Transcript ABC - HCC Learning Web

BIOL 2416
Chapter 12: Sex Determination,
Nondisjunction, X-linkage
Chapter 13: Mendelian Extensions
Chapter 14: Genetic Mapping (2
and 3-point test crosses)
Sex Determination
Complex developmental process
under genetic and hormonal control
Sex Determination Systems
• Chromosomal
– XY system (humans, fruit flies)
– WZ system (birds, some insects)
– XO system (grasshoppers, beetles)
– Haploid-diploid system (bees,
ants,wasps)
Sex Determination Systems, cont’d.
• Genic
– Small allelic differences (yeast)
• Nonexistent
– Hermaphrodites (earthworms, some
snails)
– Parthenogenesis (lizards, fish, insects)
• Environmental
– Temperature (geckos, alligators)
– Infection (some arthropods)
XY System in Humans
• 2 sets of 22 different autosomes:
homologous chromosomes / homomorphic
pairs
• Plus 2 sex chromosomes:
heteromorphic pair with homologous ends
to allow pairing during meiosis
XX in females, XY in males (sex
determined by presence / absence of Y with
SRY gene)
XY System:
Male gametes:
X or Y
Female gametes :
X
XY System in Drosophila
• Sex determined by X:A ratio (absence of 2nd X
in males),
where
X = # of X chromosomes
A = # of autosomal (haploid) sets
sets
• X:A = 1.00 or more = female
= 0.50 or less = male
• So XO fruit fly = male!
Changes in Chromosome
Number
• Euploidy
– whole sets of chromosomes
– polyploids- common in plants
• Aneuploidy
– by addition or deletion of less than a whole set
– Caused by nondisjunction = failure of chromosomes to
separate during mitosis or meiosis > causes faulty
gametes
Nondisjunction of Sex
Chromosomes in Humans
• In mother
– At meiosis I:
XX and O
– At meiosis II:
XX and O
• In father
– At meiosis I:
XY and O
– At meiosis II:
XX and O, YY
and O
Fig. 11.5 Nondisjunction in meiosis involving the X chromosome
Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings.
nondisjunction gametes
XY
XX
YY
X XXY XXX XYY
O
XX
O
XO
normal gametes
X
XXX XO
Y
XXY YO
Human X and Y Chromosome
Abnormalities
Genotype
Syndrome
Gender
Barr
bodies
45,X
Turner
female
0
47,XXX
Triple-X
female
2
47,XXY
Klinefelter
male
1
47,XYY
XYY
male
0
Dosage Compensation in
Humans
• Barr bodies
– Inactivated (scrunched-up) X chromosome (highly
condensed and methylated)
– Because human cells cannot handle more than one
active X
(se)X-linked Inheritance
• Discovered by Thomas Hunt Morgan in 1910
• A gene on X chromosome; use Xsuperscript notation
(see pedigree handout)
• Males are hemizygous
• Pseudodominance = single copy of recessive
gene determines phenotype in males
• Criss-cross inheritance pattern; not reciprocal
Mendelian Extensions may
alter expected ratios
•
•
•
•
•
•
•
•
•
•
Multiple alleles
Codominance
Incomplete dominance
Polygenic (discuss later)
Gene Interactions
– Between non-allelic genes on different loci
Lethal alleles
Environmental effects
Maternal Effect
Extranuclear Inheritance
Linkage
• Completely dominant alleles mask a recessive
allele.
– For example, wrinkled seeds in homozygous
recessive pea plants are due to the
accumulation of monosaccharides and excess
water in seeds because of the lack of a key
enzyme. The seeds wrinkle when they dry.
– Both homozygous dominants and
heterozygotes produce enough enzyme to
convert all the monosaccharides into starch
so both form smooth seeds when they dry.
• Most genes have more than two alleles in
a population.
• The ABO blood groups in humans are
determined by three alleles, IA, IB, and i.
– Both the IA and IB alleles are dominant
to the i allele
– The IA and IB alleles are codominant to
each other.
• Because each individual carries two
alleles, there are six possible genotypes
and four possible blood types.
• A clear example of incomplete dominance is
seen in flower color of snapdragons.
– A cross between a
white-flowered plant
and a red-flowered
plant will produce all
pink F1 offspring.
– Self-pollination of the
F1 offspring produces
characteristic 1:2:1
phenotypic ratio
Gene Interactions
• Resulting in new phenotypes:
– Comb shape in chickens (9:3:3:1)
– Fruit shape in summer squash (9:6:1)
• No new phenotypes (masking effect):
– Recessive epistasis; rodent coat
color, (9:3:4)
– Duplicate recessive epistasis; sweet
pea flower color, (9:7)
Gene Interaction in chickens
So as soon as you have one big R AND
one big P combination, they
interact to produce a walnut comb…
that’s because both R and P genes
control comb shape.
Gene Interaction in Summer Squash
Here the A and B genes
Both control fruit shape.
One big A OR one big B
Gives sphere-shaped.
But the combination one
Big A AND one big B
interacts to give diskshaped.
Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings.
Recessive epistasis
As soon as you have
nothing but little c’s,
they will mask what
is happening at the A
gene… you’ll be
albino regardless.
Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings.
Duplicate recessive epistasis
Here little cc will
mask whatever
goes on at the P
locus (white
flowers regardless),
AND little pp will
mask what is
happening at
the C locus (white
flowers regardless).
C and P genes
control successive
reactions in the
purple pigmentproducing pathway.
Either homozygous
recessive (cc or pp)
will block the
pathway.
Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings.
Lethal Genes
•
•
•
•
•
Mutations in essential genes
Can alter Mendelian ratios
Fatal to the individual (sometimes before birth)
In all inheritance patterns
E.g.
– AY in mice
– Tay-Sachs
– Achondroplasia
– Huntington’s
– (untreated) hemophilia
Genes Can Vary in Penetrance
and Expressivity
• Penetrance = the percentage of individuals
with a given genotype that express it at the
phenotypic level. (“100% complete” when
all AA’s=1 phenotype, all aa’s = another,
and all Aa’s are the same)
• Expressivity = the level of phenotypic
expression for a particular genotype in an
individual. (constant or variable range)
Environmental Effects on Identical genotypes
• Sex-limited traits
– Not seen in one sex (testicular cancer)
• Sex-influenced traits
– Different phenotypes due to different hormonal environment (baldness)
• Temperature
– Siamese cats
• Chemicals (phenocopy)
– Mimic heritable mutations; non-heritable phenotype with same genotype (due
to viruses, drugs, nutrients)
– PKU only seen with intake of phenylalanine
• Epigenetics
– Reversible, heritable changes in phenotype without change in
genotype/DNA.
– E.g. imprinting (methylation/silencing) re-established each generation in
animals (but heritable in plants?)
– Established as the cell differentiates
– See http://www.pbs.org/wgbh/nova/genes/mice.html
Environmental
effects:
Temperature
Nature vs. Nurture
• Genotype sets range of phenotypic potential
(“norm of the reaction”)
Maternal Effect:
Babies’ phenotype
determined by their
mother’s genotype
(nuclear genes).
Extranuclear (mitochondrial) inheritance:
Figure 3-22
Gene Linkage
• Linked genes are 2 genes stuck on the same
chromosome
• Inherited as a package deal (unless “unlinked” by a
crossover event)
• Gene linkage reduces the # of possible gametes
(CANNOT FOIL to find gametes)
• The closer two genes are linked (located) on the same
chromosome, the more likely they will turn up together
in the same offspring; can use this fact to get an idea of
relative gene location! The first step in finding the gene
you want!
• NOT the same as (a single) X-linked gene!
– an X-linked gene is one gene that happens to be located on
the X chromosome and not on the Y counterpart.
Chi-Square Test
Pheno
classes
o
e
d=
o-e
d2
d2 / e
gray
normal
black
vestigial
gray
vestigial
black
normal
Total
2 =
A single crossover event creates recombinant
chromosomes/gametes:
Crossing over only happens some of the time:
In a 2-point testcross, the X-over frequency =
Recombination frequency = # map units (cM) between
the 2 linked genes
TIP:
• In any testcross, the most numerous babies are going to be
PARENTAL: they come from – relatively common –
“package deal” gametes
• The least numerous babies are RECOMBINANT: they
come from –relatively rare - crossover gametes
• In 3-point test crosses, RECOMBINANT babies from
double-crossover gametes are going to be rarer than
RECOMBINANT babies coming from single crossover
gamete
• so IDENTIFY babies by the NUMBERS!
THREE-POINT TEST CROSSES - WHY?
•To determine if three genes are linked in any
way
•To determine the linear order of three linked
genes
•To estimate the map distance between any of
the linked genes (more accurate if genes are
close together and if large numbers of progeny
are scored)
WHAT?
Let’s keep it simple and just discuss a test cross involving three autosomal
recessive traits. One parent would be heterozygous for all three, the other
parent would be homozygous recessive for all three. Let’s say that all genes
exhibit ABC/abc CIS coupling (with all dominant alleles on one homologue
before the slash, and all recessive alleles on the other shown after the slash, as
opposed to trans), so that in the heterozygous parent, all three dominant
alleles are on one chromosome, and all recessive alleles are all on the other
chromosome.
HOW?
For the above scenario:
P genotypes:
P phenotypes:
ABC / abc
ABC
x
abc / abc
abc
From this cross, you might get the following progeny
phenotypes (baby flies):
Baby phenotype
Looks like?
Observed # of babies
ABC (“big-big-big”)
1st parent
401
abc
2nd parent
389
ABc
Recombinant type
4
abC
Recombinant type
6
AbC
Recombinant type
75
aBc
Recombinant type
65
aBC
Recombinant type
35
Abc
Recombinant type
25
Total = 1000 babies
This pattern is diagnostic for all three genes linked!
Baby phenotype
Type
ABC (“big-big-big”)
Parental type:
no crossover
Parental type:
no crossover
Recombinant type:
double crossover
Recombinant type:
double crossover
Recombinant type:
single crossover
Recombinant type:
single crossover
Recombinant type:
single crossover
Recombinant type:
single crossover
abc
ABc
abC
AbC
aBc
aBC
Abc
Observed
numbers
Relative frequency
401
many
389
many
4
very few
6
very few
75
in between
65
in between
35
in between
25
in between
By comparing the ABC and abc parentals (most
numerous) to each of the sets of recombinant babies
(less numerous), figure out which of the three genes
has been changed:
Baby phenotype
Baby type
What did not change?
What changed?
ABC (“big-big-big”)
Parental type
ABC
-
abc
Parental type
abc
-
ABc
Recombinant type
AB from ABC parent
abC
Recombinant type
ab from abc parent
AbC
Recombinant type
AC from ABC parent
aBc
Recombinant type
ac from abc parent
aBC
Recombinant type
BC from ABC parent
Abc
Recombinant type
bc from abc parent
c
C
b
B
a
A
Double crossovers
• Probability is product of two single crossovers
• So they are very rare
• And produce the least numbers of babies; use
this info to identify double-crossover progeny
• See which gene has changed in the doublecrossover babies
• This is the gene that must be in the middle!
• From there, look at the other babies and figure
out where each single crossover took place:
Baby
phenotype
Type
Observed
number
Due to what kind of
Xover?
ABC
abc
ABc
parental
Recombinant
401
389
4
abC
Recombinant
6
Double
AbC
Recombinant
75
aBc
Recombinant
aBC
Abc
What changed?
Where is (are the crossover(s)?
-
-
-
-
-
-
Around C (middle)
Single
c
C
b
65
Single
B
Between
C (middle) and B (end)
Recombinant
35
Single
a
Between
C (middle) and A (end)
Recombinant
25
Single
A
Between
C (middle) and A (end)
parental
Double
Around C (middle)
Between
C (middle) and B (end)
So the map distance between A and C is:
(# ALL double Xovers + # of all single crossovers INVOLVING A and C)(100) / (total) =
(4 + 6 + 35 + 25) (100) / 1000 = 7000 / 1000 = 7 map units or 7 cM between A and C
The map distance between C and B is:
(# ALL double Xovers + # of all single crossovers INVOLVING B and C)(100) / (total) =
(4 + 6 + 76 + 65 ) (100) / (1000) = 15000 / 1000 = 15 map units or 15 cM between C and B
Food for Thought
• What would you expect to see if there were
no linkage?
• What would you expect to see if two of the
three genes were linked, but one was
located on a different chromosome? (hint:
would it still be a 3-point test cross
problem?)
3-Point Test Crosses
• For autosomal genes = triple heterozygote x
triple homozygous recessive
• For TWO X-linked genes = triple
heterozygous female x hemizygous
recessive male
• Can order 3 genes on the same chromosome
• Maps distances between them
3-Point Test Crosses, cont’d
• See Chapter 13 iGenetics Animation on 3point test crosses
• Do iActivity Crossovers and Tomato
Chromosomes