a. sex determination environmental sex determination chromosomal sex determination
Post on 21-Jan-2016
303 Views
Preview:
DESCRIPTION
TRANSCRIPT
A. Sex Determination1.Environmental Sex Determination2.Chromosomal Sex Determination
a. Protenor sex determinationb. Lygaeus sex determinationc. Balanced sex determinationd. Human sex determination: SRY gene
A. Sex Determination1.Environmental Sex Determination2.Chromosomal Sex Determination
a. Protenor sex determinationb. Lygaeus sex determinationc. Balanced sex determinationd. 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 Determination1.Environmental Sex Determination2.Chromosomal Sex Determination
a. Protenor sex determinationb. Lygaeus sex determinationc. Balanced sex determinationd. 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 Determination1.Environmental Sex Determination2.Chromosomal Sex Determination
a. Protenor sex determinationb. Lygaeus sex determinationc. Balanced sex determinationd. Human sex determination: SRY gene
X X
X* X*X male X*X male
Y- XY- female XY- female
Human mutations demonstate it is the presence/absence of this gene, not the whole Y, that stimilates male development
Insertion of homolog in mice also changes their sex.
MALE: AAXY
FEMALE: aa XX
A X A Y
a X AaXX AaXY
a X AaXX AaXY
MALE: aa XY
FEMALE:AA XX
a X a Y
A X Aa XX Aa XY
A X Aa XX Aa XY
A. Sex DeterminationB. 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.
All offspring, regardless of sex, express the A trait in both reciprocal crosses
MALE
FEMALE
Xg Y
XG XGXg XGY
XG XGXg XGY
MALE
FEMALE
XG Y
Xg XGXg XgY
Xg XGXg XgY
A. Sex DeterminationB. 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
100% G, for all offspring 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 DeterminationB. 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 DeterminationB. Sex LinkageC. 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 DeterminationB. Sex LinkageC. 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 DeterminationB. Sex LinkageC. Dosage Compensation
Actually, in all humans and mammals, all but one X condenses, regardless of sex or number of X’s.
A. Sex DeterminationB. Sex LinkageC. 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 DeterminationB. Sex LinkageC. 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 red-green 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 DeterminationB. Sex LinkageC. Dosage Compensation
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?
HP
1 =
“h
eter
och
rom
atic
pro
tein
1”
A. Sex DeterminationB. Sex LinkageC. 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 InteractionsII. Sex Determination and Sex LinkageIII. 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.
AB
AB
ab
ab
LINKEDA a
B b
AA
B Bb b
aa
INDEPENDENT ASSORTMENT (IA)
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.
X
AABB aabbAB
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.
X
AABB aabbAB
AB
ab
ab
AB abGametes
AB
abF1
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.
X
AB
ab
Gametes
AB
ab ab
AB
F1 x F1
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.
X
AB
ab
Gametes
AB
ab ab
AB
F1 x F1
AB ab
AABB AaBb
AaBb aabb
3:1 ratio A:a
3:1 ratio B:b
3:1 ratio AB:ab
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
MORE LIKELY IN HERELESS LIKELY IN HERE
a b c
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…)
FEWER ‘CO’ GAMETES: Ab, aB
A B C
a b c
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?
III. Linkage
A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage B
a b
A b
a b
a
TEST CROSS !!!
III. Linkage
A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage
- So, since crossing-over 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.
B
a b
b
a b
a
A B
a b
A
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’
B
a b
b
a b
a
A B
a b
A
a B
A 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…
B
a b
b
a b
a
A B
a b
A
a B
A b
gamete genotype phenotype
ab aabb ab
ab AaBb AB
ab aaBb aB
ab Aabb Ab
TEST CROSS !!!
III. Linkage
A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage B
a b
b
a b
a
A B
a b
A
a B
A b
gamete genotype phenotype
ab aabb ab
ab AaBb AB
ab aaBb aB
ab Aabb Ab
TEST CROSS !!!
ALTERNATIVES
‘IA’ LINKAGE
LOTS of PARENTALS
FEWER CO’S
FR
EQ
UE
NC
IES
EQ
UA
L T
O
PR
OD
UC
T O
F I
ND
EP
EN
DE
NT
P
RO
BA
BIL
ITIE
S
III. Linkage
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
AaBb x aabb
III. Linkage
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
AaBb x aabb
III. Linkage
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
AaBb x aabb
The frequency of ‘AB’ should = f(A) x f(B) x N = 55/100 x 51/100 x 100 = 28
III. Linkage
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
AaBb x aabb
The frequency of ‘AB’ should = f(A) x f(B) x N = 55/100 x 51/100 x 100 = 28The frequency of ‘Ab’ should = f(A) x f(B) x N = 55/100 x 49/100 x 100 = 27The frequency of ‘aB’ should = f(a) x f(B) x N = 45/100 x 51/100 x 100 = 23The frequency of ‘ab’ should = f(a) x f(b) x N = 45/100 x 49/100 x 100 = 22
III. Linkage
A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage
Example:-How do we discriminate between these two alternatives?-Conduct a Chi-Square Test of Independence
This is fairly easy to do by creating a contingency table:
Offspring Number
AB 43
Ab 12
aB 8
ab 37
AaBb x aabb
B b Row Total
A 43 12
a 8 37
Col. Total
III. Linkage
A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage
Example:-How do we discriminate between these two alternatives?-Conduct a Chi-Square Test of Independence
This is fairly easy to do by creating a contingency table:
Add across and down…
This gives the totals for each trait independently.
Offspring Number
AB 43
Ab 12
aB 8
ab 37
AaBb x aabb
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
Example:-How do we discriminate between these two alternatives?-Conduct a Chi-Square Test of Independence
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
Offspring Number
AB 43
Ab 12
aB 8
ab 37
AaBb x aabb
B Exp. b Row Total
A 43 28 12 55
a 8 37 45
Col. Total
51 49 100
III. Linkage
A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage
Example:
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:
F(A) x f(B) x N
Offspring Number
AB 43
Ab 12
aB 8
ab 37
AaBb x aabb
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
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
AaBb x aabb
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
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
AaBb x aabb
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
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. If they are far apart from the expected results, then they refute that hypothesis and support the alternative: linkage.
Offspring Number
AB 43
Ab 12
aB 8
ab 37
AaBb x aabb
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
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
AaBb x aabb
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
A. ‘Complete’ Linkage B. ‘Incomplete’ Linkage
We determine this probability with a Chi-Square Test of Independence.
Phenotype 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
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
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
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
AaBb x aabb
top related