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Chpt. 14 Mendelian Genetics
Chpt. 14 Mendelian Genetics
How are traits
passed from
parent to
offspring?
Chpt. 14 Mendelian Genetics
WE … know
genes on
it is
chromosom
es…
Chpt. 14 Mendelian Genetics
MENDEL
had no
idea what
a
chromosome
even
was!!!!!
Chpt. 14 Mendelian Genetics
1843 entered
an
Augustinian
monaster
y.
Chpt. 14 Mendelian Genetics
1846 assigned to a
High School
as a teacher…
didn’t pass the
teacher test.
BACK TO THE
Chpt. 14 Mendelian Genetics
1851 - 53
Tried to
“find
himself”,
sent to Univ.
of Vienna
Chpt. 14 Mendelian Genetics
1851 1853 two
profs.
influence
d him and
he began
research.
Statistics Professor and a Botany Professor
Chpt. 14 Mendelian Genetics
1851 - 53 His
research was
“controversi
al”he
FLUNKED
OUT!
Mendel became an emotional wreck!!
Chpt. 14 Mendelian Genetics
1854 BACK to the
monastery,
HOWEVER,
this time…
Chpt. 14 Mendelian Genetics
1854 - the
Monsignor
allowed him
to research
inheritance
…
Mendel’s
Methods:
Worked with
Pea Plants
observed
several
characters
Either - Or
traits
Mendel’s
Methods:
Cross-pollinated
two
contrasting
true breeding
pea varieties
pg. 567 anatomy
P
True breeding
F1 4:0
F2 3:1
P
True breeding
hybridization
Mating two true-breeding varieties
F1 4:0
F2 3:1
White
did not
disappear
in F1!
Purple is
just
DOMINANT
over
white
We now know:
Alternative versions of a gene
Results - Summary
• In all crosses, the F1
generation shows only one of
the traits regardless of which
was male or female (4:0
ratio).
• The other trait reappeared in
the F2 at ~25% (3:1 ratio).
Mendel’s Laws =
•Alternate versions of genes
(alleles) account for
variations.
•Organisms inherit two alleles
(one from each parent)- for each
character.
•When alleles differ, the
dominant is fully expressed.
Mendel’s Laws =
•Each pair of alleles separate
during gamete formation
by chance- 50% probability that each
version will be the oneGerm cell
involved in the
Aa
fertilization.
after meiosis II
A
A
a
a
P
Law of Segregation
P
p
p
-->
VO
CA
BU
LA
RY
-->
Testcross=
Cross of a
phenotypicall
y dominant
with a
homozygote
recessive
Testcross=
ex:
T_ x tt
Cross of a
suspected
If TT: all offspring will
heterozygot be dominant
e with a
homozygote If Tt: offspring will be
recessive
1 Dominant :
1 recessive
Sorry,but College Board
who runs AP, says you
must memorize this… so
does Hobby, so does your
Freshman collegiate Bio
instructor!!
6 Mendelian
Crosses are
Possible
(these are your expected):
Cross
Genotype
TT X tt
all Tt
Tt X Tt
1TT:2Tt:1tt
TT X TT all TT
tt X tt
all tt
TT X Tt
1TT:1Tt
Tt X tt
1Tt:1tt
Phenotype
all Dominant
3 Dom: 1 Res
all Dominant
all Recessive
all Dominant
1 Dom: 1 Res
Lets try it with one trait:
COLOR
What if you are tracking
more than ONE trait???
Do the traits “travel”
together when gametes
are formed?
In other words, does
seed color & seed
shape always stay
together?
Does the “R” allele
always stay with the “Y”
allele?
Law of Independent Assortment
Mendel’s
Law of Independent
Assortment
•Each pair/types of alleles
segregates into gametes
independently.
•Color and seed shape are
not a “package deal”!`
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, tR, tr = possible gametes
How big of a Punnett Square?
Number of Gametes possible
Critical to calculating the
results of higher level crosses.
Look for the number of
heterozygous traits.
Equation to
determine number of gametes possible:
The formula 2n can be used,
where “n” equals the number
of heterozygous traits possible in
a cross.
ex: TtRr, n=2
22 or 4 different kinds of
gametes are possible.
TR, tR, Tr, tr
Dihybrid Cross
FOIL
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 of TtRr X TtRr
9
Tall, Red flowered
3 Tall, white flowered
3 short, Red flowered
1 short, white flowered
Or: 9:3:3:1
Comment
Ratio
of Tall to short is 3:1
(b/c Tt X Tt)
Ratio of Red to white is 3:1
(b/c Rr X Rr)
The
cross is really a product of
the ratio of each trait multiplied
together.
(3:1) x (3:1) FOIL
PROBABILITY
RULES OF INHERITANCE
Rule of Multiplicationused when determining
chances that
two independent events
w/ occur together
simultaneously.
Rule of Multiplicationtwo independent events w/
occur together
simultaneously.
two independent events
ex. height & flower color
Rule of Multiplication1.Determine the probability of
each independent event
(ex. allele in each sperm and allele
in each egg)
2. multiply the probabilities of
those independent events.
Rule of Multiplicationex.
P generation = Pp x Pp
“what is the probability of a
white offspring??”
(pp)
Can only get this one way…
 Rule of
Multiplication1.
generation
= Pp
Pp
2. Pchance
of Ovum
p = x1/2
chance of Sperm p = X 1/2
1/4 pp
P
p
P
p
Rule of Multiplicationused when determining
chances that
two independent events
w/ occur together
simultaneously.
Example:
TtRr X TtRr
The
probability of getting
a tall offspring is ¾ (Tt x Tt).
The
probability of getting
a red offspring is ¾ (Rr x Rr).
The
probability of getting
a TALL red offspring is
¾ x ¾ = 9/16
Rule of Multiplication1. YyRr is in a germ cell:
“What is the probability that
a gamete will be YR?”
2. 1/2 a chance for Y
x 1/2 a chance for R
1/4 YR
Rule of Multiplicationused when determining
chances that
two independent events
w/ occur together
simultaneously.
But, what about, for
example, heterozygote
parents? Yy x Yy
What is the probability that
these parents would produce a
heterozygote offspring?
Rule of AdditionUsed when determining the
probability of an event… an
event that can occur in 2 or
more different ways.
Rule of Addition1. Compute the probability for
each way in which each event
can happen.
2. Add the probabilities for
each of those ways.
Rule of AdditionP1 =
Yy x Yy
ex. “what is the probability of a
heterozygous offspring?”
Ovum Y = 1/2 Sperm y = 1/2
Ovum y = 1/2 Sperm Y = 1/2
Rule of AdditionOvum Y = 1/2 x Sperm y = 1/2
Ovum y = 1/2 x Sperm Y = 1/2
1/4 + 1/4 = 2/4 = 1/2
There is a 1/2 chance for a Yy
offspring
Rule of AdditionUsed when determining the
probability of an event… an
event that can occur in 2 or
more different ways.
Rules of
additio
n and
multiplicatio
n:
Variations on
Mendel
1.
2.
3.
4.
5.
Incomplete Dominance
Codominance
Multiple Alleles
Epistasis
Polygenic Inheritance
Law of
Incompl
ete
Dominanc
e
Law of Incomplete
Any of a set of three or
more alleles, only two
of which can be present
in a diploid organism.
Multiple alleles
clumped
clumped
clumped
clumped
clumped
clumped
clumped
Comment
• Rh blood factor is a separate
factor from the ABO blood
group.
• Rh+ = dominant
• Rh- = recessive
• A+ blood = dihybrid trait
Epistasis
• When 1 gene locus alters the
expression of a second locus.
ex:
• 1st gene: C = color, c =
albino
• 2nd gene: B = Brown, b =
“Labs”
In Gerbils too!
CcBb X CcBb
Black X Black
F1 = 9 black (C_B_)
3 brown (C_bb)
4 albino (cc__)
Result
• Ratios often altered from the
expected.
• One trait may act as a
recessive because it is
“hidden” by the second trait.
Problem
• Wife is type A
• Husband is type AB
• Child is type O
Question - Is this possible?
Comment - Wife’s boss is
type O
Bombay Effect
• Epistatic Gene on ABO
group.
• Alters the expected ABO
outcome.
• H = dominant, normal ABO
• h = recessive, no A,B,
reads as type O
Genotypes
• Wife: type A (IA IA , Hh)
• Husband: type AB (IAIB, Hh)
• Child: type O (IA IA , hh)
Therefore, the child is the offspring
of the wife and her husband (and
not the boss).
Bombay - Detection
• When ABO blood type inheritance patterns
are altered from expected.
Polygenic
Several
genes
affect one
character
ex. skin
color, height
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.
Pleiotropy
one gene, multiple
phenotypic effects
Pleiotropy
Cystic Fibrosis
one gene, multiple
phenotypic effects
Some DOMINANT genes, are
not often expressed in a
population
Some DOMINANT genes, are
not often expressed in a
population
Does the “R” allele
always stay with the “Y”
allele?
= normal
male
= normal
female
= affected
male
= affected
female
Reproductive
partners
siblings
Can determine:
1) Autosomal recessive
disorder
2) Autosomal dominant
disorder
Can determine:
3) Sex-Linked disorder
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
Carrier
Recognition
• Fetal Testing
–Amniocentesis
–Chorionic villi sampling
• Newborn Screening
Summary
• Know the Mendelian crosses and their
patterns.
• Be able to work simple genetic problems
(practice).
• Watch genetic vocabulary.
• Be able to read pedigree charts.
• Be able to recognize and work with some of
the “common” human trait examples.