)2
/E
0.01
0.042
0
0.052
11
Chi-Square (Χ2 ) Test
3. Estimate the probability of obtaining a deviation from
the expected at least this large by chance alone
•
•
•
Χ2 = 0.052
Degrees of freedom (df) = 3-1=2
0.975>p< 0.9
4. Report the chi-square value, the degrees
of freedom, and the probability and the
rejection level.
• Χ2 =0.052
• Degrees of freedom (df) = 3-1 = 2
• Probability of observing a deviation from the expected
results at least this large on the basis of chance alone, in
our example 0.975>p< 0.9 i.e. p >> 0.05
• Rejection level is p=0.05
• Conclusion:
– At the 5% rejection level we fail to reject the null hypothesis that
the observed progeny is 12 P: 3 ST: 1S, the dominant epistasis
ratio of piping to spiny tip to spiny pineapple leaf types. We
conclude that observed data supports the hypothesis of
dominant epistasis.
Testing interactions
• Hypothesis: independent
assortment
–
–
–
–
9: P/_; T/_
3 P/_ ; t/t
3 p/p ; T/_
1 p/p ; t/t
• The dihybrid ratio is
modified
– 95 P/_ ; T/_
– 25 p/p; T/_
– 8 p/p; t/t
=12 P
=3 ST
=1 S
• So it seems that only
whenever P is dominant
we can “see” the piping
phenotype
• The data support a 12:3:1
ratio typical of dominant
epistasis
– Piping P/_ (T or t irrelevant)
– Spiny tip p/p ; T/_
– Spiny p/p ; t/t
Two genes, duplicate action
• A graduate student crosses wheat plants
that are homozygous for spring growth
with plants that are homozygous for winter
growth
• The F1 all have the spring growth
phenotype
• How would you write the cross?
12
Two genes, duplicate action
• The graduate student then crossed two F1 plants and
got the following progeny ratio:
– 15 : 1
– Spring growth : Winter growth
• How would she explain what happened?
– Let S and H represent the dominant alleles
Cross:
¼ SH
¼ Sh
¼ sH
¼ sh
¼ SH
SS; HH
SS; Hh
sS; HH
sS; Hh
S/s; H/h x S/s; H/h
¼ Sh
¼ sH
SS; Hh sS; HH
SS; hh
sS; Hh
sS; Hh
Ss; Hh
sS; hh
ss; Hh
¼ sh
Ss ; Hh
Ss; hh
Ss; Hh
ss; hh
Modified ratios from the cross
AaBb x AaBb
A_ ; B _
A_ ; bb
aa ; B_
aa ; bb
Independent
assortment
9
3
3
1
Complementary
gene action
9
7
Duplicate gene
action
Recessive epistasis
Dominant epistasis
15
9
1
3
12
4
3
1
Duplicate gene action
• Spring growth is the result of duplicate gene
action between two independent genes S and H
• Spring growth: S_; H_ , S_; hh , ss; H_
• Winter growth: ss;hh
• In duplicate gene action the dominant phenotype
is evident when at least one dominant allele is
present at each locus, the recessive phenotype
is only visible when both genes are homozygous
recessive
Suppression ratio
• Suppressors cancel the expression of the
mutant allele of another gene, resulting in
normal wildtype phenotype
• Ratio typically of a suppressor is 13:3
• Consider the recessive allele that confers purple
eye colour (pd) in Drosophila
• Suppressed by a recessive allele su.
– pd /pd ; su+ / su+ X pd+/ pd+ ; su / su
• Gives all red eyed (wildtype) F1:
– pd+ /pd; su+ / su
13
recessive suppressor
• F1 X F1 red eyed
(wildtype) :
• F1 X F1 red eyed
(wildtype) :
– pd+ /pd; su+ / su X pd+
/pd; su+ / su
•
•
•
•
9
3
1
3
pd+ /_; su+ / _
pd+ /_; su / su
pd /pd; su / su
pd /pd; su +/ _
Suppressors
13
3
– i.e. 9+3+1= 13 red eyed
to 1 purple eyed
– pd+ /pd; su+ / su X
pd+ /pd; su+ / su
•
•
•
•
9
3
1
3
pd+ /_; su+ / _
pd /pd; su +/ _
pd /pd; su / su
pd+ /_; su / su
The purple eye colour can only be expressed when the suppressor is inactive
(su+/_) AND the recessive mutation purple (pd) homozygous
Text p.206
Introduction to genetic analysis
Griffiths, A., Wessler, S.R., Lewontin,R.C., Gelbart, W.M.,Suzuki, D.T.
and Miller, J.H.
Eighth Edition, W.H. Freeman and Company NY
• Part I Transmission genetic analysis
–
–
–
–
Chapter 1: all questions p. 24-26
Chapter 2: all the questions p. 62-72
Chapter 3: questions #1-12,18,19, 22, 25-27, 29, 30, 32, 40-42.
Chapter 4: sections 4.1- 4.4 and 4.6, questions # 1-4, 6-13,1522,24-43.
• Part II The relationship of DNA and phenotype
– Chapter 6: questions# 1,3-5,7,9,11,14,18,19,21,23,24,
26,29,32,33,35,39,41,44,45,47,49,51,53,55,63-66.
• Part IV The nature of heritable change
– Chapter 15: sections 15.1 and 15.3; questions #1-3,1113,19,21,22, 32, 38, 52.
Test and Tutorials
• You need to do the assigned questions at the
end of chapter 6
• As stated on the website you need to present a
doctor’s note to the instructor within 48h of
missing a test (or a quiz)
• If you have a legitimate reason for missing a
quiz email IN ADVANCE
• NB. Academic conflict is NOT a legitimate
reason for missing a quiz
• Late penalty is 20% per day for quizzes that are
to be handed in to drop boxes.
.
14