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Mendelian Genetics
Genetics
the branch of Biology that deals with
INHERITANCE
(the passing of traits from parents to offspring)
Gregor Mendel
~Austrian monk who studied pea plants in the 1850’s
~theorized that characteristics are inherited as a
result of the transmission of “factors” from parents
to offspring (he didn’t know about genes &
chromosomes!)
 Mendel
used pea plants b/c he
noticed contrasting traits:
Tall plants vs. short plants
Green
pods vs. yellow pods
Wrinkled
seeds vs.
smooth seeds
“Brother Mendel, we grow tired of peas”
~his work wasn’t valued until the early
1900’s when scientists began observing
contrasting traits in Drosophila (fruit flies)
and when better
microscopes led
to the discovery
of chromosomes
Drosophila have short life cycles
and few chromosomes and
therefore are easy to study in
genetics
Mendel is considered to be the
“Father of Genetics”
The garden at Mendel’s Monastery
II. The Gene-ChromosomeTheorychromosomes contain genes which have all
the instructions for the traits of an organism
gene- -the unit of hereditary material found on
chromosomes
- genes are made of DNA
allele- another word for gene
ex)
2 alleles/genes for flower
color: one white, one purple
loci- Latin for “place”; the place/position on a
chromosome where you find a gene/allele
Loci on fruit fly chromosome
Add drosophila loci picture
 Homologous
chromosomes:
pairs of chromosomes that have genes
for the same traits. You have 2 genes
for every trait and they’re located on
homologous chromosomes.
XX (double stranded homologs)
II (single stranded homologs)
III.Major Genetic Concepts
III.Dominant Allele -the allele (gene) that is
expressed if present
-the “stronger” allele
capital
represented by a _________letter
example: tallness=______
T
Recessive Allele- the gene that isn’t expressed if
dominant is present.
- the “weaker”, rarer allele
lower case
represented by a _________letter
t
example: lack of tallness (shortness)=_____
Genotype- the genes (letters) of a trait
described as homozygous or
heterozygous
ex) Dd; heterozygous
DD; homozygous dominant
dd; homozygous recessive
Homozygous- also called pure
ex) TT, tt, DD, dd
when both genes are the
same, either dominant or
recessive.
Heterozygous- also called ________
hybrid
-when both genes are different;
one dominant, one recessive
ex) Dd, Tt
Phenotype- the physical appearance of a trait
ex) tall, short, curly, straight
(the genotype is the musical score and the phenotype
is the music we hear)
Ataxia video link
~Mendel wondered how he could cross 2
tall pea plants and end up w/some tall &
some short
~he theorized that the recessive trait can
disappear in 1 generation and show up
in the next generation as follows:
Genes T,t are
SEGREGATED or
separated during MEIOSIS
Parents:
x
Tt
Tt
meiosis
T
t
T
t
fertilization
TT
F1
possibilities
Tt
Tt
tt
All of the genes are RECOMBINED RANDOMLY during fertilization
(this diagram is a graphical representation of a Punnet Square)
Law of Segregation and Recombination:
Mendel explained that “factors” which occur in pairs
are separated from each other during gamete
formation and recombined at fertilization.
Law of Independent Assortment:
Mendel concluded that different traits are inherited
independently of one another
(genes for different traits are separated &
distributed to gametes independently of one
another when they are on different chromosomes).
He learned this after doing
dihybrid crosses which
examine 2 traits (like pea
color & pea plant height);
yellow peas aren’t always
on tall plants…
More on dihybrids to
come…..
PUNNETT SQUAREPredicts possible outcomes of a genetic cross
T
T
SegregationOnly 1 gene
from each
parent
is passed along
due to meiosis
TT
OR
t
Tt
OR
t
Tt
tt
Results: 75% dominant
25% recessive
Filling in the
squares
represents
fertilization
Which
recombines
genes
Problems:
1.If a heterozygous freckled person marries a
non-freckled person, what is the chance that
they will have a non-freckled baby?
Freckles (F)
no freckles (f)
F
f
f
Ff
ff
f
Ff
ff
50% chance of
Non-freckled (ff)
baby
2. The gene for brown eyes (B) is dominant over the
gene for blue eyes (b). Show the results of a cross
between a hybrid brown-eyed woman and a hybrid
brown-eyed man.
B=brown
b=blue
B
b
phenotype
B
BB
Bb
b
Bb
bb
genotype
75% brown 50% Bb
25% blue
25% BB
25%bb
3.Normal skin pigmentation (A) is dominant
over albino (a). Show the cross of an albino
man with a woman who is homozygous normal.
A= normal
a=albino
A
a
a
A
Aa
Aa
Aa
Aa
phenotype
100%
normal
genotype
100%
heterozygous
Ff
tt
Bb
hh
+,+
mm
CC
freckled
non taster
brown
no disease
has blood factor
light skin
color vision
Ff
freckeld
TT
Taster
bb
blond
hh
no disease
+,-
has blood factor
mm
light skin
Cc
color vision
ff
tt
BB
Hh
no freckles
non taster
brown hair
has disease
-,MM
cc
doesn’t have blood factor
dark skin
color blind
Can’t tell if he was Ff or ff
yes
FF
Can’t tell if parents were Tt or tt
Can’t tell if parents were Bb or bb
yes
brown
dark
Can’t tell if parents were Cc or cc
Intro to dihybrid crosses (not in note pkt)
As Mendel continued his study of peas, he
noticed that different traits (pea color, pea shape)
were inherited independently of one another.
When he crossed yellow x yellow, the F2
smooth
smooth
generation had : yellow, yellow, green, green
smooth wrinkled smooth wrinkled
We know now, that during meiosis,
genes for different traits are separated
and distributed to gametes
independently of one another. (when
they’re on different chromosomes).
Genes located on the same chromosome
do NOT sort independently (aka gene
linkage).
Breeding experiments that examine 2
different traits are called dihybrid crosses
Dihybrid Crosses
In guinea pigs, black coat color (B) is dominant to albino
(b); coarse coat (R) is is dominant to smooth (r) . Two
animals are selected for breeding. Their genotypes are
BBRR and bbrr. Give the results of the following
including expected genotypes and phenotypes of:
a) the F1 generation
b) the F2 generation
c) offspring produced from a cross of an F1pig
with one having genotype BBRr
a)
BBRR x bbrr
BR
br
BbRr
100% black course
pure parents
gametes
F1 generation
b)F2: BbRr x BbRr
BR
Br
bR
br
BR
Br
bR
br
possible
gametes
BR
Br
bR
br
BR BBRR
BBRr
BbRR
BbRr
BBRr
BBrr
BbRr
Bbrr
BbRR
BbRr
bbRR
bbRr
BbRr
Bbrr
bbRr
bbrr
Br
bR
br
Key
B=black
b=albino
R=Coarse
r=smooth
Phenotype results
9 black coarse
3 black smooth
3 albino coarse
1albino smooth
9:3:3:1
c)BbRr x
BBRr
BR
Br
bR
br
BR
Br
BR
BBRR
BR
Br
BBRr
Br
BBRr
BBrr
bR BbRR BbRr
br
BbRr
possible
gametes
Bbrr
Phenotype
results
6 black coarse
2 black smooth
Statistically:
If x = the number of traits in the
cross
x
2 = # phenotypes
x
3 = # genotypes
x
4 = # individuals
Monohybrid
Results: 2 phenotypes
3 genotypes
4 individuals
Tt x Tt
Dihybrid TtBb x TtBb
2
Results: (2) = 4 phenotypes
(3)2= 9 genotypes
2
(4) = 16 individuals
Trihybrid
TtBbRr x TtBbRr
Results: (2)3= 8 phenotypes
3
(3) = 27 genotypes
3
(4) = 64 individuals
9:3:3:1
Dihybrid Problems….answers
1.
L= long
l = vestigal
G = gray
g = ebony
F1 llgg x LLGG
LG
lg
LlGg
100% LlGg; long, gray
F2 LlGg x LlGg
LG
Lg
lG
lg
LG
LLGG
LLGg
LlGG
LlGg
Lg
LLGg
LLgg
LlGg
Llgg
lG
LlGG
LlGg
llGG
llGg
lg
LlGg
Llgg
llGg
llgg
9 long gray
3 long ebony
3 vest. gray
1 vest ebony
2.
R= red
r = yellow
T = tall
t = dwarf
F1 RRtt x rrTT
Rt
rT
RrTt
100% RrTt; tall, red
F2 RrTt x RrTt
RT
Rt
rT
rt
RT
RRTT
RRTt
RrTT
RrTt
Rt
RRTt
RRtt
RrTt
Rrtt
rT
RrTT
RrTt
rrTT
rrTt
rt
RrTt
Rrtt
rrTt
rrtt
9 red tall
3 red short
3 yellow tall
1 yellow
short
3.
T = taste
t = non taste
N = normal pigment
n = albino
from his mother
from her father
Mom: ttNn
x
Dad: Ttnn
tN
tn
Tn
TtNn
Ttnn
tn
ttNn
ttnn
25% taster, normal
25% taster, albino
25% non-taster, normal
25% non-taster, albino
4.
B= bark
b = silent
E = erect ears
e = droopy ears
EeBb x eebb
eb
EB
Eb
eB
EeBb
Eebb
eeBb
eb
eebb
25% each:
erect barking
erect silent
drooping
barking
drooping silent
Intermediate Inheritance
Many genes do not follow the patterns of dominance:
may produce a trait between the 2 parentsINCOMPLETE DOMINANCE (blending)
red flowers + white flowers = pink flowers
“R”
+
“W”
= “RW”
OR
May produce a trait that expresses 2 dominant alleles
at the same time :
CODOMINANCE
red fur + white fur = red & white fur
CR+
CW
= CRCW
Ex) blood type AB: IAIB
Test Cross:
Consider a phenotypically tall organism. What is it’s genotype? Is it TT
or Tt? To answer this question, you would do a test cross:
definition: when an organism showing the dominant trait is
crossed with a pure recessive to determine if
that dominant organism is homozygous or
heterozygous
*By observing the phenotypes of the offspring, we can trace back to
the genotype of the parent:
If the genotype
of the parent was
homozygous (TT),
all the offspring are tall:
T
t
Tt
t
Tt
T
Tt
Tt
If the genotype
of the parent was
heterozygous (Tt),
50% of the offspring
are tall and 50% are
short:
T
t
t
Tt
tt
t
Tt
tt
Problem:
1. A cattle rancher buys a black bull which
is supposed to be a purebred. Black coat is
dominant to red. The new owner decides to
be sure he has gotten a good deal by mating
the bull with several red cows. Such crosses
are called test crosses. If the farmer really
got a good deal, what color calves should be
born?
B
B
b Bb
Bb
Bb
Bb
b
B= black
b= red
They should all be black
Multiple Alleles:
~when there are more than 2 (“multiple”) alleles for a
trait
~example: Human blood groups
have 3 alleles: IA
where I is dominant
IB
and i is recessive
i
~the possible combinations of these 3 blood alleles
are as follows:
Blood type
Phenotype
genotype
A
IAIA or IAi
B
IBIB or IBi
AB
IAIB
ii
O
Problems:
1. A couple preparing for marriage have their blood typed
along with the other required blood tests. Both are
type AB. They ask what types of blood their children
may have. What will you tell them?
I
I
A
I
B
A
A A
I I
A B
I I
I
B
A B
I I
B B
I I
Phenotype Genotype
50% AB
IAIB
25% A
IAIA
25% B
IBIB
2. A type A person marries a type A person. Their firstborn
has type O blood. What are the genotypes of the parents
and the child?
IA ___
i
parent ___
A
I
A
I
i
IA IA
A
I i
IA ___
i
parent ___
i ___
i
child ___
i
IAi
ii
Each parent must
have a recessive
gene to give to the
child
3. A wealthy, elderly couple die together in an accident. Soon a man
shows up to claim their fortune claiming that he is their long lost son.
Other relatives dispute the claim. Hospital records show that the
deceased couple were blood types AB and O. The person claiming to be
their son was type O. Do you think the man is an imposter?
A
i
i
B
I
I
A
B
I i
A
I i
I i
B
I i
50% A
50% B
He was an imposter!
Sex Determination:
~there are 2 types of chromosomes:
sex chromosomes (1 pair)
autosomes (all other pairs)
~in each diploid human cell this looks like:
sex chromosomes 1 pair
autosomes
22 pairs
46
23 homologous pairs or _____chromosomes
female
~The genotype XX represents a _______________
male
~The genotype XY represents a _______________
sperm
~The sex of an individual is determined by the _______________
at
fertilization
the time of________________________.
X chromosomes while the human
~The human egg contains only ______
X or _______
Y chromosomes.
sperm can contian either ______
X
X
Every time a man
X XX
XX
Y XY
XY
and woman have a
child there is a
50 / 50 chance it
________
will be a boy or girl.
Sex Linkage:
~sex-linked traits are caused by genes found on
the X chromosome
~sex-linked traits are recessive
~since they are recessive, fewer females are
afflicted with these traits because they have
another X chromosome which is dominantly
normal. Males have only one X chromosome so
when they have a sex-linked gene, they’ll display
the trait.
~the genotype of a sex-linked trait is represented
as_________
X or X (can also be represented as Xh, where h is the trait)
~females with one gene for the trait ( X X ) are
called_____________
carriers
~examples of sex-linked traits: hemophilia, colorblindness
1. A man normal for blood clotting marries
a woman who is a carrier for hemophilia.
What are the chances they will have a child
with hemophilia?
X
X
Y
XX
H
or X X
X
XY
H
XX
H
or X X
H
X Y
XY
h
h
X Y
25% chance of a child with
hemophilia:
25% normal female (XX)
25% carrier female (XX)
25% normal male (XY)
25% hemophila male (XY)
2.What is the probability that a woman
with normal vision who marries a
colorblind man will have a colorblind
child?
X
X XX
Or XHXh
Y
XY
H
X Y
X
XX
XHXh
XY
H
X Y
0% chance:
50% carrier females
50% normal males
3. A man with a normal vision and a woman
with normal vision have 3 sons. Two of the
sons have normal color vision and one of
them is color-blind. What are the probable
genotypes of the parents?
X
X
XX
or XhXH
X
XX
XHXH
Y
XY
XhY
XY
XHY
XY and XX ;
The gene for color
blindness must
be hidden/carried
by mother to be
passed on
The Y chromosome
QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.
Pedigree charts: used to show he presence or absence of
a certain trait in families through
several generations.
**they’re kind of like a “genetic family tree”
Bb
bb
bb
Bb
Bb
bb
Bb
1
X
X
X
Y
XX XY
XX XY
3
Human Pedigree
A diagram showing the transmission of a trait through several generations of a
family is called a pedigree. In Figure 1, generation 1 is made up of
grandparents, generation II is their children, and generation III is their
grandchildren.
1. Study the pedigree
diagram and the key
in Figure 1 to learn the
symbols.
3
1. How many generations are shown?_____
5 women?_____
9
2. How many men are shown?_____
3. How many women have cystic fibrosis?_____
2
1
4. How many men have cystic fibrosis?_____
5. How many marriages are shown?_____
4
5
6. How many single women are in the family?_____
4
7. How many children are there?_____
5
8. How many grandchildren are there?_____
9. Do you think the gene for cystic fibrosis is dominant or
recessive?_____________________________________________________
Recessive b/c it remains “hidden” in several people
Use A to represent the allele for the ability to taste PTC, a
dominant allele. Use aa for the PTC nontaster, who
exhibits the recessive trait.
Aa
aa
Aa
aa
aa
aa
aa
aa
Aa
aa
aa
aa
aa
aa
Aa
Aa