Transcript (G YY )(G YY ) = (G YY )

Friday’s Class 3 Extra Credit Questions
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•
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3 Questions, Multiple Choice
10 Minutes
12 Extra Points Possible
Not Required to Participate; if you are
happy with your grade, come to class 10
minutes later at 9:15 on Friday.
Our Goal: To understand the “population consequences”
Of Mendel’s Laws.
Question 1: How do we describe a Mendelian population?
Genes in Populations With NO
Natural Selection
Gametes
(Sperm and eggs)
Mating
System
Generation Begins
Zygotes
(fertilized eggs)
{GYY, GYy, Gyy}
NO Natural
Selection
Generation Ends
Reproducing Adults
{GYY, GYy, Gyy}
Life cycle
Natural
Selection
Natural
Selection
Juveniles
Answer: We describe a diploid Mendelian Population in
two ways:
1. List all the different genetic kinds of individuals,
i.e., all the genotypes.
Calculate the genotype frequencies.
2. List all the different kinds of genes,
i.e., all the alleles at all the genes.
Calculate the allele frequencies.
Can ALWAYS take individual genotypes apart into alleles
150 YY
300 Y Alleles = 2 x 150
400 Y Alleles = 1 x 400
400 Yy
400 y Alleles = 1 x 400
90 yy
180 y Alleles = 2 x 90
PY = {(2)(#YY) + (1)(#Yy)}/{(2)(Total # Genotypes)}
PY = {(2)(#YY)}/{2N} + {(1)(#Yy)}/{2N}
PY = (1){(#YY)}/{N} + (1/2){(#Yy)}/{N}
PY = GYY + (1/2)GYy
Can ALWAYS take individual genotypes apart into alleles
150 YY
300 Y Alleles = 2 x 150
400 Y Alleles = 1 x 400
400 Yy
400 y Alleles = 1 x 400
90 yy
180 y Alleles = 2 x 90
Py = {(2)(#yy) + (1)(#Yy)}/{(2)(Total # Genotypes)}
Py = {(2)(#yy)}/{2N} + {(1)(#Yy)}/{2N}
Py = (1){(#yy)}/{N} + (1/2){(#Yy)}/{N}
Py = Gyy + (1/2)GYy
Alleles
Unique
Genotypes
There is More than One Way to package alleles into
genotypes.
Knowing {PY, Py} CANNOT ALWAYS calculate the
One and only, unique genotype frequency distribution:
{GYY, GYy, Gyy}
The Two ways to genetically describe a Mendelian
Population are only partially interchangeable
Genotypes
Always Unpackage
Alleles
Alleles
Unique
Genotypes
Cannot
Always REpackage
Genes in Populations With NO
Natural Selection
Generation Ends
Reproducing Adults
Generation Begins
Diploid Zygotes
{GYY, GYy, Gyy}
Parents
{GYY, GYy, Gyy}
(fertilized eggs)
??
Offspring
How are these Genotype Frequency
Distributions related to one another?
Under What Circumstances are the Two ways to
genetically describe a Mendelian Population
interchangeable ???
Genotypes
Always Unpackage
Alleles
When CAN we
Repackage????
Alleles
Unique
Genotypes
THE ANSWER: is something you need to know
Random Mating is a mating system in which the
frequency of a Mating Type or Family Type
equals the PRODUCT of the
Genotype Frequencies of the Parents.
Examples:
Family Type
Family Type Frequency
Male x Female
Parents
YY x YY
(GYY)(GYY) = (GYY)2
YY x Yy
(GYY)(GYy)
yy x Yy
(Gyy)(GYy)
Female Parents in Population
YY
M
a
l
e
s
Gametes: 1 Y
YY
Yy
yy
Gametes: 1 Y
Gametes: ½ Y, ½ y
Gametes: 1 y
Frequency: GYY
Frequency: GYy
Frequency: Gyy
YY
½ YY, ½ Yy
Yy
(GYY)2
(GYY)(GYy)
(GYY)(Gyy)
Gametes: ½ Y, ½ y
½ YY, ½ Yy
½ Yy, ½ yy
Frequency: GYy
(GYY)(GYy)
¼ YY
½ Yy
¼ yy
Frequency: GYY
Yy
(GYy)(Gyy)
(GYy)2
yy
Gametes: 1 y
Yy
½ Yy, ½ yy
yy
Frequency: Gyy
(GYY)(Gyy)
(GYy)(Gyy)
(Gyy)2
Adding up YY Offspring
YY
M
a
l
e
s
Gametes: 1 Y
Female Parents in Population
YY
Yy
yy
Gametes: 1 Y
Gametes: ½ Y, ½ y
Gametes: 1 y
Frequency: GYY
Frequency: GYy
Frequency: Gyy
YY
½ YY, ½ Yy
Yy
(GYY)2
(GYY)(GYy)
(GYY)(Gyy)
Gametes: ½ Y, ½ y
½ YY, ½ Yy
½ Yy, ½ yy
Frequency: GYy
(GYY)(GYy)
¼ YY
½ Yy
¼ yy
Frequency: GYY
Yy
(GYy)(Gyy)
(GYy)2
yy
Gametes: 1 y
Yy
½ Yy, ½ yy
yy
Frequency: Gyy
(GYY)(Gyy)
(GYy)(Gyy)
(Gyy)2
Parental Genotype Frequencies: GYY , GYy, Gyy
Parental Allele Frequencies: pY = GYY + (½)Gyy
Offspring Genotype Frequencies: GYY , GYy, Gyy
GYY = (1)(GYY)2 + (½)(GYY)(GYy)
+ (½)(GYY)(GYy) + (¼) (GYy)2
= (GYY +[½][GYy])2 = (pY)2
Note: Genotype frequency in offspring, GYY, equals
square of the gene frequency, pY, in the parents.
Adding up Yy Offspring
YY
M
a
l
e
s
Gametes: 1 Y
Female Parents in Population
YY
Yy
yy
Gametes: 1 Y
Gametes: ½ Y, ½ y
Gametes: 1 y
Frequency: GYY
Frequency: GYy
Frequency: Gyy
YY
½ YY, ½ Yy
Yy
(GYY)2
(GYY)(GYy)
(GYY)(Gyy)
Gametes: ½ Y, ½ y
½ YY, ½ Yy
½ Yy, ½ yy
Frequency: GYy
(GYY)(GYy)
¼ YY
½ Yy
¼ yy
Frequency: GYY
Yy
(GYy)(Gyy)
(GYy)2
yy
Gametes: 1 y
Yy
½ Yy, ½ yy
yy
Frequency: Gyy
(GYY)(Gyy)
(GYy)(Gyy)
(Gyy)2
Parental Genotype Frequencies: GYY , GYy, Gyy
Parental Allele Frequencies: pY = GYY + (½)Gyy
Offspring Genotype Frequencies: GYY , GYy, Gyy
GYy = (½)(GYY)(GYy) + (½)(GYY)(GYy) + (½)(GYy)2
+ (½)(GYy)(Gyy) + (½)(GYy)(Gyy) + (2)(GYY)(Gyy)
= (2)(GYY +[½][GYy])(Gyy +[½][GYy])
GYy = (2)(pY)(py)
Note: Genotype frequency in offspring, GYY, equals
product of gene frequencies, pY and pY in the parents.
Hardy – Weinberg Equilibrium
• Describes a population that is NOT evolving, because
there is NO Natural Selection or any other Evolutionary
Force acting on the population.
• Allele frequencies do not change from parents to
offspring under Hardy-Weinberg conditions!
• Genotype frequencies {GYY, GYy, Gyy} in the offspring
population at fertilization are a simple function of the
allele frequencies {p, q} in the parent generation.
2 , 2P P , P 2 }
{GYY, GYy, GFreq.
}
=
{
P
yyof A
Y y
y
Freq. ofY
a
allele
allele
and parents PY = offspring PY
The Hardy – Weinberg Equilibrium is one of the
Population consequences of Mendel’s Laws
NECESSARY ASSUMPTIONS for H-W
• Large Mendelian population
• Random mating
• No mutation
• No migration
• No natural selection
Under these assumptions there is no change in allele
frequency from one generation to the next (i.e. no
evolution)! parents P = offspring P
Y
Y
The Hardy – Weinberg Equilibrium is one of the
Population consequences of Mendel’s Laws
It is an Equilibrium that is achieved in one
generation of random mating.
When a population deviates away from the
Hardy-Weinberg Equilibrium it means
EITHER:
(1) Mating is not random in the population;
Or
(2) Some Evolutionary Force is acting in the
population!
Mutation as an Evolutionary Force
1. It occurs when errors are made in
duplicating alleles in producing the gametes.
2. It is one of the weaker evolutionary forces,
because errors are relatively rare. The error
rate or mutation rate, m, in copying an allele
of a nuclear gene is ~ 1 x 10-6 to 1 x 10-9.
3. It changes allele frequencies in a population
and this change in the genetic composition
of a population from parents to offspring is
what we mean by evolution.
No Mutation
AA Parents produce only ‘A’ bearing gametes.
Aa Parents produce ½ ‘A’ and ½ ‘a’ bearing gametes
aa Parents produce only all ‘a’ bearing gametes.
With Mutation
AA Parents produce some ‘a’ bearing mutant gametes.
Aa Parents produce ½ ‘A’ and ½ ‘a’ gametes
aa Parent produce some ‘A’ bearing mutant gametes.
= A alleles
= a alleles
Parent
population
Reproduction
With Mutation
Offspring
population
How strong is mutation as an
evolutionary force?
Calculate how much the frequency of an
allele changes in the population as a result
of mutation.
Mechanism of
Mutation
μ
A
Allele in the
Parent
a
Allele in the
Parent
u
a
Mutant Allele in the
Gamete and then
In the Offspring
A
Mutant Allele in the
Gamete and then
In the Offspring
Change in allele frequency, DPa, as a
result of mutation
Mechanism of
Mutation
μ
A
Parent Frequencies:
{PA, Pa}
a
u
Reproduction
With Mutation
Offspring Frequencies:
{PA’, Pa’}
How similar are PA’ and PA?
The change in allele frequency, DPa,
caused by mutation
Parent Frequencies:
{PA, Pa}
Freq of a
allele in
offspring
after
mutation
Pa’
Reproduction
With Mutation
Offspring Frequencies:
{PA’, Pa’}
Mutation rate
from A to a times
the Freq of A
before mutation
Non-Mutation
rate times the
Freq of a
before mutation
=
(1- v) Pa
+
μPA
ΔPa = Pa’– Pa = μ – (u + m)Pa
Change in allele frequency, DPa, as a
result of mutation
Parent Frequencies:
{PA, Pa}
Reproduction
With Mutation
Offspring Frequencies:
{PA’, Pa’}
ΔPa = Pa’– Pa = μ – (u + m)Pa
At the Mutation Equilibrium, ΔPa = 0.
0 = μ – (u + m)P*a
P*a = μ/(u + m) =
The Equilibrium Allele Frequency =
Rate at which A is wrongly copied as a,
Relative to all errors at that gene.