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
onwards with:
GSD: genotypic sex determination
Segregation of alleles (genes) determines sex
best for generating 1:1 sex ratios
Amazing variety of mechanisms
(sex determination & sexual dimorphism is
the most rapidly evolving aspect of developmental programming)
For fruit flies:
XX (±Y) females X(±Y) males
What about X-chromosome number matters?
absolute number:
1=male, 2or more = female
odd vs. even (paired?)
XX X=male?
number relative to ploidy (non-sex chromosomes)? X:A ratio
X AA male, but X A female?
…again, genetic exceptions to the rule provide the answer
For fruit flies:
X(±Y) AA
X:A = 0.5, male
XX(±Y) AA
X:A = 1, female
Progeny from Bridges' "exceptional" female:
XXX AAA X:A= 1, female (jumbo)
XX(±Y) AAA
X:A = 0.67, intersex (phenotypic mosaic)
What about:
XA
X:A=1,
(dead) female
how could he know???
GENETIC MOSAICS revealed the sex of X A cells
(nonmosaic XA animals never reach the sexually dimorphic (adult) stage)
recall that:
X-chromosome loss generates
“gynandromorphs”
XX AA zygote -->
XXAA cells / X AA cells
(XXAA)
(X AA)
Female
Male
XXAA zygote -->
XXAA cells/XA cells
(loss of an entire haploid set)
(XXAA)
(X A)
Female
Female
(genetic "markers"
allowed him to infer
the genotypes)
X AA
X:A = 0.5, male
XX AA
X:A = 1, female
XX(±Y) AAA
X:A = 0.67, intersex
XXX AAA X:A= 1, female (large)
XA
X:A=1,
(dead) female
GSD by X:A ratio (balance)
GSD by X:A ratio (balance)
Balance between what?
What is it about ploidy changes that affects the way that
X-chromosome dose determines sex?
"Obviously": female-determining genes on the X
and
male-determining genes on the autosomes
"numerator genes"
:
"denominator genes"
Obviously:
female-determining genes on the X
("numerator genes")
and
male-determining genes on the autosomes
("denominator genes")
Pure speculation
on Bridges' part
(1921)
…but what else could it be?
speculation turned into
textbook "fact"
Fig. 18.21
(your text fudges whether
hypothesis vs. fact but many other
texts present it as fact)
Fig. 18.21
Obviously:
female-determining genes on the X
("numerator genes") O.K.
and
male-determining genes on the autosomes
not O.K.
("denominator genes")
1921
finally
2007 the answer!
and not what Bridge's guessed
or what the textbooks say)
female-determining genes on X exist
(including the master, Sxl)
vs.
maternal gene products
…and
ploidy alters the
TIME of counting by Sxl
and hence the concentration
of the relevant female-determining gene products
The (round) worm (Caenorhabditis elegans):
XX self-fertilizing hermaphrodite
mostly more
hermaphrodites
XO male (heterogametic sex)
Two ways to get males:
(1) Spontaneous X-chromosome nondisjunction (rare)
to make “O” eggs in the hermaphrodite
(+ X self sperm)-> XO male
(2) Mating (outcross) of hermaphrodite to XO male:
X eggs join with X or O male sperm -> 50:50
Worm X-chromosome counting is somewhat like that in the fly,
but different in important ways:
XX self-fertilizing hermaphrodite
XO male (heterogametic sex)
XX AAA X:A= 0.67 = male (not a mosaic)
XXX AAAA X:A = 0.75 = hermaphrodite
again:
GSD by X:A ratio
…and there do appear to be "male-determining
genes on the autosomes against which
"hermaphrodite-determining genes on X" are measured
GSD by X:A ratio
is a remarkably rare sex-determination mechanism
among species
HUMANS:
XX female
XY male
XXY Kleinfelter Syndrome
sterile male (1:1000 men)
XO Turner Syndrome
sterile female (1:2000-5000)
GSD by Active Y
dominant masculinizer
HOUSE FLIES:
m/m female
M/m male
GSD by dominant masculinizing allele M
(no heteromorphic sex chromosomes)
(…actually, only one of three different GSD systems
that operate in different races of the same species!)
Birds, moths and butterflies:
ZZ male
ZW female
female is
the heterogametic sex
(compare: XY males)
GSD by
feminizing W or Z:A ?
>>20% of all animals use a very different GSD system:
Eggs fertilized --> Queens (females) or workers (sterile)
Diploid (± royal jelly)
Eggs not fertilized --> Drones (males)
Haploid
GSD by “haplodiploid” system
But is the relevant variable ploidy?
Let’s encourage inbreeding (incest) among the honeybees:
increased homozygosity
suddenly: DIPLOID MALES!
(fertilization)
a1/a2 Queen X a1 Drone --> a1/a1 & a2/a1 & a1 & a2
diploid
haploid
diploid
(perhaps
workers
drones
her brother) drones
& queens
a1/a2 heterozygotes: females (queens and workers)
a1 or a1/a1 hemizygotes and homozygotes: males
a2 or a2/a2
GSD by a multiple allele system
--- highly “polymorphic” sex gene (many alleles)
GSD via a Maternal Effect system:
(for a blowfly)
Genotype of mother
determines (or at least influences) the
Phenotype of the progeny
Male-producing mothers
f/f
X males f / f
f / f sons
Female-producing mothers F / f
1:1
sex ratio
X males f / f
F / f & f / f daughters
A potential problem with many GSD systems (including our own):
fruit flies:
honeybees:
XX
AA
XY
AA
female
male
PROBLEM
AA
female
A
male
NO PROBLEM
fact: gene output is generally proportional to gene dose in metazoans
fact: 20% of all fly genes are on X
few of these are on fly Y
hence: males are monosomic
for 1/5 of their genome
(even 1% is rarely tolerated)
potentially genetically unbalanced
males are monosomic
haploid for entire genome
not genetically
unbalanced
XX
XY
How eliminate the anticipated X-linked gene expression
difference between the sexes?
= X-chromosome dosage compensation
(1) increase X-linked gene expression 2x in males
fruit flies (“the fly”)
(2) decrease X-linked gene expression in females by 1/2
2a: reduce each X by 50%
2b: inactivate one X
the worm
us mammals
Recall that Muller observed X-linked gene dose effect within a sex,
but not between the sexes
o
+
wa/wa > (darker, more “wildtype”) wa/Df(w)
(or wa/w-)
>
o
wa/Y < (lighter, less “wildtype”) wa/Y; Dp(wa)/+
YET:
wa/wa = (same color as) wa/Y
“leaky” (hypomorphic) mutant alleles twice as leaky in males vs. females
It must follow that:
wa/Y; Dp(wa)/+ > (darker, more “wildtype”) wa/wa
Infer: wildtype alleles twice as active in males vs. females to achieve balance
wildtype (normal) X-linked alleles
work twice as hard in males
as they do in females
XX
XY
= X-chromosome dosage compensation
Are the male genes working twice as hard, or instead are the
female genes working half as hard? (is the glass half full or half empty)
Can actually answer the question.
But first:
Are the alleles on both the female’s X chromosomes even working?
YES
Muller knew from gynandromorphs: white gene functioning is
“cell autonomous”:
Gynandromorph:
w+/w-
w-
(XXAA)
(X AA)
Female
Male
a cell’s phenotype reflects
its genotype with respect
to the particular gene
w+/w- non-mosaic eye is solid red,
not mosaic red and white
…alleles on both X’s must be active
Are male X-linked genes turned UP
or are
female X-linked genes turned DOWN?