GENETICS
Genetics
• The study of heredity.
heredity
• Gregor Mendel (1860’s) discovered the
fundamental principles of genetics by breeding
garden peas.
peas
Genetics
• Alleles
1. Alternative forms of genes.
2. Units that determine heritable traits.
3. Dominant alleles (TT - tall pea plants)
plants
a. homozygous dominant
4. Recessive alleles (tt - dwarf pea plants)
plants
a. homozygous recessive
5. Heterozygous (Tt - tall pea plants)
plants
Phenotype
• Outward appearance
• Physical characteristics
• Examples:
1.
2.
tall pea plant
dwarf pea plant
Genotype
• Arrangement of genes that produces the
phenotype
• Example:
1. tall pea plant
TT = tall (homozygous dominant)
2. dwarf pea plant
tt = dwarf (homozygous recessive)
3. tall pea plant
Tt = tall (heterozygous)
Punnett square
• A Punnett square is used to show the
possible combinations of gametes.
gametes
Breed the P generation
• tall (TT) vs. dwarf (tt) pea plants
T
t
t
T
tall (TT) vs. dwarf (tt) pea plants
T
T
t
Tt
Tt
produces the
F1 generation
t
Tt
Tt
All Tt = tall
(heterozygous tall)
Breed the F1 generation
• tall (Tt) vs. tall (Tt) pea plants
T
T
t
t
tall (Tt) vs. tall (Tt) pea plants
T
T
t
TT
Tt
t
Tt
tt
produces the
F2 generation
1/4 (25%) = TT
1/2 (50%) = Tt
1/4 (25%) = tt
1:2:1 genotype
3:1 phenotype
Monohybrid Cross
• A breeding experiment that tracks the inheritance
of a single trait.
• Mendel’s “principle of segregation”
a. pairs of genes separate during gamete
formation (meiosis).
b. the fusion of gametes at fertilization pairs
genes once again.
Homologous Chromosomes
eye color locus
B = brown eyes
eye color locus
b = blue eyes
This person would
have brown eyes (Bb)
Paternal Maternal
Meiosis - eye color
B
sperm
B
B
Bb
haploid (n)
b
diploid (2n)
b
b
meiosis I
meiosis II
Monohybrid Cross
• Example:
Example
Cross between two heterozygotes
for brown eyes (Bb)
BB = brown eyes
Bb = brown eyes
bb = blue eyes
B
b
B
Bb x Bb
b
female gametes
male
gametes
Monohybrid Cross
B
b
B
BB
Bb
b
Bb
bb
Bb x Bb
1/4 = BB - brown eyed
1/2 = Bb - brown eyed
1/4 = bb - blue eyed
1:2:1 genotype
3:1 phenotype
Dihybrid Cross
• A breeding experiment that tracks the inheritance
of two traits.
• Mendel’s “principle of independent assortment”
a. each pair of alleles segregates independently
during gamete formation (metaphase I)
b. formula: 2n (n = # of heterozygotes)
Independent Assortment
• Question: How many gametes will be produced
for the following allele arrangements?
• Remember:
2n (n = # of heterozygotes)
1.
RrYy
2.
AaBbCCDd
3.
MmNnOoPPQQRrssTtQq
Answer:
1. RrYy: 2n = 22 = 4 gametes
RY Ry rY ry
2. AaBbCCDd: 2n = 23 = 8 gametes
ABCD ABCd AbCD AbCd
aBCD aBCd abCD abCD
3. MmNnOoPPQQRrssTtQq: 2n = 26 = 64 gametes
Dihybrid Cross
• Example:
R
r
Y
y
= round
= wrinkled
= yellow
= green
cross between round and yellow
heterozygous pea seeds.
RrYy x RrYy
RY Ry rY ry x RY Ry rY ry
possible gametes produced
Dihybrid Cross
RY
RY
Ry
rY
ry
Ry
rY
ry
Dihybrid Cross
RY
Ry
RY RRYY RRYy
Ry RRYy
RRyy
rY
RrYY
RrYy
ry
RrYy
Round/Yellow:
9
Round/green:
3
Rryy
wrinkled/Yellow: 3
rY RrYY
RrYy
rrYY
rrYy
ry
Rryy
rrYy
rryy
RrYy
wrinkled/green:
1
9:3:3:1 phenotypic ratio
Test Cross
• A mating between an individual of unknown genotype
and a homozygous recessive individual.
• Example: bbC__ x bbcc
BB = brown eyes
Bb = brown eyes
bb = blue eyes
CC = curly hair
Cc = curly hair
cc = straight hair
bC
bc
b___
Test Cross
• Possible results:
bc
bC
b___
C
bbCc
bbCc
or
bc
bC
b___
c
bbCc
bbcc
Incomplete Dominance
• F1 hybrids have an appearance somewhat in
between the phenotypes of the two parental
varieties.
• Example: snapdragons (flower)
• red (RR) x white (rr)
R
RR = red flower
rr = white flower
r
r
R
Incomplete Dominance
R
R
r
Rr
Rr
produces the
F1 generation
r
Rr
Rr
All Rr = pink
(heterozygous pink)
Codominance
• Two alleles are expressed (multiple alleles)
alleles
in heterozygous individuals.
individuals
• Example: blood
1.
2.
3.
4.
type A
type B
type AB
type O
=
=
=
=
IAIA or IAi
IBIB or IBi
IAIB
ii
Codominance
homozygous male B (IBIB)
x
heterozygous female A (IAi)
• Example:
IB
IB
IA
I AI B
I AI B
i
I Bi
IBi
1/2 = IAIB
1/2 = IBi
Codominance
• Example: male O (ii) x female AB (IAIB)
IA
IB
i
I Ai
IBi
i
I Ai
IBi
1/2 = IAi
1/2 = IBi
Codominance
• Question:
Question
If a boy has a blood type O and
his sister has blood type AB,
what are the genotypes and
phenotypes of their parents.
• boy - type O (ii) X girl - type AB (IAIB)
Codominance
• Answer:
IA
i
I B I AI B
i
ii
Parents:
genotypes = IAi and IBi
phenotypes = A and B
Sex-linked Traits
• Traits (genes) located on the sex
chromosomes
• Example:
fruit flies
(red-eyed
male) X (white-eyed
female)
red
white
Sex-linked Traits
Sex Chromosomes
fruit fly
eye color
XX chromosome - female
Xy chromosome - male
Sex-linked Traits
• Example:
fruit flies
(red-eyed male) X (white-eyed female)
• Remember: the Y chromosome in males
does not carry traits.
RR = red eyed
Rr = red eyed
rr = white eyed
Xy = male
XX = female
XR
Xr
Xr
y
Sex-linked Traits
Xr
XR
y
XR Xr
Xr y
1/2 red eyed and female
1/2 white eyed and male
Xr
XR Xr
Xr y
Population Genetics
• The study of genetic changes in populations.
populations
• The science of microevolutionary changes in
populations.
populations
• Hardy-Weinberg equilibrium:
the principle that shuffling of genes that occurs
during sexual reproduction, by itself, cannot
change the overall genetic makeup of a population.
• Hardy-Wienberg equation:
1 = p2 + 2pq + q2
Question:
• How do we get this equation?
Answer:
“Square” 1 = p + q
↓
12 = (p + q)2
↓
1 = p2 + 2pq + q2
Hardy-Wienberg equation
• Five conditions are required for Hardy-Wienberg
equilibrium.
1. large population
2. isolated population
3. no net mutations
4. random mating
5. no natural selection
Important
• Need to remember the following:
p2 = homozygous dominant
2pq = heterozygous
q2 = homozygous recessive
Question:
• Iguanas with webbed feet (recessive trait) make
up 4% of the population. What in the population
is heterozygous and homozygous dominant.
dominant
Answer:
1. q2 = 4% or .04
q2 = .04
2. then use 1 = p + q
1 = p + .2
q = .2
1 - .2 = p
.8 = p
3. for heterozygous use 2pq
2(.8)(.2) = .32 or 32%
4. For homozygous dominant use p2
.82 = .64 or 64%
Hardy-Wienberg equation
1 = p2 + 2pq + q2
•
•
•
•
64% = p2
32% = 2pq
04% = q2
100%
= homozygous dominant
= heterozygous
= homozygous recessive