Download Chapter 11: Genetics

Chapter 11:
Genetics
When told to, put the PTC paper on your tongue
PTC, or phenylthiourea, is an organic
compound having the unusual property of
either tasting very bitter, or being virtually
tasteless, depending on the genetic makeup
of the taster.
• The ability to taste PTC is a dominant genetic
trait.
T = Taster t = non-taster
– If you can taste, you are either TT or Tt.
– Those who can not taste are tt
• About 70% of people can taste PTC, varying
from a low of 58% for Aborginal people of
Australia to 98% for Native Americans
• Ever wonder why people resemble their
parents &/or siblings?
• How do farmers select the best plants
or animals for breeding purposes?
• How’d you like that PTC paper?
• Why did some people in your
class/family taste it and others didn’t?
Genetics
• Studies the
transmission of
traits or
characteristics from
1 generation to the
next
Gregor Mendel
(1822 – 1884)
The Father of Genetics
• Central European monk (Now the
Czech Republic) discovered the basic
underlying principles of heredity.
• Work completed in 1865
• His work did not get recognized though
until 1900 (he was dead by then).
• He used pea plants and selectively
crossed them over many
generations
• Saw that certain traits show up in
offspring plants without any blending
of parent characteristics.
X
Why all purple?
Mendel's work
1.
Removed the stamen/anther (male parts) to prevent selfpollination
2. Used the stamen from a tall plant and pollinated only
flowers from other tall plants
3. Observed generations for 2 years to be certain of
purebreds
Tall X
Tall
Short X Short
Yielded all Talls
Yielded all shorts
•
Crossed plants with contrasting traits
Tall
X
Short
Short X Tall
• P = Parental generation
• F1 = 1st Filial generation (offspring)
• F2 = 2nd Filial generation (next set of
offspring off the F1)
• P1: Pure Tall
X
Pure Short
T = Tall stems t = short stems
TT
X
tt
• F1:
Tall (Tt)
Only one trait showed
• F1 cross:
F1 Tall
X
F1 Tall
Tt
Tt
• F2:
3 Tall
1 Short
TT Tt Tt
tt
• The “lost” trait reappears!
• To get these results for the
F2, do
FOIL
– (firsts, outers, inners, lasts)
• *******Always got 75% Tall
and 25% short or 3:1
whenever this type of cross
occurred.
Mendel’s 4 conclusions (hypotheses)
from his experiments
1. Concept of Unit Characters: The
inheritance of each trait is determined
by “units” or “factors” that are passed
on.
• We now know these “units” as GENES
• These “units” occur in pairs. One from
each parent
2. Law of Dominance: –
One factor “masks” the appearance of
another factor. It prevents it from
showing
• Dominant – trait that must show if
present (TT or Tt)
• Recessive – trait that will only show if
in the pure form (tt)
• Hybrid – Contains a dominant & a
recessive trait (Tt)
• Allele – alternate genes
for a given trait
(Tall allele or short allele)
3. Law of Segregation
• For any trait, pairs of alleles are separated
in forming the gametes (during Anaphase I
of meiosis).
• Only 1 gene (allele) from the pair goes
into the gamete during meiosis.
More genetic vocabulary to learn
• Genotype: Shows actual genetic
makeup. (Use symbols for genes)
– TT or Pure dominant or homozygous
dominant
– Tt or Hybrid or heterozygous
– tt or Pure recessive or homozygous
recessive
Homozygous vs Heterozygous: Pure vs
Hybrid (mixed)
• Phenotype:
– Tells appearance (describes the trait)
Tall (if TT or Tt)
Short (if tt)
– NO HYBRIDS HERE!!
– Describe what they look like
– For Eye color:
– Blue
– Green
– Brown
– For Hair color; Red, Blond or Brown
Widows peak
Dimples
Mid – digit hair
Free earlobe
Hitchhikers thumb
Chin Dimple
2nd toe larger than
1st
Straight pinkie finger
Punnett Squares
• Shows possible gene
pairing & probability of
each pairing.
• Checkerboard method
• Foiling
• During meiosis, when the
gametes are formed, the
genes of the father and
mother separate from each
other. This is the law of
segregation. genes of 1 parent
T
T genes of 1 parent
t
Tt
Tt
t
Tt
Tt
Phenotype ratio : 100% tall
Genotype ratio : 100% Tt (hybrid)
Possible crosses
• Do the crosses (Punnett Sqs) & determine
the :
•
Phenotype and Genotype ratios
•
Trait: T = Tall stem length
•
t = short stem length
•
1. TT x TT
•
2. TT x Tt
•
3. TT x tt
•
4. Tt x Tt
•
5. Tt x tt
•
6. tt x tt
TT X TT
TT X Tt
TT X tt
Tt x Tt
Tt x tt
tt x tt
1. In humans, having a chin dimple is dominant to not having dimples.
a.Show the cross of two parents who are both hybrid for having dimples.
b.Give the expected Phenotype ratio and genotype ratio for this cross.
____ = Dimple gene
____ = Non dimple gene
Parent cross ______ x ______
Phenotype ratio :
Genotype ratio :
2. For the same trait, cross a hybrid with an individual who does not have dimples
Parent cross ______ x ______
Phenotype ratio :
Genotype ratio :
Probability
• Likelihood of an event occurring
• Shown by Punnett Square
• Shows how often a gene pairing may occur.
**Need large numbers to get accurate
predictions**
Coin Toss Lab:
Flip one coin 10x. Keep track of the # of
Heads and Tails you get.
50:50 chance of getting heads or tail if you toss
one coin
What happens when you toss two coins?
• If you toss 2 coins 100 times, you should get:
•
25 Heads/Heads
•
50 Heads/Tails
•
25 Tails/Tails
• As the # of trials increases, the ratios
predicted by the laws of probability get
closer
• Actual outcomes get closer to calculated
predictions
A/A
1
2
5/6
3
4
5
6
7
8
9
10
Total/1000
Genotype Ratio
Phenotype Ratio
A/a
a/a
A
Offspring
Phenotype
Offspring
Genotype
Yellow
AA
Aa
Green
aa
Total Total Total
10 Tosses
100 Tosses
1000
Tosses
Total
Number
of
Yellow
seeds
(AA +
Aa)
B
Expected
Genotypic
Ratio
C
Expected
Phenotypic
Ratio
D
Experimental
Genotype
Ratio
E
Experimental
Phenotype
Ratio
A/A
1
2
7/8
3
4
5
6
7
8
9
10
Total/1000
Genotype Ratio
Phenotype Ratio
A/a
a/a
A
Offspring
Phenotype
Offspring
Genotype
Yellow
AA
Aa
Green
aa
Total Total Total
10 Tosses
100 Tosses
1000
Tosses
Total
Number
of
Yellow
seeds
(AA +
Aa)
B
Expected
Genotypic
Ratio
C
Expected
Phenotypic
Ratio
D
Experimental
Genotype
Ratio
E
Experimental
Phenotype
Ratio
A/A
1
2
9
3
4
5
6
7
8
9
10
Total/1000
Genotype Ratio
Phenotype Ratio
A/a
a/a
A
Offspring
Phenotype
Offspring
Genotype
Yellow
AA
Aa
Green
aa
Total Total Total
10 Tosses
100 Tosses
1000
Tosses
Total
Number
of
Yellow
seeds
(AA +
Aa)
B
Expected
Genotypic
Ratio
C
Expected
Phenotypic
Ratio
D
Experimental
Genotype
Ratio
E
Experimental
Phenotype
Ratio
A
Offspring
Phenotype
Offspring
Genotype
Yellow
AA
Aa
Green
aa
Total Total Total
10 Tosses
100 Tosses
1000
Tosses
Total
Number
of
Yellow
seeds
(AA +
Aa)
B
Expected
Genotypic
Ratio
C
Expected
Phenotypic
Ratio
D
Experimental
Genotype
Ratio
E
Experimental
Phenotype
Ratio
A/A
1
2
7/8
3
4
5
6
7
8
9
10
Total/1000
Genotype Ratio
Phenotype Ratio
A/a
a/a
So you are a right handed person. How do you know if
you are pure or a hybrid for the right handed trait?
Could you ever have a lefty child?
• Test Cross:
•
Method used to find out if something
(or someone) is pure dominant or a
hybrid for a given trait.
•
Cross them with a pure recessive for
that trait. If the results come up with a
recessive individual, then the parent
was a hybrid
Is your black guinea pig pure for its coat
color?
In Guinea pigs: B = Black coat
b = white coat
Cross the (BB) guinea pig with a pure recessive (bb)
Cross the (Bb) guinea pig with a pure recessive (bb)
If: BB x bb
If: Bb x
bb
If any of the offspring are white coated, then we
knew the black guinea pig was BB
Monohybrid crosses:
• Cross two hybrids. Bb x Bb
Tt x Tt
Rr x Rr
• Always get a 3:1 phenotypic ratio!!!
Genetic Corn lab
• Cover page must be present with a COLOR
graphic and the names of all the people in the
group along with the date.
• Formal Lab write-up so typed & in a specific
order
– Purpose: What are we trying to determine from our
corn plants?
• The purpose of this experiment with genetic corn seeds is to
determine the Phenotype and Genotype of the parent corn
plants from observations of the offspring
– Procedure: Explain clearly what we did to set up the
lab and how we take care of the plants and collect our
data
– Data Tables: 2 tables, one for your group only and the
other one for the class data
– Calculations/Analysis: Determine the F1 Phenotype
ratios for your group and the entire class or 2 classes
together. Show all your work and calculations.
– Questions: You will answer the 10 questions
thoroughly and in complete sentences with the
question incorporated into the answer
Calculations for Genetic Corn Lab
Calculations:
Determine the F1 Phenotype ratios of
your data and the class data. To
determine this, divide the # of each color
plant by the # of albino plants
If you obtained 7 green and 5 white:
7:5 or 1.4:1
Analysis Questions
Questions: You are to answer these questions thoroughly and in
complete sentences
1. Could you tell before the seeds germinated which would have
the green pigment and which would be albino? Explain
2. What are the three requirements for seeds to germinate?
3. Which plants after 1 week were larger?
4. Why do you think the above was so?
5. Why do the seeds with albino traits die before the green
plants?
6. Use the results shown in your data tables and your
calculations to determine the genotypes and phenotypes of
the parents. These are not ratios!!
7. Show the genetic cross of the parents using a Punnett Square
with a key to your two alleles for pigmentation
8. Using the Punnett Square from question #7, determine the F1
genotypic ratio and phenotypic ratio and percentages.
9. If we wanted more accurate results, what could be done to
bring the results closer to the expected phenotype ratios?
10. Why couldn’t one of the parents be pure recessive for chlorophyll
production?
Table #2 – Class
Results
Group #
# of Green Plants
# of White Plants
# of Plants
counted
A
21
9
30
B
9
3
12
C
9
2
11
D
9
1
10
E
11
1
12
F
8
4
12
G
7
5
12
H
8
2
10
I
8
3
11
90
30
120
J
Other Class Data
TOTALS
% Green Plants
% White Plants
Table #2 – Class
Results
Group #
A
B
C
D
E
F
G
H
I
J
Other Class Data
TOTALS
# of Green Plants
# of White Plants
# of Plants
counted
% Green Plants
% White Plants
What are your chances of having a
blue eyed, blond haired child if
one of your parents is brown
haired and eyed and the other is a
blond with blue eyes?
4. Law of Independent Assortment
• Different pairs of alleles are passed to
offspring independent of each other.
B = Brown
b = White
S = Short tail
s = Long tail
Dihybrid crosses
• Crossing two hybrids, that are hybrid for two
different traits
• T = Tall stem
Y = Yellow seed
• t = short stem
y = green seed
•
TtYy x
TtYy
• Dihybrid cross
• Phenotype ratio for a dihybrid is always:
– 9:3:3:1 = (3:1)(3:1)
9 – Tall Yellow
3 – Tall green
3 – Short yellow
1 – Short green
• Trihybrid would be: 27:9:9:3:9:3:3:1 (3:1)(3:1)(3:1)
F1: TtYy
x
TtYy
Foil each parent to get gamete
combinations
TY
Ty
tY
ty
TY
Ty
____ = Tall Yellow
____ = Tall green
tY
____ = short Yellow
____ = short green
ty
1. A normal toed, hybrid, dark haired person mates with a hybrid
for webbed toes and has light
hair. What could be the phenotypes of their children?
Parent Cross _________x _______
_____ = Webbed toes
_____ = Normal toes
_____ = Dark hair
_____ = Light hair
2. Cross an individual who is heterozygous for both curly hair and
freckles with a
homozygous curly haired, non-freckled person. What will be
the possible phenotypes of
this cross.
Parent Cross _________x _______
_____ = Curly hair
_____ = Straight hair
_____ = Freckles
_____ = No freckles
In fruit flies, red eyes are dominant to sepia eyes and
having normal wings is dominant to vestigial wings.
Cross a dihybrids for eye and wings
____ = Red eyes
___ = Normal wings
____ = Sepia eyes
___ = Vestigial wings
4. Law of Independent Assortment
• Different pairs of alleles are passed to
offspring independent of each other.
B = Brown
b = White
S = Short tail
s = Long tail
F1 Cross:
Cross 2 plants from the F1 generation
RrYy x RrYy
Now FOIL RrYy to figure out the possible
gametes for each parent plant
RrYy
RY
Ry
rY
ry
RY
RY
RY
RY
ry
RrYy
RrYy
RrYy
RrYy
ry
RrYy
RrYy
RrYy
RrYy
ry
RrYy
RrYy
RrYy
RrYy
ry
RrYy
RrYy
RrYy
RrYy
Phenotype ratio = 100% Round/Yellow
Genotype ratio = 100% RrYy
RY
Ry
rY
ry
RY
RRYY RRYy
RrYY
RrYy
Ry
RRYy
RRyy
RrYy
Rryy
rY
RrYY
RrYy
rrYY
rrYy
ry
RrYy
Rryy
rrYy
rryy
Phenotype ratio:
Round/Yellow
Round/Green
Wrinkled/ Yellow
Wrinkled/Green
=
=
=
=
9/16
3/16
3/16
1/16
P = Purple stem
P = Green stem
G = Green leaves
g = yellow leaves
1. A normal toed, hybrid dark haired person mates with a hybrid
for webbed toes and has light
hair. What could be the phenotypes of their children?
Parent Cross _________x _______
_____ = Webbed toes
_____ = Normal toes
_____ = Dark hair
_____ = Light hair
2. Cross an individual who is heterozygous for both curly hair and
freckles with a
homozygous curly haired, non-freckled person. What will be
the possible phenotypes of
this cross.
Parent Cross _________x _______
_____ = Curly hair
_____ = Straight hair
_____ = Freckles
_____ = No freckles
In rabbits, grey hair is dominant to white hair and black eyes are dominant to red eyes.
_____ Grey hair
_____ Black eyes
_____ White hair
_____ Red eyes
1.
What are the phenotypes of rabbits that have the following genotypes?
a. ggBb __________________________________________________________________
b. Ggbb __________________________________________________________________
1.
A male rabbit is a hybrid for grey hair & is pure for black eyes mates with a female rabbit that is a hybrid for both
grey hair and black eyes. Fill in the Punnett square below and then answer the following questions:
Parent cross: ______________ x _________________
a.
b.
c.
d.
What is the likelihood of one of the offspring having grey hair
and black eyes? ___________
What is the likelihood of an offspring being grey hair and red
eyes?
___________
How about the likelihood of one having white hair and black
eyes?
__________
Now what about white hair and red eyes?
___________
2. A male rabbit is a homozygous for grey hair & is hybrid for black eyes mates with a female
rabbit that is a hybrid for both grey hair and black eyes. Fill in the Punnett square below
and then answer the following questions:
Parent Cross: _______________ x ____________
a.
b.
c.
d.
What is the likelihood of one of the offspring having grey hair and black eyes? ___________
What is the likelihood of an offspring being grey hair and red eyes?
__________
How about the likelihood of one having white hair and black eyes
__________
Now what about white hair and red eyes?
___________
Dihybrid cross lab
• In corn:
R = Rough Seed shape r = smooth seeds
Y = Yellow seed color y = white seed color
If P:
RRYY x rryy
RY
ry
F1:
RrYy
We will do the F1 cross of RrYy x RrYy
Foil RrYy to get the possible gamete
combinations from each parent and then do
theDataPunnett
Square
Table 1
Female gametes
RY
RY
Ry
Male gametes
rY
ry
Ry
rY
ry
From the F1 cross Punnett Square, fill in Data Table 2
Data Table 2
Genotype
RRYY
RrYY
RrYy
RRYy
RRyy
Rryy
rrYY
rrYy
rryy
Genotype ratio
(X:16)
X:16
Phenotype
Phenotype
ratio
On Data Table 3, using Data Table 2, fill in the Expected
ratios for 16 offspring and then multiply those by 3 to
get the Expected ratio out of 48 offspring
Now flip 4 labeled coins, 48 times and record
on Data table 3.
Data Table 3
Phenotypes
Genotypes
Rough, yellow
seeds
RRYY
RrYy
RRYy
RrYY
Rough, white
seeds
RRyy
Rryy
Smooth,
yellow seeds
rrYY
rrYy
Smooth, white
seeds
rryy
Number of
Expected for
16 Offspring
Number
Expected for
48 Offspring
Toss Results
Total
Number
Observed
Coins
F1:
RrYy
R
x
Y
RrYy
R
Y
r
R
y
X
r
y
Gene Expressions in
Humans:
Chapter 11
Incomplete Dominance
• Blending
• Cross Red carnations with White
carnations.
• You would expect to get Red carnations.
• WRONG!!!
• All will be Pink carnations.
How did this happen?
P1:
Red carnations (rr)
•
F1:
F1:
Pink carnation (rw)
x
White carnations (ww)
Pink carnations (rw)
F2: 25% Red (rr)
x
Pink carnation (rw)
50% Pink (rw)
25% White (ww)
1:2:1 phenotype ratio!!
1:2:1 genotype ratio!!
rr
rw
ww
Incomplete Dominance
Hair color
Eye color
Height
Face shape
Wavy hair
Pitch of male’s voice (tt – tenor, tb baritones, bb – low bass)
Tay-Sachs Disease •
Inability to produce the enzyme that breaks down
•
•
•
lipids
Causes fluid pressure on brain then breakdown of
brain. Starts a ~ 6 months w/ death by 2 – 3 years.
Most common among the descendents of Eastern
European Jews (Ashkenazi Jews).
Tt individuals produce 40-60% of enzyme
Crosses:
1. Cross a Round
faced person with
an square faced
person
2. Cross a Black with
a white horse
3. Cross a Wavy
haired person with a
curly haired person
Codominance
•
•
•
•
Condition in which both alleles of a gene are
expressed or active
Roan cattle or horses: Red is codominant with
white.
Results in a Cattle or horse with some red and
some white hair. (they look pink from a distance)
Also a blue roan (black & white hairs)
Andalusian chickens – have both
black and white feathers.
Roans
Blood types in humans
A is codominant with B resulting in AB blood
Animation
Sickle Cell Anemia
• Sickle Cell anemia gene is codominant to the normal
red blood cell gene. Caused by a point mutation
• Normal red blood cells are round like doughnuts,
move easily through small blood vessels to deliver
oxygen.
• Sickle red blood cells become hard, sticky and sickle
shaped. When they go through small blood vessels,
they clog the flow and break apart. This can cause
pain, damage and a low blood count, or anemia.
Complications from the sickle cells
blocking blood flow and early breaking
apart include:
•
•
•
•
•
•
•
•
•
•
•
•
•
pain episodes
strokes
increased infections
leg ulcers
bone damage
yellow eyes or jaundice
early gallstones
lung blockage
kidney damage & dehyration
blood blockage in the spleen or liver
eye damage
low red blood cell counts (anemia)
delayed growth
Inheritance of Sickle cell anemia
•
•
•
•
•
•
Normal hemoglobin allele - (N)
Sickle cell hemoglobin allele – (S)
Normal RBC’s (NN)
Carrier for Sickle cell anemia (NS)
Sickle Cell Anemia (SS)
Normal RBC and Sickle RBC genes are
codominant. IF the gene is present, it
will show
Where is it found?
• Basically found in people of African decent.
• 19% of Americans of African ancestry.
• 40% of population of Africa carry the trait.
• Carriers of SC anemia (NS) have a partial
resistance to MALARIA
Malaria
Sickle Cell Anemia
Distribution of
Malaria
Distribution of
SCA
Problems:
1. A woman who is a
carrier for SCA has
children with a man who
does not have an allele
for SCA. What could their
children be?
2. What are the chances of
a carrier having a child
without the SCA allele
with a person who has
SCA?
Multiple Alleles
• For some traits, there
are more than 2 alleles
for a given trait.
– Ie. Blood types There
is the A, B and o alleles
– Hair color: Red, blond
and brown
– Eye color: Blue, green
and brown
Blood type
alleles
For simplicity,
we call these
IA
A
IB
B
i
o
Blood type
A
B
AB
O
Possible genotypes
AA
Ao
BB
Bo
AB
oo
• With the two blood antigens (proteins located on
the RBC):
A=B>o
• A type blood contains the A antigen (blood
protein) on its RBC’s.
Produces anti-B antibodies
• B type blood has the B antigen on its RBC’s.
Produces anti – A antibodies
• AB blood has both A & B antigens
• Produces no antibodies against the 2 types of
blood proteins
• O blood contains no blood antigens
Produces both anti –A & anti – B antibodies
• When a person with type A is given type B
blood in a transfusion, the person’s blood
starts producing anti – B antibodies,
causing the blood cells to agglutinate
(clump up).
• This is a type of transplant rejection.
Animation
What mixes with what?
Blood type
Can donate to
•
A
A, AB
•
B
B, AB
•
AB*
AB
•
O**
A, B, AB, O
• ** Universal donor
Can receive from
A, O
B, O
A, B, AB, O
O
* Universal receiver
Problems
1.
Figure out what blood type
two parents must be in
order to have 4 children,
each with a different blood
type.
2. Who could be the father? A
woman with O type blood
has a child with B type
blood
Man #1 is A type blood
Man #2 is AB blood
3. What could be your blood
type?
What is with the + and – with blood types?
• Rh factor – Additional blood antigens –
named after the Rhesus monkey
• Rh+ - has the Rh antigens
• Rh- doesn’t have the Rh antigens
Produces anti – Rh+ antibodies
• Rh + is dominant over the Rh- allele.
Mother
RhRh-
Father
Rh+
Rh-
Child
Rh+
Rh-
• Problem may arise with a woman who is Rhhaving a child with a man who is Rh+.
– Since Rh+ is dominant, the child she is carrying most
likely is Rh+.
• At birth, there is a mixing of maternal and fetal
blood (at the placenta). Immediately, the
mother’s blood starts to build up anti – Rh+
antibodies.
• Since #1 child is born and away from the mother,
there is no problem for this child. With each
additional birth though, the mother’s blood has
antibodies to the child’s blood. Fetal blood
agglutinates, possibly causing still births.
• Problem solved by giving the mother a Rogam
shot. Actually, this is an anti- anti – Rh+ shot.
Makes it like she never had a child.
Polygenic Inheritance
• Some human traits are controlled by
more than one set of genes that
determine the expression of the traits.
• Ie. Eye color (controlled by 3 genes), hair
color, height, body weight, skin color,
etc…
Light blue
0 dominate alleles
Blue
1 dominate allele
Blue-green
2 dominate alleles
hazel
3 dominate alleles
Light brown
4 dominate alleles
Brown
5 dominate alleles
Dark brown / black
6 dominate alleles