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A second practice problem set (with answers) is on the course website. The review session for the second midterm is on Thursday evening, April 10, from 7-9pm in ROOM 141 GIANNINI HALL. determines . G enotype of mother. progeny phenotype mutant.

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A second practice problem set with answers is on the course website

A second practice problem set

(with answers)

is on the course website.

The review session for the second midterm

is on Thursday evening, April 10, from 7-9pm

in ROOM 141 GIANNINI HALL


A second practice problem set with answers is on the course website

determines

Genotype of mother

progeny

phenotype

mutant

determines (or at least influences) the

m/m

vs.

m/+

progeny: m/m, m/+, or +/+

mother:

Phenotype of the progeny

progeny

phenotype

wildtype

progeny: m/m, m/+, or +/+

Strict maternal effect (only mother can supply the needed + gene product)

(for a blowfly)

mentioned GSD via a Maternal Effect system


A second practice problem set with answers is on the course website

influences

Genotype of mother

progeny

phenotype

mutant

determines (or at least influences) the

m/m

vs.

m/+

m/m

vs.

m/+

progeny: m/m, m/+, or +/+

mother:

mother:

Phenotype of the progeny

progeny

phenotype

wildtype

progeny

phenotype

wildtype

progeny: m/m, m/+, or +/+

Strict maternal effect (only mother can supply the needed + gene product)

progeny phenotypemutant

m/m

or

m/+, or +/+

progeny:

progeny phenotypewildtype

progeny: m/m, m/+, or +/+

Rescuable maternal effect (either mother or progeny can supply + gene product)

(for a blowfly)

mentioned GSD via a Maternal Effect system


A second practice problem set with answers is on the course website

X chromosome

2800 genes

X-chromosome dosage compensation in Drosophila:

Are male X-linked genes turned UP

or are

female X-linked genes turned DOWN?

Giant Polytene Salivary-Gland Chromosomes

measure rate of RNA precursor incorporation

into “nascent” transcripts (during interphase)

…average transcription rates (per unit DNA)


A second practice problem set with answers is on the course website

average transcription rates (per unit DNA):

female X = femaleautosomes = maleautosomes< male X

X chromosome

2800 genes

transcription rate for Male X-linked genes are turned UP

relative to autosomal or female X-linked genes


A second practice problem set with answers is on the course website

Female: XX

Male: XY

noX hyperactivation

X hyperactivation

Phenotypic consequences

of loss by mutation:

Normal gene function:

What phenotype would one expect for mutations

that disrupted genes that encode the machinery

for X-chromosome dosage compensation?

needed (only) for hyper

male (X:A=0.5)-specific lethal

needed (only) to preventhyper

female (X:A=1)-specific lethal

That is how the relevant genes are recognized

(MSLs encode protein complex on male X)


A second practice problem set with answers is on the course website

Fig. 18.23 (p675)

Sxlnullis female-specific lethal

SxlConstitutiveis male-specific lethal

also dosage

compensation

Among the genetic pathways that control development,

those controlling sexual development are perhaps the best understood.

Sxl controls sex determination; dsx controls sexual dimorphism


A second practice problem set with answers is on the course website

XX AA

X AA

hermaphrodites

(females that make sperm)

males

What about worms? (C. elegans)

(1) both hermaphrodite X’s active

(like the fly)

(2) male X twice as active as each hermaph. X

(like the fly)

(3) at the transcriptional level

(like the fly)

(4) hermaph.-specific lethal genes encode protein complex

on hermaphrodite X’s that turns transcription down

fly male-specific lethal genes encode protein complex on male X that turns it up


A second practice problem set with answers is on the course website

XY

AA

XX

AA

First clue:

“sex chromatin”

females

males

Barr Body rule:

One

Barr Body

No

Barr Body

#BB = #X-1

XO

AA

Turner females

XXY

AA

Kleinfelter males

XXXX

AA

(mentally retarded) females

How do we mammals dosage compensate?

NoBarr Body

OneBarr Body

ThreeBarr Bodies


A second practice problem set with answers is on the course website

G6PD+/G6PD-:

heterozygote

Another clue:

Odd behavior of an X-linked mammalian gene:

Individual blood cells are

phenotypically either

G6PD+or G6PD-

only one or the other X-linked allele

seems to be active

in any given blood cell

not what we saw with the eye of the w+/w-fly


A second practice problem set with answers is on the course website

XY

AA

Xx

AA

females

males

mosaic c+ expression whenc+ on X

(translocation of autosomal coat color gene c to X)

(1) Barr Body = inactivated X chromosome

Xxxx

AA

x

females

Barr Body

Geneticist Mary Lyon:

#BarrBodies= #X-1

mosaic expression of G6PD+(on X)

Observations:

Hypothesis:

(2) Dosage compensation by inactivation

of all but one X chromosome


A second practice problem set with answers is on the course website

X-chromosome inactivation:

XmaternalXpaternal

(1) initiated very early in development

(at ~500 cell stage in humans)

(2) generally random in embryo proper

(paternal = maternal)

(often paternal in extra-embryonic)

(3) once initiated, stably inherited

an epigenetic phenominon

(4) reactivation of inactivated X

occurs in germ cells during oogenesis


A second practice problem set with answers is on the course website

hemizygous males (EDA-/Y) & homozygous females (EDA-/EDA-)

no sweat glands (incl. breasts)

missing & abnormal teeth/hair

EDA+/EDA-

PHENOTYPIC

MOSAICS

Identical twins:

Patchiness signifies

little skin cell mixing during development

Striking human example of X inactivation in action:

Anhidrotic Ectodermal Dysplasia (EDA):

cell autonomous trait


A second practice problem set with answers is on the course website

Need to know for gene a:

how is a phenotype related to a+ gene expression?

(2) perhaps cell autonomous, but deleterious early

--- abnormal cells selected against

(they may be outcompeted by normal cells)

For X-linked genes:

If a+/a-mammals are functional mosaics of a+ & a- cells

…are all non-functional X-linked alleles (a-) semi-dominant?

(dominance depends on how phenotype is operationally defined)

NO

(1) perhaps not cell autonomous

(and 50% a+ function is sufficient for normal phenotype)

consider hemophilias

Most animals compensate well for cells lost during development


A second practice problem set with answers is on the course website

X1matX1pat

50:50mat vs. pat active

the genetics of the X controlling element

X2matX2pat50:50 mat vs. pat active

65:35mat(1) vs. pat(2) active

X1matX2pat

X2matX1pat

35:65mat(2) vs. pat(1) active

Mapping the source of the inactivation bias definedXce


A second practice problem set with answers is on the course website

c+

almost all

genes in cis

shut off

X

Study of variations in “X inactivation strength” (in whole mice)

defined the

Xce (controlling element)

(determines chromosomal inactivation competativeness)

Study of T(X,A)s in mouse tissue-culture cells defined the

Xic (inactivation center)

(“source” of inactivation in cis)

Xic = Xce


A second practice problem set with answers is on the course website

(s)

(number of)

Xist

X-inactivation-specific-transcript:

X

X

X

X

X

X

Xic = Xce

The source of a very odd RNA :

…a non-coding RNA from the inactive X that coats

the inactive X chromosome in cis

one of the first examples of a regulatory RNA


A second practice problem set with answers is on the course website

X

X

Xist RNA

X

Xist RNA

X

Consequences of deleting Xic (source of Xist):

always the active X


A second practice problem set with answers is on the course website

Xist transgene (inducible) on autosome

will coat autsome with Xist and silence it

…but only during an early window of time


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