Download CH 14 Mendel Notes

Chp 14
Mendel and the Gene Idea
Blending Model
vs.
Particulate Model
•genetic material mixes like paint
•Discreet inheritable units
•Parent’s traits inseparable
•Traits retain separate identities
Figure 14-01
Gregor Mendel
•Austrian Monk
•Did not know about DNA,
genes, or chromosomes!
•Tried to prove particulate
model of inheritance using
pea plants
Web Lab: Mendel and his Peas
Advantages of working with pea plants:
•Many different varieties
•Character = inheritable feature (GENE)
•Trait = variation of that character (ALLELE)
•Can control plant matings (paintbrush)
LE 14-2
Removed stamens
from purple flower
Transferred spermbearing pollen from
stamens of white
flower to eggbearing carpel of
purple flower
Parental
generation
(P)
Carpel
Stamens
Pollinated carpel
matured into pod
Planted seeds
from pod
First
generation
offspring
(F1)
Examined
offspring:
all purple
flowers
•Started with true-breeding varieties = HOMOZYGOUS
•If allowed to self-pollinate, all offspring had the same
traits as their parent plant
•Then, cross-pollinated two different true-breeding varieties
= HYBRIDIZATION
Hybridization
LE 14-5_1
P Generation
Appearance:
Genetic makeup:
Gametes:
Purple
flowers
PP
White
flowers
pp
P
p
F1 Generation
Appearance:
Genetic makeup:
Purple flowers
Pp
hybrid
Monohybrid Cross – tracks a single character/gene
Mendel’s Law of Segregation: Parental
alleles separate (segregate) during gamete
formation
• occurs in Metaphase I of Meiosis I
• only one allele per gamete
LE 14-5_2
P Generation
Appearance:
Genetic makeup:
Purple
flowers
PP
White
flowers
pp
P
p
Gametes
F1 Generation
Appearance:
Genetic makeup:
Purple flowers
Pp
1
Gametes:
2
1
P
p
2
gametes
F1 sperm
P
p
PP
Pp
Pp
pp
F2 Generation
P
gametes
F1 eggs
p
3
:1
Punnett Square
shows possible
offspring after
fertilization
LE 14-6
3
Phenotype
Genotype
Purple
PP
(homozygous
Purple
Pp
(heterozygous
1
2
1
Purple
Pp
(heterozygous
White
pp
(homozygous
Ratio 3:1
Ratio 1:2:1
1
r
R
F1
Rr
F2
R
r
Rr
Rr rr
3 round: 1 wrinkled
1 RR : 2 Rr : 1 rr
Law of Dominance: The dominant allele
(trait) is fully expressed and the recessive
allele has no noticeable effect
Recessive ≠ Bad
Test Cross
•Used to
determine the
genotype of an
individual with
a dominant
phenotype
•Cross it with a
homozygous
recessive and
observe
offspring ratios
100% tall
100% Tt
1 tall : 1 dwarf
T
T
T
1 Tt : 1 tt
t
t
Tt
Tt
t
Tt
tt
t
Tt
Tt
t
Tt
tt
Law of Independent Assortment: Each allele pair
segregates independently of other allele pairs during
Meiosis (Metaphase I).
Dihybrid Cross: tracks two genes (characters)
TP
TP
Tp
tP
tp
Tp
tP
tp
Tall purple
TP
TTPP TTPp TtPP
TtPp
TtPp
Tp
TTPp TTpp TtPp
Ttpp
tP
TtPP TtPp
ttPP
ttPp
tp
TtPp Ttpp
ttPp
ttpp
9 tall purple
3 tall white
3 dwarf purple
1 dwarf white
12:4 or 3:1
12:4 or 3:1
Rules of Probability & Genetics
•
•
Probability ranges from 0 – 1
Rule of Multiplication: Used to determine the
chance of two or more independent events
occurring together
1. Determine probability of each independent event
2. Multiply probabilities together
Ex: What is the probability (chance) that when 2
coins are flipped they will both end up heads?
½ x ½ = ¼ (or .25)
½
½
¼
¼
¼
¼
½
½
Multiplication Rule
Probability Problems
• What is the probability (chance) that 2 dice
will both show a 4 when rolled?
1/
6
x 1/6 = 1/36
Ex (F1):
F2:
TtPp
x TtPp
½ ½
½ ½
t p
t
t
T
TT
Tt
t
Tt
tt
P
p
P
PP
Pp
p
Pp
pp
p
½ x ½ x ½ x ½ = 1/16
ttpp
T
Rule of Addition
• What is the probability (chance) that the sum
of the numbers shown on 2 dice will equal 5?
• Rule of Addition: used to determine the
probability of an event that can occur in two or
more different ways/combinations
– Calculate probability for each possible
combination/way using Rule of Multiplication
– Add the probabilities for each separate
combination to get the total probability
Rule of Addition
• What is the probability (chance) that the sum
of the numbers shown on 2 dice will equal 5?
• Possible combinations:
1+4
2+3
1/
1/
x
6
6
1/
36
1/
+
1/
x
6
6
1/
36
3+2
1/
1/
x
6
6
+ 1/36 +
4/
1/ )
(
36
9
4+1
1/
1/
x
6
6
1/
36
=
= 3/8
x
¾
A
a
A
AA
Aa
a
Aa
aa
1
x
½
b
b
B
Bb
Bb
b
bb
bb
C
C
c
Cc
Cc
c
Cc
Cc
A__ B__ C__ =
A__ B__ c c =
A__ b b C__ =
a a B__ C__ =
A
a
A
AA
Aa
a
Aa
aa
b
b
B
Bb
Bb
b
bb
bb
C
C
c
Cc
Cc
c
Cc
Cc
A__ B__ C__ = 3/8
A__ B__ c c = 0
7
Sum = /8
A__ b b C__ = 3/8
a a B__ C__ = 1/8
A
a
A
AA
Aa
a
Aa
aa
b
b
B
Bb
Bb
b
bb
bb
C
C
c
Cc
Cc
c
Cc
Cc
Text book: pages 272 - 273
• All problems (#1-17) due on
• Tonight, work on problems:
# 2, 3, 4, 7, 8, & 10
Types of Dominance
• Complete Dominance – one allele complete
masks the other; the heterozygous (Rr) and
homozygous dominant (RR) have the same
phenotype
–The dominant allele usually
codes for some protein/enzyme
and the recessive allele codes for a
defective protein/enzyme. Both
are “expressed”, but only the
dominant allele is functional and
observable
Types of Dominance
• Codominance = two alleles expressed
separately & both affect phenotype
– Ex: Red & White rhododendron
“Roan” color in cattle
CW
CR
CR
CR CW CR CW
CW CR CW CR CW
Types of Dominance
• Incomplete Dominance =
F1 hybrids (heterozygotes)
have an intermediate
phenotype
– Results in a THIRD
phenotype
– NOT the same as blending
(F2 show all phenotypes)
– 1:2:1 phenotypic and
genotypic ratios in F2
Multiple Alleles = more than two alleles/varieties
IA & I B
are
codominant
over
i
Fill in Interactive Question 14.6
Blood Type practice…
• Suppose a father of blood type B and a mother
of blood type A have a child of type O. What
are the chances that their next child will be:
–
–
–
–
Type O?
Type B?
Type A?
Type AB?
Blood Type practice…
• Suppose a father of blood type B and a mother
of blood type A have a child of type O. What
are the chances that their next child will be:
–
–
–
–
Type O?
Type B?
Type A?
Type AB?
¼
¼
¼
¼
IB
i
IA
IAIB
IAi
i
IBi
ii
Pleiotropy
• Pleiotropy = one gene has multiple phenotypic
effects
– Ex: cystic fibrosis & sickle-cell disease
Epistasis =
(“force upon”)
one gene affects
the expression
of another gene
BbCc
Sperm
1
1
1
Ex: coat color in
mice
BbCc
1
1
4
BC
1
4
bC
1
4
1
Bc
4
bc
4
BC
BBCC
BbCC
BBCc
BbCc
4
bC
BbCC
bbCC
BbCc
bbCc
4
Bc
BBCc
BbCc
BBcc
Bbcc
4
bc
BbCc
bbCc
Bbcc
bbcc
9
16
3
16
4
16
¾ * ¾ = 9/16
¾ * ¼ = 3/16
mm __ __ ¼ * 1 = 4/16
M__B__
brown
white
(¼)
Polygenic =
additive effect
of 2 or more
genes on one
phenotype
aabbcc
Aabbcc
AaBbCc
AaBbcc AaBbCc AABbCc AABBCc AABBCC
20/64
15/64
Fraction of progeny
Ex: skin color
height
AaBbCc
6/64
# alleles + 1 = # phenotypes
1/64
AaBbCc = 25 cm
Seven (6 alleles + 1 = 7 phenotypes)
Nature vs. Nurture….
Acidic soil
Basic soil
Pedigree
• Family tree showing relationships between
family members and the pattern of inheritance
across the generations
Figure 14-16
Aa
AA
aa
aa
Aa
Aa A_
A_
A_
Aa Aa Aa Aa aa Aa Aa A_ A_ A_ A_ A_
Aa
AA
Aa
Aa
Aa
No – either he or his wife is AA, but can’t be certain
Recessive Disorders
• Carriers = heterozygotes (Aa) have the
defective allele, but are not affected by
disorder
– “carry” allele, but do not express it
– Can pass it on to offspring (50% chance)
¼
2/
3
A
a
A
a
AA Aa
Aa aa
Recessive Disorders
• Cystic Fibrosis
• Tay-Sachs Disease
• Sickle-cell Disease
2/ Aa
3
x 2/3 Aa x ¼ aa = 4/36 = 1/9
Aa
Aa
Aa
aa
A_
A_
?
Aa
aa
A
a
A
a
AA Aa
Aa aa
Dominant Disorders
• RARE – nature will weed out (no carriers!)
• Achondroplasia Dwarfism
• Huntington’s Disease
LE 14-17a
Amniocentesis
Amniotic
fluid
withdrawn
Fetus
A sample of
amniotic fluid can
be taken starting at
the 14th to 16th
week of pregnancy.
Centrifugation
Placenta
Uterus
Cervix
Fluid
Fetal
cells
Biochemical tests can be
performed immediately on
the amniotic fluid or later
on the cultured cells.
Fetal cells must be cultured
for several weeks to obtain
sufficient numbers for
karyotyping.
Biochemical
tests
Several
weeks
Karyotyping
LE 14-17b
Chorionic villus sampling (CVS)
A sample of chorionic villus
tissue can be taken as early
as the 8th to 10th week of
pregnancy.
Fetus
Suction tube
inserted through
cervix
Placenta Chorionic villi
Fetal
cells
Biochemical
tests
Several
hours
Karyotyping
Karyotyping and biochemical
tests can be performed on
the fetal cells immediately,
providing results within a day
or so.