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Chapter 2
classical Mendelian genetics
© 2006 Jones & Bartlett Publishers
Gregor Mendel
monk
gardener
careful observer
experimenter
http://history.nih.gov/exhibits/nirenberg/popup_htm/01_mendel.htm
http://biology.clc.uc.edu/Fankhauser/Travel/Berlin/for_web/Mendel_in_Brno.html
Mendel
at that time (1850’s-1860’s) it was thought
that the traits passed on by the parents
were blended in the offspring
pink
red + white = ?
purple + white = purple or white
Mendel
parents contributed to offspring
factors remained unchanged
factors
set out to trace their movements
looked at phenotypeappearance
looked at ratio’s
Mendel
used peas
access to different varieties
usually self-pollinated
could manipulate pollinization
female
male
Mendel
started with true-breeding plants
parent 1
round
x
parent 1
offspring
round
round
wrinkled x wrinkled
wrinkled
Mendel
started with true-breeding plants
seed shape round vs. wrinkled
seed color yellow vs. green
Mendel
seed shape round vs. wrinkled
Ppollen
Pflower
F1
round
x
wrinkled
round
wrinkled
x
round
round
Fig. 2.2. Reciprocal crosses of true-breeding pea plants
© 2006 Jones & Bartlett Publishers
Mendel
seed shape round vs. wrinkled
Ppollen
Pflower
F1
round
x
wrinkled
round
wrinkled
x
round
round
dominant vs. recessive
hybrids
recessive
dominant
Fig. 2.3. Traits studied in
peas by Mendel
© 2006 Jones & Bartlett Publishers
F1 cross
round
hybrid
X
?
round
hybrid
Fig. 2.7.
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F1 cross
round
hybrid
X
round
hybrid
each hybrid has two parents
therefore
each hybrid has two “factors”
F1 cross
round
hybrid
X
round
hybrid
To solve:
define terms (use consistently)
determine parent genotype
determine gamete genotype
Punnett square
To solve:
R
define terms
= round
capital letter=dominant
?r
= wrinkled
lower case letter-recessive
R and r are alleles:
alternative forms of a gene
To solve:
determine parent genotypes
original cross:
round
vs. wrinkled
(pure-breeding)
parents:
RR
(homozygous)
rr
To solve:
determine gamete genotypes
original cross:
round
vs. wrinkled
(pure-breeding)
parents:
RR
rr
gametes:
R or R
r or r
Punnett square
To solve:
R
R
r
Rr
Rr
r
Rr
Rr
all F1 would be Rr
(heterozygous)
back to the
F1 cross
round
hybrid
X
round
hybrid
parents:
Rr
Rr
gametes:
R or r
R or r
#1
Punnett square
R
r
R
RR
Rr
r
Rr
rr
What would the ratio of round to wrinkled be?
Table 2.1. Results of Mendel’s
monohybrid experiments
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Mendel’s first law…
…The Law of Segregation
The two “factors” in the adult
separate from each other during the
production of the gametes.
#2
homologous pairs of chromosomes
separate during meiosis I.
A problem:
Suppose you have a plant with
purple flowers, but unknown
ancestry. What is its’ genotype?
Is it homozygous dominant
or
heterozygous?
PP
Pp
2 P’s or not…..
A solution:
cross it with homozygous recessive
P? x
pp
do it!
A solution:
if PP x pp
all offspring would be Pp
(heterozygous)
(purple flowers)
if Pp x pp
some offspring would be Pp
(purple)
some offspring would be pp
(white)
The solution:
#3
Geneticists call this a “test cross”
(pg. 44)
pause for ?
Mendel also looked at two traits at once
round
yellow
vs wrinkled
vs
green
RR, rr
YY, yy
pure-breeding round and yellow
pure-breeding green and wrinkled
pure-breeding round and yellow
pure-breeding green and wrinkled
parental
genotype
RRYY
rryy
gametes
genotype
RY
?
ry
?
ry
RY
RrYy
all heterozygous; phenotype =?
all F1 were round, yellow hybrids RrYy
(dihybrids)
Mendel then did a dihybrid cross (F1 cross)
parental
genotype
RrYy
RrYy
gametes
genotype
?
?
split
class
R
Y
R
y
r
y
r
Y
metaphase I cell
RY
ry
Ry
rY
RY RRYY
RrYy
Ry RRyy
RrYy
ry
rryy
rY RrYy
rrYY
RrYy
3 round, yellow
1 wrinkled, green
1 round, green
2 round, yellow
1 wrinkled, yellow
Mendel’s results:
(dihybrid cross)
round, yellow
315 9
round, green
108 3
wrinkled, yellow 101 3
wrinkled, green 32 1
556
#4
Fig. 2.11. Independent segregation of the Ww and Gg allele pairs
© 2006 Jones & Bartlett Publishers
Fig. 2.12.
Dihybrid cross
Mendel’s second law…
…The Law of Independent
Assortment (in modern language)
#5
How a pair of homologous
chromosomes align at Metaphase I
is independent of how all the other
pairs of chromosomes align at
metaphase I
RY
Ry
R
Y
R
y
r
y
r
Y
rY
ry
metaphase I cell
2.4 probability
a fraction (ratio; like 1/4)
between 0 will never happen
and 1 will always happen
yellow yellow
Yy x Yy
Y
y
Y
YY
Yy
y
Yy
yy
yellow yellow
Yy x Yy
What is the probability of getting
plants with green seed ?
Y
y
Y
YY
Yy
y
Yy
yy
2001 green
6022 yellow
2001
(6022 + 2001)
=
1
4.01
yellow yellow
Yy x Yy
Y
y
Y
YY
Yy
y
Yy
yy
What is the probability of
getting plants with yellow peas?
YY
Yy
1/4
2/4
prob [YY]
[YY] ++ prob
prob [Yy]
[Yy]
prob [YY or Yy] = prob
= 1/4 + 2/4 = 3/4
#6
R
r
R
RR
Rr
r
Rr
rr
RrYy
x
RrYy
Y
y
Y
YY
Yy
y
Yy
yy
What is the probability of getting plants
with round, yellow peas?
YY or Yy
and
RR or Rr
3/4
3/4
prob [round and yellow] = prob [round] x prob [yellow]
= 3/4 x 3/4 = 9/16
#7
?
2.6 dominance ?
RR
same phenotype ?
Rr
R gene codes for
starch branching enzyme 1 (SBEI)
dominance is
not necessarily
all or none
Fig. 2.20. Three attributes of phenotype affected by Mendel's alleles W and w, which
determine round versus wrinkled seeds.
© 2006 Jones & Bartlett Publishers
RR and Rr same phenotype in peas,
but really different at the molecular level
Are there some cases where
homozygous dominant (RR)
looks different than
heterozygous (Rr) ?
Fig. 2.21. Incomplete dominance in
snapdragons.
© 2006 Jones & Bartlett Publishers
Incomplete dominance
When the heterozygote is intermediate
between the homozygous phenotypes
Seen with traits that are quantitative
(can be measured on a continuous scale)
as opposed to a discrete trait
(appears to be all or none)
Blood typing (ABO)
Codominance
both traits are expressed
Human blood types:
A
B
AB
O
Blood typing (ABO)
three alleles (multiple alleles)
IA
IB
i
make “A” carbohydrate
make “B” carbohydrate
make neither carbohydrate
Fig. 2.22. The
ABO antigens
on the surface
of human red
blood cells are
carbohydrates.
Blood typing (ABO)
Human blood types:
phenotype:
A
B
AB
O
genotype:
IAIA, IAi
IBIB, IBi
codominant
IAIB
ii
Blood typing (ABO)
Our immune system makes proteins
called antibodies
to attack foreign molecules
called antigens.
Blood typing (ABO)
For someone with type A blood
(someone with the IA allele):
A carbohydrate
“self”
B carbohydrate
“non-self” or foreign
antigen
Blood typing (ABO)
For someone with type B blood
(someone with the IB allele):
A carbohydrate
“non-self” or foreign
antigen
B carbohydrate
“self”
Blood typing (ABO)
For someone with type O blood
(someone with ii alleles):
A carbohydrate
“non-self” or foreign
antigen
B carbohydrate
“non-self” or foreign
antigen
Blood typing (ABO)
For someone with type AB blood
(someone with IA and IB alleles):
A carbohydrate
“self”
B carbohydrate
“self”
Blood typing (ABO)
What kind of antibodies would they make?
Type
A
B
AB
O
- B antibodies
- A antibodies
- neither A nor B antibodies
- A and B antibodies
Table 2.3. Genetic control of the ABO blood groups
AB universal recipient
O universal donor
© 2006 Jones & Bartlett Publishers
Fig. 2.23. Antibody against typeA antigen will agglutinate red
blood cells carrying the type-A
antigen.
© 2006 Jones & Bartlett Publishers
Other reasons why Mendel
rules aren’t always observed
incomplete dominance
multiple alleles
polygenic
traits ?
variable expressivity
same mutation-different results
penetrance (phenotype = expected)
complete
100%
lung
incomplete < 100%
cancer
example
2.7 Epistasis
non-allelic genes interacting
to affect the same trait
e.g.,
When the expected 9:3:3:1 ratio
of a dihybrid cross is altered
Merriam Webster:
suppression of the effect
of a gene by a nonallelic gene
2.7 Epistasis
Given:
Two different genes C and P for flower color
C
c
purple flowers
white flowers
P
p
purple flowers
white flowers
plants with C- and P- genotypes have
purple flowers (wt)
plants with cc or pp have white
? flowers
2.7 Epistasis
aside:
How could that happen?
(hypothetically)
E1
A
E2
B
mutation
#1
E3
C
E4
D
mutation
#3
purple pigment
2.7 Epistasis
CC pp
phenotype
gamete
genotype
F1 genotype
F1 phenotype
x
cc PP
white flowers
Cp
cP
CcPp
all purple
F1 cross
what gametes ?
Fig. 2.24. A cross showing epistasis
in the determination of flower color
in peas.
© 2006 Jones & Bartlett Publishers
CP
Cp
cP
cp
CP CCPP CCPp CcPP CcPp
Cp CCPc CCpp CcPp
Ccpp
cP CcPP CcPp ccPP
ccPp
cp CcPp Ccpp
ccpp
ccPp
epistasis
When the expected
9:3:3:1 ratio of a
dihybrid cross is altered
9:7 ratio
purple to white
Fig. 2.25. Modified F2
dihybrid ratios.
© 2006 Jones & Bartlett Publishers
2.8 Complementation
mutations c and p show
complementation
CCpp
white
x
CcPp
purple
ccPP
white
Fig. 2.26. Complementation reveals whether two
recessive mutations are alleles of different genes.
© 2006 Jones & Bartlett Publishers
2.8 Complementation
means that mutations
affect different genes
Lack of complementation
means that mutations
affect the same gene
Fig. 2.27. Results of complementation tests among six mutant strains of peas.
© 2006 Jones & Bartlett Publishers
Fig. 2.28. A method for interpreting results of complementation tests.