Transcript III. Linkage

A. Sex Determination
1. Environmental Sex Determination
2. Chromosomal Sex Determination
a. Protenor sex determination
b. Lygaeus sex determination
c. Balanced sex determination
d. Human sex determination: SRY gene
A. Sex Determination
1. Environmental Sex Determination
2. Chromosomal Sex Determination
a. Protenor sex determination
b. Lygaeus sex determination
c. Balanced sex determination
d. Human sex determination: SRY gene
The presence of the Y, regardless of the number of X’s, determines maleness
Klinefelter’s Male
Turner’s Female
A. Sex Determination
1. Environmental Sex Determination
2. Chromosomal Sex Determination
a. Protenor sex determination
b. Lygaeus sex determination
c. Balanced sex determination
d. Human sex determination: SRY gene
SRY gene produces
the protein called the
testis determining
factor, which stimulates
the undifferentiated
gonadal tissue to
become a testis. It is
probably a transcription
factor that binds to
other genes,
stimulating their
expression.
A. Sex Determination
1. Environmental Sex Determination
2. Chromosomal Sex Determination
a. Protenor sex determination
b. Lygaeus sex determination
c. Balanced sex determination
d. Human sex determination: SRY gene
Human mutations demonstate it is
the presence/absence of this gene,
not the whole Y, that stimilates male
development
X
X
X*
X*X male
X*X male
Y-
XY- female
XY- female
Insertion of homolog in mice also
changes their sex.
A. Sex Determination
B. Sex Linkage: Genes of interest are one of the sex chromosomes (X or Y)
1. For Comparison –heredity for sex (as a trait) and an autosomal dominant trait.
MALE: AAXY
AX
FEMALE:
aa XX
MALE: aa XY
AY
aX
AaXX AaXY
aX
AaXX AaXY
aX
FEMALE:
AA XX
aY
AX
Aa XX Aa XY
AX
Aa XX Aa XY
All offspring, regardless of sex, express the A trait in both reciprocal crosses
A. Sex Determination
B. Sex Linkage
1. For Comparison –heredity for sex (as a trait) and an autosomal dominant trait.
2. Sex Linkage example: red-green coloblindness in humans
MALE
FEMALE
Xg
Y
XG
XGXg
XGY
XG
XGXg
XGY
100% G, for all offspring
MALE
FEMALE
XG
Y
Xg
XGXg
XgY
Xg
XGXg
XgY
50% G daughters, 50% g sons
Now, the sex of the parent that expresses the G trait matters; the transmission of
this gene correlates with the sex of the offspring, because this trait and ‘sex’ are
influenced by the same chromosome.
A. Sex Determination
B. Sex Linkage
1. For Comparison –heredity for sex (as a trait) and an autosomal dominant trait.
2. Sex Linkage example: red-green coloblindness in humans
Queen Victoria of England
Her
daughter
Alice
X-linked recessive traits are expressed in
males more than females, because females
get a second X that may carry the dominant
allele.
A. Sex Determination
B. Sex Linkage
C. Dosage Compensation
- Females have two ‘doses’ of X-linked genes, while males have
one ‘dose’. Since protein concentration is often important in protein
function, how is this imbalance corrected?
A. Sex Determination
B. Sex Linkage
C. Dosage Compensation
- Females have two ‘doses’ of X-linked genes, while males have
one ‘dose’. Since protein concentration is often important in protein
function, how is this imbalance corrected? In human females, one X in each
cell condenses.
Barr Body
A. Sex Determination
B. Sex Linkage
C. Dosage Compensation
Actually, in all humans and
mammals, all but one X
condenses, regardless of
sex or number of X’s.
A. Sex Determination
B. Sex Linkage
C. Dosage Compensation
Which X condenses is random. So, in heterozygous female cats (XOXo), when
the X with the gene for orange color condenses, the ‘non-orange’ allele allows
genes for other colors at other loci to be expressed (black, brown, ‘blue’). The X
that is inactivated is determined randomly, early in development. This
inactivation is imprinted on that X, such that descendants of those cell
inactivate that X. White is due to another gene that influences melanocyte
migration to skin surface, and also affects the size of patches from tortoiseshell
to calico.
A. Sex Determination
B. Sex Linkage
C. Dosage Compensation
This happens in humans, too – so that females are really a ‘mosaic’, with
some cells in a tissue expressing one X (and it’s X-linked traits) and some
cells in that tissue expressing the other X. Females heterozygous for redgreen colorblindness have patches of retinal cells that can’t distinguish red
from green.
Anhidrotic ectodermal
dysplasia
Females heterozygous
for this X-linked
condition have
patches of skin that
lack sweat glands
A. Sex Determination
B. Sex Linkage
C. Dosage Compensation
HP1 = “heterochromatic protein 1”
How?
- each X has a gene – the Xic (X-inactivation center).
- this is ‘on’ in inactivated X’s… it produces an RNA (Xist) that binds with the
chromosomes, making it inaccessible to transcription enzymes.
- this RNA is NOT translated – it is functional as an RNA molecule.
- of course, this just pushes the question one step ‘upstream’ – what
determines why Xic is only active in one X chromosome?
A. Sex Determination
B. Sex Linkage
C. Dosage Compensation
How?
- each X has a gene – the Xic (X-inactivation center).
- this is ‘on’ in inactivated X’s… it produces an RNA that binds with the X
chromosomes, making it inaccessible to transcription enzymes.
- this RNA is NOT translated – it is functional as an RNA molecule.
- of course, this just pushes the question one step ‘upstream’ – what
determines why Xic is only active in one X chromosome?
When?
- It seems to be an ‘imprinted’ phenomenon, so that daughter cells have
the same X inactivated. However, this seems to happen at different points in
development for different tissues.
I. Allelic, Genic, and Environmental Interactions
II. Sex Determination and Sex Linkage
III. Linkage
- Overview:
‘Linkage’ is a pattern of correlated inheritance between traits governed by
genes on the same chromosome. Because the genes are part of the same physical
entity, they are inherited together rather than independently.
INDEPENDENT ASSORTMENT (IA)
a
A
B
A
B
A
b
LINKED
AB
ab
b
a
B
a
b
AB
ab
III. Linkage
- Overview:
‘Linkage’ is a pattern of correlated inheritance between traits governed by
genes on the same chromosome. Because the genes are part of the same physical
entity, they are inherited together rather than independently. Only ‘crossing-over’ can
cause them to be inherited in new combinations.
Cross-over
products
III. Linkage
A. ‘Complete’ Linkage
- if genes are immediate neighbors, they are almost never separated by
crossing over and are ‘always’ inherited together. The pattern mimics that of a single
gene.
AABB
aabb
AB
ab
X
AB
ab
III. Linkage
A. ‘Complete’ Linkage
- if genes are immediate neighbors, they are almost never separated by
crossing over and are ‘always’ inherited together. The pattern mimics that of a single
gene.
AABB
aabb
AB
ab
X
ab
AB
Gametes
AB
ab
ab
F1
AB
Double Heterozygote in F1;
no difference in phenotypic
ratios compared to IA
III. Linkage
A. ‘Complete’ Linkage
- if genes are immediate neighbors, they are almost never separated by
crossing over and are ‘always’ inherited together. The pattern mimics that of a single
gene.
ab
ab
F1 x F1
X
AB
AB
Gametes
AB
AB
ab
ab
III. Linkage
A. ‘Complete’ Linkage
- if genes are immediate neighbors, they are almost never separated by
crossing over and are ‘always’ inherited together. The pattern mimics that of a single
gene.
ab
ab
3:1 ratio A:a
F1 x F1
X
AB
AB
3:1 ratio B:b
3:1 ratio AB:ab
Gametes
AB
AB
ab
ab
AABB
AaBb
AaBb
aabb
III. Linkage
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
III. Linkage
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
- Crossing over in a region is rare
III. Linkage
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
- Crossing over in a region is rare
- Crossing over events increase as the distance between genes increases
A
B
C
a
b
c
LESS LIKELY
IN HERE
MORE LIKELY IN HERE
III. Linkage
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
- Crossing over in a region is rare
- Crossing over events increase as the distance between genes increases
- So, the frequency of crossing over (‘CO’) gametes can be used as an index of
distance between genes! (Thus, genes can be ‘mapped’ through crosses…)
A
B
C
a
b
c
FEWER ‘CO’
GAMETES:
Ab, aB
MORE ‘CO’
GAMETES:
bC, Bc
III. Linkage
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
- Crossing over in a region is rare
- Crossing over events increase as the distance between genes increases
- So, the frequency of crossing over (‘CO’) gametes can be used as an index of
distance between genes! (Thus, genes can be ‘mapped’ through crosses…)
- How can we measure the frequency of recombinant (‘cross-over’) gametes?
Is there a type of cross where we can ‘see’ the frequency of different types of gametes
produced by the heterozygote as they are expressed as the phenotypes of the
offspring?
TEST CROSS !!!
III. Linkage
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
a
A
b
B
a
a
b
b
TEST CROSS !!!
III. Linkage
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
- So, since crossingover is rare (in a particular
region), most of the time it
WON’T occur and the
homologous chromosomes will
be passed to gametes with
these genes in their original
combination…these gametes are
the ‘parental types’ and they
should be the most common
types of gametes produced.
a
A
b
a
B
a
b
A
B
a
b
b
TEST CROSS !!!
III. Linkage
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
- Sometimes, crossing
over WILL occur between these
loci – creating new
combinations of genes…
This produces the ‘recombinant
types’
a
A
b
a
B
a
b
A
B
a
B
A
b
a
b
b
TEST CROSS !!!
III. Linkage
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
As the other parent only
contributed recessive alleles,
the phenotype of the offspring
is determined by the gamete
received from the
heterozygote…
a
A
b
a
a
B
a
b
A
B
b
b
gamete
genotype
phenotype
ab
aabb
ab
ab
AaBb
AB
a
B
ab
aaBb
aB
A
b
ab
Aabb
Ab
TEST CROSS !!!
III. Linkage
a
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
A
b
a
a
B
b
b
ALTERNATIVES
FREQUENCIES EQUAL TO
PRODUCT OF INDEPENDENT
PROBABILITIES
‘IA’
LINKAGE
a
b
A
B
gamete
genotype
phenotype
ab
aabb
ab
ab
AaBb
AB
LOTS of
PARENTALS
a
B
ab
aaBb
aB
A
b
ab
Aabb
Ab
FEWER
CO’S
III. Linkage
AaBb x aabb
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
Example:
- How do we discriminate
between these two alternatives?
- Conduct a Chi-Square Test of
Independence
Offspring
Number
AB
43
Ab
12
aB
8
ab
37
III. Linkage
AaBb x aabb
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
Example:
- How do we discriminate
between these two alternatives?
- Conduct a Chi-Square Test of
Independence
- Compare the observed
results with what you would expect
if the genes assorted independently
Offspring
Number
AB
43
Ab
12
aB
8
ab
37
III. Linkage
AaBb x aabb
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
Example:
- How do we discriminate
between these two alternatives?
- Conduct a Chi-Square Test of
Independence
Offspring
Number
AB
43
Ab
12
aB
8
ab
37
- Compare the observed
results with what you would expect
if the genes assorted independently
The frequency of ‘AB’ should = f(A) x f(B) x N = 55/100 x 51/100 x 100 = 28
III. Linkage
AaBb x aabb
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
Example:
- How do we discriminate
between these two alternatives?
- Conduct a Chi-Square Test of
Independence
Offspring
Number
AB
43
Ab
12
aB
8
ab
37
- Compare the observed
results with what you would expect
if the genes assorted independently
The frequency of ‘AB’ should = f(A) x f(B) x N = 55/100 x 51/100 x 100 = 28
The frequency of ‘Ab’ should = f(A) x f(B) x N = 55/100 x 49/100 x 100 = 27
The frequency of ‘aB’ should = f(a) x f(B) x N = 45/100 x 51/100 x 100 = 23
The frequency of ‘ab’ should = f(a) x f(b) x N = 45/100 x 49/100 x 100 = 22
III. Linkage
AaBb x aabb
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
Example:
- How do we discriminate
between these two alternatives?
- Conduct a Chi-Square Test of
Independence
Offspring
Number
AB
43
Ab
12
aB
8
ab
37
This is fairly easy to do by creating
a contingency table:
B
b
A
43
12
a
8
37
Col.
Total
Row
Total
III. Linkage
AaBb x aabb
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
Example:
- How do we discriminate
between these two alternatives?
- Conduct a Chi-Square Test of
Independence
Offspring
Number
AB
43
Ab
12
aB
8
ab
37
This is fairly easy to do by creating
a contingency table:
B
b
Row
Total
A
43
12
55
a
8
37
45
Col.
Total
51
49
100
Add across and down…
This gives the totals for each trait
independently.
III. Linkage
AaBb x aabb
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
Example:
- How do we discriminate
between these two alternatives?
- Conduct a Chi-Square Test of
Independence
Offspring
Number
AB
43
Ab
12
aB
8
ab
37
This is fairly easy to do by creating
a contingency table:
Then, to calculate an expected
value based on independent
assortment (for ‘AB’, for example),
you multiple ‘Row Total’ x ‘Column
Total’ and divide by ‘Grand Total’.
55 x 51 / 100 = 28
B
Exp.
b
Row
Total
A
43
28
12
55
a
8
37
45
Col.
Total
51
49
100
III. Linkage
AaBb x aabb
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
Offspring
Number
Example:
AB
43
Repeat to calculate the other expected
values… (This is just an easy way to set
it up and do the calculations, but you
should appreciate it is the same as the
product rule:
Ab
12
aB
8
ab
37
F(A) x f(B) x N
B
Exp.
b
Exp.
Row
Total
A
43
28
12
27
55
a
8
23
37
22
45
Col.
Total
51
49
100
III. Linkage
AaBb x aabb
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
Compare our observed results with what
we would expect if the genes assort
independently.
Offspring
Number
AB
43
Ab
12
aB
8
ab
37
B
Exp.
b
Exp.
Row
Total
A
43
28
12
27
55
a
8
23
37
22
45
Col.
Total
51
49
100
III. Linkage
AaBb x aabb
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
Compare our observed results with what
we would expect if the genes assort
independently.
. If our results are close to the
expectations, then they support the
hypothesis of independence.
Offspring
Number
AB
43
Ab
12
aB
8
ab
37
B
Exp.
b
Exp.
Row
Total
A
43
28
12
27
55
a
8
23
37
22
45
Col.
Total
51
49
100
III. Linkage
AaBb x aabb
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
Offspring
Number
AB
43
Ab
12
aB
8
ab
37
Compare our observed results with what
we would expect if the genes assort
independently.
. If our results are close to the
expectations, then they support the
hypothesis of independence. If they are
far apart from the expected results, then
they refute that hypothesis and support
the alternative: linkage.
B
Exp.
b
Exp.
Row
Total
A
43
28
12
27
55
a
8
23
37
22
45
Col.
Total
51
49
100
III. Linkage
AaBb x aabb
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
Typically, we reject the hypothesis of
independent assortment (and accept the
hypothesis of linkage) if our observed results
are so different from expectations that
independently assorting genes would only
produce results as unusual as ours less than
5% of the time…
Offspring
Number
AB
43
Ab
12
aB
8
ab
37
B
Exp.
b
Exp.
Row
Total
A
43
28
12
27
55
a
8
23
37
22
45
Col.
Total
51
49
100
B
Exp.
b
Exp.
Row
Total
A
43
28
12
27
55
a
8
23
37
22
45
Col.
Total
51
III. Linkage
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
We determine this probability with a
Chi-Square Test of Independence.
49
Obs
Exp
(o-e)
(o-e)2/e
AB
43
28
15
8.04
Ab
12
27
-15
8.33
aB
8
23
-15
9.78
Ab
37
22
15
10.23
X2 =
36.38
Phenotype
100
III. Linkage
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
Our X2 = 36.38
First, we determine the ‘degrees of
freedom’ = (r-1)(c-1) = 1
B
b
Row
Total
A
43
12
55
a
8
37
45
Col.
Total
51
49
100
III. Linkage
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
Our X2 = 36.38
First, we determine the ‘degrees of
freedom’ = (r-1)(c-1) = 1
Now, we read across the first row in
the table, corresponding to df = 1.
The column headings are the
probability that a number in that
column would occur at a given df.
In our case, it is the probability that
our hypothesis of independent
assortment (expected values are
based on that hypothesis) is true.
III. Linkage
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
Our X2 = 36.38
Note that larger values have a lower
probability of occurring by chance…
This should make sense, and the
value increases as the difference
between observed and expected
values increases.
III. Linkage
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
Our X2 = 36.38
So, for instance, a value of 2.71 will
occur by chance 10% of the time.
III. Linkage
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
Our X2 = 36.38
So, for instance, a value of 2.71 will
occur by chance 10% of the time.
But a value of 6.63 will only occur
1% of the time... (if the hypothesis is
true and this deviation between
observed and expected values is
only due to chance).
III. Linkage
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
Our X2 = 36.38
So, for instance, a value of 2.71 will
occur by chance 10% of the time.
For us, we are interested in the 5%
level. The table value is 3.84.
Our calculated value is much greater
than this; so the chance that
independently assorting genes
would yield our results is WAY LESS
THAN 5%. Our results are REALLY
UNUSUAL for independently
assorting genes.
III. Linkage
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
Our X2 = 36.38
Our results are REALLY UNUSUAL for
independently assorting genes.
So, either our results are wrong, or
the hypothesis of independent
assortment is wrong. If you did a
good experiment, then you should
have confidence in your results;
reject the hypothesis of IA and
conclude the alternative – the genes
are LINKED.
III. Linkage
AaBb x aabb
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
OK… so we conclude the genes are linked…
NOW WHAT?
Offspring
Number
AB
43
Ab
12
aB
8
ab
37