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Today: some things Mendel did not tell us…
plus Mapping and Epigenetics
–Exam #3 W 7/30 in class (bonus #2 due)–
Single genes controlling a single trait are unusual.
Inheritance of most genes/traits is much more
complex…
Dom.
Rec.
Rec.
Dom.
Genotype
Phenotype
Genes code for
proteins (or RNA).
These gene
products give rise
to traits…
It is rarely this
simple.
Fig 4.4
Sickle-cell anemia is caused by a point mutation
Fig
4.7
Sickle and normal red blood cells
Fig
4.7
Sickle-Cell Anemia:
A dominant or recessive allele?
Mom = HS
Dad
H or S
H HH
Mom or
S HS
HS
SS
S=sickle-cell
H=normal
Fig
4.7
Dad = HS
possible offspring
75% Normal
25% Sickle-cell
Coincidence of malaria
and sickle-cell anemia
Fig 24.14
Sickle-Cell Anemia:
A dominant or recessive allele?
Mom = HS
Dad
H or S
H HH
Mom or
S HS
HS
SS
S=sickle-cell
H=normal
Fig
4.7
Dad = HS
possible offspring
Oxygen transport:
75% Normal
25% Sickle-cell
Malaria resistance:
75% resistant
25% susceptible
The relationship between genes
and traits is often complex
Complexities include:
• Complex relationships between alleles
Fig 3.18
Sex determination is
normally inherited by
whole chromosomes or
by number of
chromosomes.
X/Y chromosomes in humans
The X
chromosome
has many
genes; the Y
chromosome
only has genes
for maleness.
Human sex chromosomes
(includes Mic2 gene)
Fig 4.14
Sex-linked traits are genes located on the X
chromosome
Color Blind Test
Sex-linked traits: Genes on the X chromosome
A= normal;
a= colorblind
normal
similar to Fig 4.13
colorblind
No one affected,
female carriers
Sex-linked traits: Genes on the X chromosome
A= normal;
a= colorblind
normal
normal
50% of males
affected,
0 % females
affected
similar to Fig 4.13
Sex-linked traits: Genes on the X chromosome
A= normal;
a= colorblind
normal
similar to Fig 4.13
colorblind
50% males
affected,
50% females
affected
Sex-linked traits: Genes on the X chromosome
A= normal ; a= colorblind
No one affected,
female carriers
similar to Fig 4.13
50% of males
affected, 0 %
female affected
50% males
affected, 50%
females affected
Fig 3.18
males and females may
have different numbers
of chromosomes
Tbl 7.1
dosage
compensation
Fig 7.4
The epithelial cells
derived from this
embryonic cell will
produce a patch of
white fur
While those from
this will produce a
patch of black fur
At an early stage of
embryonic development
Mammalian X-inactivation involves the
interaction of 2 overlapping genes.
Promotes compaction
Prevents compaction
Barr body compaction is heritable within an
individual
The Barr body is
replicated and both
copies remain
compacted
• A few genes on the inactivated X
chromosome are expressed in the somatic
cells of adult female mammals
– Pseudoautosomal genes
(Dosage compensation in this case is
unnecessary because these genes are
located both on the X and Y)
– Up to a 25% of X genes in humans may escape
full inactivation
• The mechanism is not understood
Lamarck was right? Sort of…
Epigenetics:
http://www.pbs.org/wgbh/nova/sciencenow/3411/02.html
Image from: http://www.sparknotes.com/biology/evolution/lamarck/section2.rhtml
Genomic Imprinting
• Genomic imprinting is a phenomenon in which
expression of a gene depends on whether it is
inherited from the male or the female parent
• Imprinted genes follow a non-Mendelian pattern
of inheritance
– Depending on how the genes are “marked”, the
offspring expresses either the maternallyinherited or the paternally-inherited allele
**Not both
Genomic Imprinting:
Methylation of genes during
gamete production.
A hypothetical example of imprinting
a
B*
A=curly hair
A*
b
a=straight hair
B=beady eyes
b=normal
*=methylation
A* in males
B* in females
a
B*
A*
b
A hypothetical example of imprinting
a
B*
A=curly hair
A*
b
a=straight hair
B=beady eyes
b=normal
*=methylation
A* in males
B* in females
A*a
bB*
a
B*
A*a
bB*
A*
b
A hypothetical example of imprinting
a
B*
A=curly hair
A*
b
a=straight hair
B=beady eyes
A*a
bB*
a
B*
A*a
bB*
b=normal
*=methylation
A* in males
B* in females
A*a
bB
Aa
bB*
A*
b
A hypothetical example of imprinting
a
B*
A=curly hair
A*
b
a=straight hair
B=beady eyes
A*a
bB*
similar to Fig 7.10
a
B* A*
b
A*a
bB*
b=normal
*=methylation
A*a
bB
Aa
bB*
A* in males
B* in females
A*b, A*B,
ab, aB
Ab, AB*,
ab, aB*
Thus genomic imprinting is permanent in the
somatic cells of an animal
– However, the marking of alleles can be
altered from generation to generation
Imprinting and DNA Methylation
• Genomic imprinting must involve a marking
process
• At the molecular level, the imprinting is
known to involve differentially methylated
regions
– They are methylated either in the oocyte or
sperm
• Not both
• For most genes, methylation results in
inhibition of gene expression
–However, this is not always the case
Changes in methylation during gamete
development alter the imprint
Haploid female gametes transmit
an unmethylated gene
Fig 7.11
Haploid male gametes transmit
a methylated gene
To date, imprinting has been identified in dozens of
mammalian genes
Tbl 7.2
Tbl 7.2
Imprinting plays a role in the inheritance of
some human diseases: Prader-Willi syndrome
(PWS) and Angelman syndrome (AS)
–PWS is characterized by: reduced motor function,
obesity, mental deficiencies
–AS is characterized by: hyperactivity, unusual seizures,
repetitive muscle movements, mental deficiencies
Usually, PWS and AS involve a small deletion
in chromosome 15
–If it is inherited from the mother, it leads to AS
–If it is inherited from the father, it leads to PWS
• AS results from the lack of expression of
UBE3A (encodes a protein called EA-6P that transfers small
ubiquitin molecules to certain proteins to target their degradation)
– The gene is paternally imprinted (silenced)
• PWS results (most likely) from the lack of
expression of SNRNP (encodes a small nuclear
ribonucleoprotein that controls gene splicing necessary for the
synthesis of critical proteins in the brain)
– The gene is maternally imprinted (silenced)
The deletion is
the same in
males and
females, but the
expression is
different
depending on
who you
received the
normal version
from.
Fig 7.12
The relationship between genes
and traits is often complex
Complexities include:
• Multiple genes controlling one trait
Two genes control
coat color in mice
Fig 4.21
The interaction of these two proteins explains
their affect on a single trait (in fruit flies).
Variation in Peas
Fig 3.2
Fig 2.8
Inheritance of 2 independent
genes
Approximate position of seed color and shape genes
in peas
Y
y
Gene for seed color
r
Chrom. 1/7
R
Chrom. 7/7
Gene for
seed shape
There must be a better way…
Fig 2.9
Inheritance
can be
predicted by
probability
Section 2.2, pg 30-32
Sum rule
• The probability that one of two or more mutually
exclusive events will occur is the sum of their
respective probabilities
• Consider the following example in mice
• Gene affecting the ears • Gene affecting the tail
– De = Normal allele
– de = Droopy ears
– Ct = Normal allele
– ct = Crinkly tail
• If two heterozygous (Dede Ctct) mice are crossed
• Then the predicted ratio of offspring is
–
–
–
–
9 with normal ears and normal tails
3 with normal ears and crinkly tails
3 with droopy ears and normal tails
1 with droopy ears and crinkly tail
• These four phenotypes are mutually exclusive
– A mouse with droopy ears and a normal tail cannot have
normal ears and a crinkly tail
• Question
– What is the probability that an offspring of the above
cross will have normal ears and a normal tail or have
droopy ears and a crinkly tail?
• Applying the sum rule
– Step 1: Calculate the individual probabilities
P(normal ears and a normal tail) = 9 (9 + 3 + 3 + 1) = 9/16
P(droopy ears and crinkly tail) = 1 (9 + 3 + 3 + 1) = 1/16
– Step 2: Add the individual probabilities
9/16 + 1/16 = 10/16
• 10/16 can be converted to 0.625
– Therefore 62.5% of the offspring are predicted to have
normal ears and a normal tail or droopy ears and a
crinkly tail
Product rule
• The probability that two or more independent
events will occur is equal to the product of
their respective probabilities
• Note
– Independent events are those in which the
occurrence of one does not affect the probability
of another
• Consider the disease congenital analgesia
– Recessive trait in humans
– Affected individuals can distinguish between sensations
• However, extreme sensations are not perceived as painful
– Two alleles
• P = Normal allele
• p = Congenital analgesia
• Question
– Two heterozygous individuals plan to start a family
– What is the probability that the couple’s first three children will all have
congenital analgesia?
• Applying the product rule
– Step 1: Calculate the individual probabilities
• This can be obtained via a Punnett square
P(congenital analgesia) = 1/4
– Step 2: Multiply the individual probabilities
1/4 X 1/4 X 1/4 = 1/64
• 1/64 can be converted to 0.016
– Therefore 1.6% of the time, the first three offspring of a
heterozygous couple, will all have congenital analgesia
Different
genes are not
always
independent
Crossingover
Meiosis I
(Ind. Assort.)
Meiosis II
4 Haploid cells, each unique
Fig 5.1
The haploid cells contain
the same combination of
alleles as the original
chromosomes
The arrangement of linked
alleles has not been altered
Fig 5.1
These haploid cells contain a
combination of alleles NOT
found in the original
chromosomes
These are
termed
parental or
nonrecombinant
cells
This new combination of
alleles is a result of
genetic recombination
These are termed
recombinant cells
Linked alleles tend to be inherited together
Crossing over produces new allelic combinations
Recombinants are produced by crossing over
For linked
genes,
recombinant
frequencies are
less than 50
percent
Homologous
pair of chromosomes
Does this pedigree show recombination or linkage?
Does this pedigree show recombination or linkage?
Longer regions
have more
crossovers and
thus higher
recombinant
frequencies
Some crosses
do not give
the expected
results
=25%
42% 41%
9%
8%
These two genes are on the same chromosome
By comparing recombination frequencies, a linkage
map can be constructed
= 17 m.u.
Linkage map of
Drosophila
chromosome 2:
This type of map, with
mapping units more
than 50, can only be
put together by
making comparisons
of linked genes.
Today: some things Mendel did not tell us…
plus Mapping and Epigenetics
–Exam #3 W 7/30 in class (bonus #2 due)–
Lecture ended here, but I am leaving
in the following material so you can
get a preview of the mapping
problem we will work on to start
class on M 7/28.
Fig 5.2
A much greater proportion
of the two types found in
the parental generation
The probability of crossing over can be used to
determine the spatial relationship of different genes
Double recombinants arise from two crossovers
Recombinant
Double recombinants can show gene order
What is the relationship between
these 3 genes? What order and
how far apart?
similar to Fig 5.3,
also see Fig 5.9,
and pg 115-117
What is the
relationship
between these 3
genes?
What order and
how far apart?
similar to Fig 5.3
Double crossover
Which order produces the double crossover?
Which order produces the double crossover?
We have the order.
What is the distance?
Recombinants
between st and ss:
(50+52+5+3)/755
=14.6%
Recombinants
between ss and e:
(43+41+5+3)/755
=12.2%
Put it all together…
26.8 m.u.
st
ss
14.6 m.u.
e
12.2 m.u.
Drosophila linkage map
Linkage map of
Drosophila
chromosome 2
Recombination is not
completely random.
physical
distance
Yeast chromosome 3
linkage
map
Alignment of physical and recombination maps
Genotype
Phenotype
Genes code for
proteins (or RNA).
These gene
products give rise
to traits…
It is rarely this
simple.
{Meiosis:
producing gametes}
For life to exist,
the information
(genes) must be
passed on.
{Mitosis:
producing more cells}