|
|
|
|||
Chapter
2. Mendelian Inheritance
Odd
principles of inheritance all the way back to 400 B.C.
1761-1766
Joseph Kolreuter- started studying genetic crosses of tobacco plants.
Found
that offspring had an intermediate appearance.
Consistent
with the blending theory of inheritance.
The
seeds of traits can blend together.
Mendel.
Published
his findings in The proceedings of the Brunn Society of Natural
History.
Died
before the importance of his discoveries was realized (1884).
1900 Three
scientists rediscovered his work.
Mendels
Laws of Inheritance:
Mendel had
an interest in ornamental flowers.
Many
people had crossed plants with different flower colors to obtain new
flower colors.
When two
plants of the same species with different characteristics are mated (crossed) it is called hybridization.
Ex: a purple
flowered vs. white flowered plant.
Mendel was intrigued
by the regular patterns of flower color resulting from these crosses.
His foundation in
math and physics made him believe that this natural phenomena may be
explained mathematically.
He chose
the garden pea plant Pisum sativum.
Several
properties of the species are advantages for studying plant hybridization.
1) Many different physical
characteristics in the species.
Seeds,
pods, flowers and stems.
2) The ease of crossing
these plants was also an issue.
Reproduction
occurs during a pollination event.
Male
gamete is pollen grain and female is the egg cell. »
Fertilization
occurs after the the sperm nucleus travels down the stigma and fuses with
the egg cell.
The
pea plant has a flower structure that favors self-fertilization
which was important for some of Mendels experiments. »
The
Keel is a modified petal that covers the stamen and ovules.
Mendel
also needed to cross-fertilize plants which requires pollen from
one plant be place on the stigma from another. »
Mendel
pried open premature flowers and remove the stamen so the flower would not
self-fertilize. »
Mendel
would then apply mature pollen from another plant with a paintbrush to the
flowers stigma.
This
allowed him to do any crosses he wanted.
Mendel
Studied seven traits:
After testing 34
varieties of peas Mendel settled on 7 characters or traits
that bred true.
Bred true?
The
trait did not vary from generation to generation.
Ex:
yellow seeded pea plants offspring always produce yellow offspring that
produce yellow seeds. A variety that continues producing the same characteristics after several generations of self-fertilization is a
Analysis of clearly
distinguished characteristics.
»
Ex:
flower color purple and white or tall and dwarf height.
A
cross in which one trait is followed is called a monohybrid cross. »
This
produces single-trait hybrids.
Experiment:
The Monohybrid Cross:
Designed
his experiment to determine the quantitative
relationships from which the laws could be discovered.
Not to explain
hybrid formation.
Terms:
P generation-
Parental generation.
P cross-
The cross of the parental plants.
F1
generation (first
filial)- Offspring of the P cross
F2
generation- Offspring
of an F1 cross.
Hypothesis:
Careful quantitative analysis of plant hybrids to uncover mathematical
relationships that provide a basis for deducing natural law.
Testing
the Hypothesis:
Interpreting
The Data:
Results argued
against a blending mechanism of heredity.
The F1
generation showed the traits of only one parent not an intermediate
phenotype.
Recessive
describes the trait that is masked by the presence of a dominant trait.
Ex:
purple over white flowers and yellow over green seeds. »
Which
are dominant? Recessive?
Dominant
traits are visualized in the F1 generation.
F2
generation most offspring show the dominant trait with a small portion
showing the recessive trait.
No intermediate
traits.
Concluded that
genetic determinants are passed along in an unaltered way generation to
generation.
Traits
inherited as discrete units that remain unchanged as they are passed from
parent to offspring.
By applying a
quantitative approach Mendel observed a 3:1 ratio between dominant
and recessive traits.
This
allowed Mendel to conclude that the genetic determinants segregate from
each other during gamete formation.
3:1 ratio
is consistent with the segregation of genes: Mendels Law of
Segregation.
At this time noone
knew what the genetic material was or the mode of transmission during
gamete formation.
Wilhelm Johannsen
(1909) was the first to coin the term gene.
A
discrete unit of heredity responsible for influencing an organisms traits.
Many eukaryotes have
genetic material organized into pairs of chromosomes.
Therefore
there are two copies of every gene.
Allele
refers to different versions of the same gene.
This
is consistent with the idea that only one gene (allele) is
transmitted per parent.
This is Mendels
law of Segregation: two copies of a gene segregate from each other
during transmission to offspring. During gamete formation the copies segregate so that only one copy is found in each.
Conceptual level:
T =
the dominant allele (tall plants)
t =
the recessive allele (dwarf plants).
The
parents are TT (homozygous dominant) and tt (homozygous
recessive). »
Homozygous
means having identical copies of the gene. »
TT
and tt are the genotype of the individual or the genetic
composition. »
Outward
appearance of an organism is denoted as phenotype.
The
F1 generation is heterozygous having the genotype Tt. »
Carries
one copy of the tall allele (T) and one for the short allele (t). »
All
tall because T is the dominant allele.
The
prediction for the F2 generation is that there will be a 3:1
ratio tall to short. »
Since
the parents are heterozygotes they can transmit either a T or a t in the
gametes. »
Makes
the possible combinations TT, Tt and tt. »
tT
= Tt so you get a 3:1 phenotype ratio. »
Genotypic
ratio is 1:2:1: TT: Tt and Tt: tt.
The Punnett
square:
Allows you to
predict the outcome of crosses.
Proposed by Reginald
Punnett in the early 1900s.
First:
Write
the genotypes of the parents (Ex: Tt X Tt). »
Think
of it as Male Tt X Female Tt.
Second:
Write
the possible gametes that each parent can generate.
Ex:
Male T and t; Female T and t.
Third:
Create
an empty Punnett square.
The
# of columns equals the number of male gametes.
The
# of rows equals the number of female gametes.
Our
example has how many rows? columns?
Experiment:
The Dihybrid Cross.
The
monohybrid cross showed Mendel how variants of a single trait are
segregated.
Moreover
it showed us how alleles for a particular gene are segregated into
gametes.
Mendel
knew more complicated experiments might lead to more information about the
laws of inheritance. Round yellow X wrinkled green.
What could be the
inheritance patterns for two different traits?
Linked-
The two different traits are inherited as a single unit.
Independent-
Not linked, the traits are assorted separately into gametes.
Do
we know any information that may point to one over the other?
Use the symbols R=
round, r = wrinkled, Y = yellow and y = green.
Schematic
of the two possibilities:
Interpreting
the results:
P generation RRYY
and rryy.
F1
generation RrYy.
F2
generation supports the independent assortment hypothesis and
refutes the linkage model. In the F2 generation there are round green seeds and wrinkled yellow seeds, these are called non-parentals, because this combo did not occur in the parental generation.
This
supports the independent assortment model.
The
linkage model would have only produced RY and ry gametes leading to
offspring with RRYY, RrYy and rryy in a 1:2:1 ratio.
Mendel
saw a 9:3:3:1 phenotypic ratio.
Independent
assortment held up for all seven traits.
The
Law of Independent Assortment:
The law states that
two different genes will randomly assort their alleles during gamete
formation.
The heterozygous
RrYy F1 generation plants could produce four gametes RY, Ry, rY
and ry.
With
self-fertilization any two of these gametes can come together this allows
42 or 16 combinations.
The F2
numbers from the example were:
315
Round Yellow seeds.
101
Wrinkled Yellow seeds.
108
Round Green seeds.
32
Wrinkled Green seeds.
Divide
each number by 32 and you get
a phenotypic ratio of 9.8: 3.2: 3.4: 1.0.
This
is within the experimental error of Mendels predicted 9:3:3:1.
Is
there a situation you can think of were independent assortment wouldnt
hold up?
The
Punnett square can also be used to solve independent assortment problems.
Ex: If you want to
solve for height and seed color of two heterozygous plants, TtYy X TtYy.
Phenotypes
are? »
Tall
and Dwarf; Yellow and Green
The
four possible gametes are? »
TY »
Ty »
tY »
ty
Relationship
between gene expression and traits.
Genes
encode for proteins which are involved in the traits of the cell and/or
organism.
Most often
geneticists try and identify defective copies of genes to analyze the
affect on phenotype.
This gives the
geneticist important information about the function of this gene.
Many of Mendels
traits that he followed were the normal and a LOF allele. It is common for LOF alleles to be recessive.
Ex:
The gene affecting flower color (purple vs. white) we expect encodes for a
protein involved in pigment production. »
May
be an enzyme in pigment biosynthesis. »
The
recessive white allele may encode for a non-functional pigment enzyme. »
By
using the non-functional allele we may be able to better study the pathway
or even recover the gene product. »
Then
with the gene we can do structure function studies.
Probability
and Statistics:
Powerful
application of Mendels work is that the Laws of inheritance can be used
to predict the outcomes of genetic crosses.
Important for animal
and plant breeders as well as human couples with genetic disorders. EX: For the couple with a genetic disorder what is the probability that my child will have that genetic disorder.
This
section is how probability calculations can be used to answer genetic
problems.
We will use the 3
mathematical operations: Sum rule, product rule and the binomial
expansion formula.
Will
allow us to determine the probability that a certain cross between two
individuals will produce a particular offspring.
Statistical
methods like the chi square test can be used to evaluate how well a
genetic hypothesis fits the observed data from the cross.
Probability
is the likelihood that an event will occur.
Ex: If you flip a
coin the probability is 0.5 or 50% that it will land on heads.
Probability is
dependent on the number of outcomes.
For the example
there are two outcomes, that are both equally likely to occur.
This
allows us to predict the 50% chance of any flip of a coin resulting in
heads.
Probability = # of
times an event occurs/ total # of events.
Pheads
= 1 head / (1 heads + 1 tails) = ½ = 50%
In genetic terms we
are interested in the probability of a particular type of offspring.
Ex:
Cross of Tt plants the result is 3 tall : 1 dwarf. »
Ptall
= 3 tall/(3 tall + 1 dwarf) = Ύ = 0.75 = 75%. »
Pdwarf
= 1 dwarf/(3 tall + 1 dwarf) = Ύ = 0.25 = 25%. »
When
the probability of both outcomes are added together it = 100% or 1.
A probability
calculation allows us to predict that an event will occur.
The
accuracy of this prediction depends on the sample size.
It
is easy to imagine if we only flipped the coin 6 times that our prediction
of 50% heads may not be accurate. »
Why?
Because the chance of heads or tails is random. »
The
larger the sample size the smaller the random sampling error.
The sum
rule states the probability that one or two mutually exclusive events will
occur is equal to the sum of the individual probabilities of the events.
Ex: Cross
of mice with heterozygous alleles affecting ears and tails.
de is the recessive
allele for droopy ears (vs. De). ct is the recessive allele for crinkly tail (Ct). A heterozygous cross (Dede/Ctct X Dede/Ctct) would yield 9 normal ears normal tails, 3 normal ears crinkly tails, 3 droopy ears normal tails and 1 droopy ears crinkly tail.
We can ask the
question: What is the probability that an offspring will have normal ears
and a normal tail or droopy ears and crinkly tail from a Dede/Ctct X Dede
Ctct cross?
Probability
of normal ears and a normal tail = 9/(9+3+3+1) = 9/16.
Probability
of droopy ears and crinkly tail = 1/(9+3+3+1) = 1/16.
If
you want to know for both just add the probabilities together 9/16 + 1/16
= 10/16.
10/16
= 0.625 = 62.5% have normal ears and tail or droopy ears and crinkly tail.
The
product rule states that the probability that two or more independent
events will occur in a particular order is equal to the product of their
probabilities.
Ex: Congenital
analgesia (Ca) is a rare recessive genetic disorder in which those
affected can not feel extremes of pain. Question: What is the probability that the first three offspring a couple heterozygous (Pp) for the Ca allele have, will be homozygous recessive (pp) for the disorder?
What
is the probability of a homozygous recessive?
Multiply
the individual probabilities: »
1/4
X 1/4 X 1/4 = 1/64 = 0.016 = 1.6%
What about if you
wanted to find the probability of the first being normal, second having Ca
and third being normal?
3/4
is the probability of normal and 1/4 for affected.
3/4
X 1/4 X 3/4 = 9/64 = 0.14 = 14% probability.
The
binomial expansion allows us to determine the probability that certain
proportion of offspring will be produced with certain characteristics.
Ex: Cross of two
tall heterozygous pea plants (Tt).
Question: What is
the probability that 2 out of 5 plants will be dwarf?
Do
not care the order only care that 2 out of 5 are dwarf.
To
solve these problems the binomial expansion equation can be used.
Where: »
P =
»
n =
»
x =
»
p =
»
q =
Calculate
the probabilities p = 1/4 and q = 3/4.
Number
of events vs. total events x = 2 and n = 5.
Chi square
test is used to test the validity of a hypothesis.
Hypothesis
testing- we must
determine if data generated in a genetic cross is consistent with a
particular pattern of inheritance.
Ex:
A geneticist who is studying fruit fly genetics over two generations wants
to know if his F2 generation is consistent with Mendels laws
of segregation and independent assortment.
May
make a tentative hypothesis from the observed outcome.
Hypothesis
testing provides an objective, statistical method to evaluate whether the
data actually agree with the hypothesis.
Must
evaluate the goodness of fit between observed data and the data
predicted by the hypothesis.
If
the observed and predicted data are closely related then we say the data
is consistent with the observed outcome. »
Reasonable
to accept the hypothesis.
May
not be a good fit and will lead to rejection of the hypothesis. »
Hopefully
this will lead to an alternative hypothesis.
The chi square test
(χ2) is often used to evaluate population data in which
members of the population fall into different categories.
O = Observed data in
each category.
E = Expected data in
each category based on experimenters hypothesis.
Σ = Sum this
calculation for each category.
If
there were two categories it would look like:
We can use this to
determine if a genetic hypothesis is consistent with the observed outcome.
Ex: Determine if a
dihybrid cross is obeying the laws of Mendel.
Organism: Drosophila
melanogaster, two traits wing shape and body color.
Normal
and curved wing shape c+ and c respectively.
Normal
and ebony body color are e+ and e respectively. Why the
letters? The cross: straight wing gray body (c+c+ e+e+) X curved wing ebony body (cc ee).
The outcome:
F1
generation = All strait winged and gray bodied offspring (c+c e+e).
F2
generation = 193 straight and gray, 69 straight and ebony, 64 curved and
gray, 26 curved and ebony for a total of 352 offspring.
Step 1: Propose a
hypothesis that allows us to calculate the expected values based on
Mendels laws.
F1
generation suggests that straight is dominant over curved and gray is
dominant over ebony.
F2
generation appears to have a 9:3:3:1 ratio and in doing so be consistent
with independent assortment.
Step 2: calculate
the expected values of each of the four phenotypes. »
9/16
straight wings, gray bodies. »
3/16 »
3/16
crooked wings, gray bodies. »
1/16
Calculate
the expected numbers of each type of offspring when the total = 352. »
Multiply
each by 352. »
9/16
X 352 = 198 »
3/16
X 352 = 66 »
3/16
X 352 = 66 »
1/16
X 352 = 22 »
These
are the expected numbers.
Degrees of
freedom (DOF) is a
measure of the number of categories that are independent of each other .
Typically
n-1, where n is the total number of categories.
n = 4 straight wings and gray body, straight wings and ebony
body, curved wings gray bony and curved wings ebony body. »
Degrees
of freedom = 3 or 4-1.
Now have χ2 = 1.06 and DOF = 3
Look at the chi
square table.
Our value fits
between a probability of 0.9 and 0.5
in actuality it is approximately 0.8.
What
does this mean? »
Means
that values equal to ours are expected to occur 80% of the time based on
random chance alone. »
Means that deviations between observed and expected are likely
to occur due to random sampling error and we can tentatively accept
our hypothesis. »
To
reject the hypothesis it would have had to have a value greater than 7.82.
Must remember this
does not prove a hypothesis correct it evaluates whether the data and the
hypothesis have a good fit.
|
