a. sex determination environmental sex determination chromosomal sex determination

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



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.



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?


- 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


Actually, in all humans and mammals, all but one X condenses, regardless of sex or number of X’s.


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.


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


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 =

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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



X

AABB aabbAB

AB

ab

ab

AB abGametes

AB

abF1

Double Heterozygote in F1; no difference in phenotypic ratios compared to IA

III. Linkage



X

AB

ab

Gametes

AB

ab ab

AB

F1 x F1

AB ab

III. Linkage



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


- Crossing over in a region is rare

III. 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


- 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


- 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


- 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


- 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


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


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



- 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




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




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



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




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




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


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


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


A 43 28 12 27 55

a 8 23 37 22 45

Col. Total

51 49 100

III. 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


A 43 28 12 27 55

a 8 23 37 22 45

Col. Total

51 49 100

III. 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


A 43 28 12 27 55

a 8 23 37 22 45

Col. Total

51 49 100

III. 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


A 43 28 12 27 55

a 8 23 37 22 45

Col. Total

51 49 100

III. 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


A 43 28 12 27 55

a 8 23 37 22 45

Col. Total

51 49 100

III. 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


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


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


Our X2 = 36.38

So, for instance, a value of 2.71 will occur by chance 10% of the time.

III. Linkage


Our X2 = 36.38


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


Our X2 = 36.38


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


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


OK… so we conclude the genes are linked… NOW WHAT?

Offspring Number

AB 43

Ab 12

aB 8

ab 37

AaBb x aabb

a. sex determination environmental sex determination chromosomal sex determination

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