Download Mendel and the Gene Idea

Mendel and the
Gene Idea
Inheritance
The
passing of traits from
parents to offspring.
Humans have known about
inheritance for thousands of
years.
Genetics
The
scientific study of the
inheritance.
Genetics is a relatively “new”
science (about 150 years).
Genetic Theories
1. Blending Theory traits were like paints and mixed
evenly from both parents.
2. Incubation Theory only one parent controlled the
traits of the children.
Ex: Spermists and Ovists
3. Particulate Model parents pass on traits as
discrete units that retain their
identities in the offspring.
Gregor Mendel
Father
of Modern Genetics.
Mendel’s
paper published in
1866, but was not recognized
by Science until the early
1900’s.
Reasons for
Mendel's Success
Used
an experimental
approach.
Applied mathematics to the
study of natural phenomena.
Kept good records.
Mendel
was a
pea picker.
He used peas
as his study
organism.
Why Use Peas?
Short
life span.
Bisexual.
Many traits known.
Cross- and self-pollinating.
(You can eat the failures).
Cross-pollination
Two
parents.
Results in hybrid offspring
where the offspring may be
different than the parents.
Self-pollination
One
flower as both parents.
Natural event in peas.
Results in pure-bred
offspring where the offspring
are identical to the parents.
Mendel's Work
Used
seven characters, each
with two expressions or
traits.
Example:
Character - height
Traits
- tall or short.
Monohybrid or
Mendelian Crosses
Crosses
that work with a
single character at a time.
Example - Tall X short
P Generation
The
Parental generation or
the first two individuals used
in a cross.
Example - Tall X short
Mendel used reciprocal
crosses, where the parents
alternated for the trait.
Offspring
F1
- first filial generation.
F2 - second filial generation,
bred by crossing two F1
plants together or allowing a
F1 to self-pollinate.
Another Sample Cross
P1
F1
F2
Tall X short (TT x tt)
all Tall (Tt)
3 tall to 1 short
(1 TT: 2 Tt: 1 tt)
Results - Summary
In
all crosses, the F1
generation showed only one
of the traits regardless of
which was male or female.
The other trait reappeared in
the F2 at ~25% (3:1 ratio).
Mendel's Hypothesis
1. Genes can have alternate
versions called alleles.
2. Each offspring inherits two
alleles, one from each parent.
Mendel's Hypothesis
3. If the two alleles differ, the
dominant allele is expressed.
The recessive allele remains
hidden unless the dominant
allele is absent.
Comment - do not use the terms
“strongest” to describe the
dominant allele.
Mendel's Hypothesis
4. The two alleles for each trait
separate during gamete
formation. This now called:
Mendel's Law of Segregation
Law of Segregation
Vocabulary
Phenotype
- the physical
appearance of the organism.
Genotype - the genetic
makeup of the organism,
usually shown in a code.
T
= tall
t = short
Helpful Vocabulary
Homozygous
- When the two
alleles are the same (TT/tt).
Heterozygous- When the two
alleles are different (Tt).
6 Mendelian Crosses
are Possible
Cross
Genotype
Phenotype
TT X tt
Tt X Tt
TT X TT
tt X tt
TT X Tt
Tt X tt
all Tt
1TT:2Tt:1tt
all TT
all tt
1TT:1Tt
1Tt:1tt
all Dom
3 Dom: 1 Res
all Dom
all Res
all Dom
1 Dom: 1 Res
Test Cross
Cross
of a suspected
heterozygote with a
homozygous recessive.
Ex: T_ X tt
If TT - all dominant
If Tt - 1 Dominant: 1 Recessive
Dihybrid Cross
Cross
with two genetic traits.
Need 4 letters to code for the
cross.
Ex:
TtRr
Each
Gamete - Must get 1
letter for each trait.
Ex.
TR, Tr, etc.
Number of Kinds
of Gametes
Critical
to calculating the
results of higher level
crosses.
Look for the number of
heterozygous traits.
Equation
The formula 2n can be used,
where “n” = the number of
heterozygous traits.
Ex: TtRr, n=2
22 or 4 different kinds of
gametes are possible.
TR, tR, Tr, tr
Dihybrid Cross
TtRr X TtRr
Each parent can produce 4
types of gametes.
TR, Tr, tR, tr
Cross is a 4 X 4 with 16
possible offspring.
Results
9
Tall, Red flowered
3 Tall, white flowered
3 short, Red flowered
1 short, white flowered
Or: 9:3:3:1
Law of Independent
Assortment
The
inheritance of 1st genetic
trait is NOT dependent on the
inheritance of the 2nd trait.
Inheritance of height is
independent of the
inheritance of flower color.
Comment
Ratio
of Tall to short is 3:1
Ratio of Red to white is 3:1
The cross is really a product
of the ratio of each trait
multiplied together.
(3:1) X (3:1)
Probability
Genetics
is a specific
application of the rules of
probability.
Probability - the chance that
an event will occur out of the
total number of possible
events.
Genetic Ratios
The
monohybrid “ratios” are
actually the “probabilities” of the
results of random fertilization.
Ex: 3:1
75% chance of the dominant
25% chance of the recessive
Rule of Multiplication
The
probability that two
alleles will come together at
fertilization, is equal to the
product of their separate
probabilities.
Example: TtRr X TtRr
The
probability of getting a
tall offspring is ¾.
The probability of getting a
red offspring is ¾.
The probability of getting a
tall red offspring is
¾ x ¾ = 9/16
Comment
Use
the Product Rule to
calculate the results of
complex crosses rather than
work out the Punnett Squares.
Ex: TtrrGG X TtRrgg
Solution
“T’s” = Tt X Tt = 3:1
“R’s” = rr X Rr = 1:1
“G’s” = GG x gg = 1:0
Product is:
(3:1) X (1:1) X (1:0 ) = 3:3:1:1
Tips for Dihybrid
Problems
Identify
all of the alleles that
can be identified from the
phenotypes of the parents or
kids.
Work from the monohybrid
ratios to solve for the missing
alleles.
Variations on Mendel
1.
2.
3.
4.
5.
Incomplete Dominance
Codominance
Multiple Alleles
Epistasis
Polygenic Inheritance
Incomplete Dominance
When
the F1 hybrids show a
phenotype somewhere between
the phenotypes of the two
parents.
Ex. Red X White snapdragons
F1 = all pink
F2 = 1 red: 2 pink: 1 white
Result
No
hidden Recessive.
3 phenotypes and
3 genotypes
(Hint! – often a “dose” effect)
= CR CR
Pink = CRCW
White = CWCW
Red
Another example
Another example

The color of fruit for plant "X" is determined by two
alleles. When two plants with orange fruits are crossed the
following phenotypic ratios are present in the offspring:
25% red fruit, 50% orange fruit, 25% yellow fruit. What are
the genotypes of the parent orange-fruited plants?
answer

Again, it comes in really handy if you can recognize right
off the bat that we have three phenotypes & just 2
alleles. That means we are dealing with either incomplete
or codominance. Since orange is a blend of red & yellow,
it's incomplete dominance. So the "in-between" phenotype
is the hybrid, orange in this example. We'll use RR = red,
YY = yellow, & our orange fruits are RY.
Another example

1. A cross between a blue blahblah bird & a white blahblah
bird produces offspring that are silver. The color of
blahblah birds is determined by just two alleles.
a) What are the genotypes of the parent blahblah birds in
the original cross?
b) What is/are the genotype(s) of the silver offspring?
c) What would be the phenotypic ratios of offspring
produced by two silver blahblah birds?
answer

a) Since there are only 2 alleles & three phenotypes (blue,
white, & silver), we must be dealing with incomplete
dominance. So the blue parent is homozygous blue (BB) &
the white parent is homozygous white (WW).

b) The silver offspring are hybrids (BW), one blue allele &
one white allele, neither one dominating the other. Instead,
we get a blending of blue & white, i.e. silver.


c) silver x silver = BW x BW The p-square would look like
what you see here.
As you can see, 25% (1/4) of the offspring are homozygous
white (WW), 25% (1/4) are homozygous blue (BB), & 50%
(2/4) are hybrid & therefor have the silver phenotype.
Codominance
Both
alleles are expressed
equally in the phenotype.
Ex. MN blood group
MM
MN
NN
Codominance




The genetic gist to codominance is pretty much the same
as incomplete dominance.
A hybrid organism shows a third phenotype --- not the
usual "dominant" one & not the "recessive" one ... but a
third, different phenotype.
With incomplete dominance we get a blending of the
dominant & recessive traits so that the third phenotype is
something in the middle (red x white = pink).
In COdominance, the "recessive" & "dominant" traits
appear together in the phenotype of hybrid organisms.
Result
No
hidden Recessive.
3 phenotypes and
3 genotypes
(but not a “dose” effect)
Example
Question
Predict
the phenotypic ratios
of offspring when a
homozygous white cow is
crossed with a roan bull.
Answer

Step #1 --- recognize that "roan" is a codominance
trait. Homozygous white = WW, & roan = RW (a hybrid
cow).
So our cross is WW x RW & the punnett square should look
something like what you see here.

The results:
2/4 offspring (50%) will be roan (RW), & 50% will be white
(WW).
Question
What
should the genotypes &
phenotypes for parent cattle
be if a farmer wanted only
cattle with red fur?
Answer


Well, the only way to have red fur is to be homozygous red
(RR). In order to get that genotype in all the offspring both
parents must be "RR". A parent with one or more "W"
alleles will cause the inheritence of roan fur in some
offspring.
Go ahead & work out all the punnett squares if you don't
believe me.
Only RR x RR gives you 100% RR.
RR x RW would produce 50% roan, 50% red,
RW x RW produces 25% red, 50% roan & 25% white,
WW x RW would produce 50% roan, 50% white,
& WW x RR would produce 100% roan (RW).
Question
A
cross between a black cat
& a tan cat produces a tabby
pattern (black & tan fur
together).
a) What pattern of inheritence
does this illustrate?
Question
What
percent of kittens
would have tan fur if a tabby
cat is crossed with a black
cat?
Answer


Tabby cats are the hybrids (because they have both colors)
& a black cat must be homozygous black.
So the cross for this problem is BB (black) x BT (tabby).
The p-square is at the right.
The results show that 50% of the offspring will be BB
(black) & 50% will be tabby (BT). So to answer the
question, 0% of the kittens will be tan.
Multiple Alleles
When
there are more than 2
alleles for a trait.
Ex. ABO blood group
IA - A
type antigen
IB - B type antigen
i - no antigen
Result
Multiple
genotypes and
phenotypes.
Very common event in many
traits.
Alleles and
Blood Types
Type
A
B
AB
O
Genotypes
IA IA or IAi
IB IB or IBi
IAIB
ii
Comment
Rh
blood factor is a separate
factor from the ABO blood
group.
Rh+ = dominant
Rh- = recessive
A+ blood = dihybrid trait
Problem
Wife
is type O
Husband is type AB
Child is type O
Question - Is this possible?
Comment - Wife’s boss is type O
Genotypes
type A (IA IA )
Husband: type AB (IAIB)
Child: type O (ii)
Therefore, the child is the
offspring of the wife and her
boss.
Wife:
Polygenic Inheritance
Factors
that are expressed as
continuous variation.
Lack clear boundaries
between the phenotype
classes.
Ex: skin color, height
Genetic Basis
Several
genes govern the
inheritance of the trait.
Ex: Skin color is likely
controlled by at least 4
genes. Each dominant gives
a darker skin.
Result
Mendelian
ratios fail.
Traits tend to "run" in
families.
Offspring often intermediate
between the parental types.
Trait shows a “bell-curve” or
continuous variation.
Genetic Studies in
Humans
Often
done by Pedigree
charts.
Why?
Can’t
do controlled breeding
studies in humans.
Small number of offspring.
Long life span.
Sex Linkage
X-
linked recessive
X-linked dominant
Y-linked
X-linked recessive
Haemophelia
Red-green
color blindness
White eyed male fruit flies
Affects males more than
females
Why??
X-linked dominant
Fragile
X syndrome
Affected males always have
affected daughters but
unaffected sons
Why??
Y linkage
Rare
to have mutated genes
on Y
Why??
Pedigree Chart
Symbols
Male
Female
Person with trait
Sample Pedigree
Dominant Trait
Recessive Trait
Human Recessive
Disorders
Several
thousand known:
Albinism
Sickle
Cell Anemia
Tay-Sachs Disease
Cystic Fibrosis
PKU
Galactosemia
Sickle-cell Disease
Most
common inherited disease
among African-Americans.
Single amino acid substitution
results in malformed
hemoglobin.
Reduced O2 carrying capacity.
Codominant inheritance.
Tay-Sachs
Eastern
European Jews.
Brain cells unable to metabolize
type of lipid, accumulation of
causes brain damage.
Death in infancy or early
childhood.
Cystic Fibrosis
Most
common lethal genetic
disease in the U.S.
Most frequent in Caucasian
populations (1/20 a carrier).
Produces defective chloride
channels in membranes.
Recessive Pattern
Usually
rare.
Skips generations.
Occurrence increases with
consaguineous matings.
Often an enzyme defect.
Affects males and females
equally.
Human Dominant
Disorders
Less
common then recessives.
Affects males and females
equally.
Ex:
Huntington’s
disease
Achondroplasia
Familial Hypercholesterolemia
Inheritance Pattern
Each
affected individual had
one affected parent.
Doesn’t skip generations.
Homozygous cases show
worse phenotype symptoms.
May have post-maturity onset
of symptoms.
Genetic Screening
Risk
assessment for an
individual inheriting a trait.
Uses probability to calculate
the risk.
General Formal
R=FXMXD
R = risk
F = probability that the female
carries the gene.
M = probability that the male
carries the gene.
D = Disease risk under best
conditions.
Example
Wife
has an albino parent.
Husband has no albinism in
his pedigree.
Risk for an albino child?
Risk Calculation
Wife
= probability is 1.0 that
she has the allele.
Husband = with no family
record, probability is near 0.
Disease = this is a recessive
trait, so risk is Aa X Aa = .25
R = 1 X 0 X .25
R = 0
Risk Calculation
Assume
husband is a carrier,
then the risk is:
R = 1 X 1 X .25
R = .25
There is a .25 chance that
every child will be albino.
Common Mistake
If
risk is .25, then as long as
we don’t have 4 kids, we
won’t get any with the trait.
Risk is .25 for each child.
It is not dependent on what
happens to other children.
Carrier Recognition
Fetal
Testing
Amniocentesis
Chorionic
Newborn
villi sampling
Screening
Fetal Testing
Biochemical
Tests
Chromosome Analysis
Multifactorial Diseases
Where
Genetic and
Environment Factors interact
to cause the Disease.
Ex. Heart Disease
Genetic
Diet
Exercise
Bacterial
Infection