friday’s class 3 extra credit questions

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Friday’s Class 3 Extra Credit Questions • 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.

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Friday’s Class 3 Extra Credit Questions. 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. - PowerPoint PPT Presentation

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Page 1: Friday’s Class 3 Extra Credit Questions

Friday’s Class 3 Extra Credit Questions

• 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.

Page 2: Friday’s Class 3 Extra Credit Questions

Our Goal: To understand the “population consequences”Of Mendel’s Laws.

Question 1: How do we describe a Mendelian population?

Page 3: Friday’s Class 3 Extra Credit Questions

Genes in Populations With NO Natural Selection

Generation EndsReproducing Adults

Gametes(Sperm and eggs)

Juveniles

Generation BeginsZygotes

(fertilized eggs)

Life cycle

MatingSystem

{GYY, GYy, Gyy}

{GYY, GYy, Gyy}

NaturalSelection

NaturalSelection

NO NaturalSelection

Page 4: Friday’s Class 3 Extra Credit Questions

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.

Page 5: Friday’s Class 3 Extra Credit Questions

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

300 Y Alleles = 2 x 150150

Yy400400 y Alleles = 1 x 400

400 Y Alleles = 1 x 400

yy 90 180 y Alleles = 2 x 90

YY

Can ALWAYS take individual genotypes apart into alleles

Page 6: Friday’s Class 3 Extra Credit Questions

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

300 Y Alleles = 2 x 150150

Yy400400 y Alleles = 1 x 400

400 Y Alleles = 1 x 400

yy 90 180 y Alleles = 2 x 90

YY

Can ALWAYS take individual genotypes apart into alleles

Page 7: Friday’s Class 3 Extra Credit Questions

There is More than One Way to package alleles intogenotypes.

Knowing {PY, Py} CANNOT ALWAYS calculate theOne and only, unique genotype frequency distribution:

{GYY, GYy, Gyy}

UniqueGenotypes

Alleles

Page 8: Friday’s Class 3 Extra Credit Questions

Genotypes Alleles

UniqueGenotypes

Alleles

Always Unpackage

The Two ways to genetically describe a Mendelian Population are only partially interchangeable

CannotAlways REpackage

Page 9: Friday’s Class 3 Extra Credit Questions

Genes in Populations With NO Natural Selection

{GYY, GYy, Gyy}

Generation BeginsDiploid Zygotes

(fertilized eggs)

{GYY, GYy, Gyy}

Generation EndsReproducing Adults

??

How are these Genotype FrequencyDistributions related to one another?

Parents Offspring

Page 10: Friday’s Class 3 Extra Credit Questions

Genotypes Alleles

UniqueGenotypes

Alleles

Always Unpackage

Under What Circumstances are the Two ways to genetically describe a Mendelian Population

interchangeable ???

When CAN weRepackage????

THE ANSWER: is something you need to know

Page 11: Friday’s Class 3 Extra Credit Questions

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)

Page 12: Friday’s Class 3 Extra Credit Questions

YYGametes: 1 Y

Frequency: GYY

YyGametes: ½ Y, ½ y

Frequency: GYy

yyGametes: 1 y

Frequency: Gyy

YYGametes: 1 Y

Frequency: GYY

YY

(GYY)2

½ YY, ½ Yy

(GYY)(GYy)

Yy

(GYY)(Gyy)

YyGametes: ½ Y, ½ y

Frequency: GYy

½ YY, ½ Yy

(GYY)(GYy)

¼ YY ½ Yy ¼ yy(GYy)2

½ Yy, ½ yy

(GYy)(Gyy)

yyGametes: 1 y

Frequency: Gyy

Yy

(GYY)(Gyy)

½ Yy, ½ yy

(GYy)(Gyy)

yy

(Gyy)2

Female Parents in Population

Males

Page 13: Friday’s Class 3 Extra Credit Questions

YYGametes: 1 Y

Frequency: GYY

YyGametes: ½ Y, ½ y

Frequency: GYy

yyGametes: 1 y

Frequency: Gyy

YYGametes: 1 Y

Frequency: GYY

YY

(GYY)2

½ YY, ½ Yy

(GYY)(GYy)

Yy

(GYY)(Gyy)

YyGametes: ½ Y, ½ y

Frequency: GYy

½ YY, ½ Yy

(GYY)(GYy)

¼ YY ½ Yy ¼ yy(GYy)2

½ Yy, ½ yy

(GYy)(Gyy)

yyGametes: 1 y

Frequency: Gyy

Yy

(GYY)(Gyy)

½ Yy, ½ yy

(GYy)(Gyy)

yy

(Gyy)2

Female Parents in Population

Males

Adding up YY Offspring

Page 14: Friday’s Class 3 Extra Credit Questions

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.

Page 15: Friday’s Class 3 Extra Credit Questions

YYGametes: 1 Y

Frequency: GYY

YyGametes: ½ Y, ½ y

Frequency: GYy

yyGametes: 1 y

Frequency: Gyy

YYGametes: 1 Y

Frequency: GYY

YY

(GYY)2

½ YY, ½ Yy

(GYY)(GYy)

Yy

(GYY)(Gyy)

YyGametes: ½ Y, ½ y

Frequency: GYy

½ YY, ½ Yy

(GYY)(GYy)

¼ YY ½ Yy ¼ yy(GYy)2

½ Yy, ½ yy

(GYy)(Gyy)

yyGametes: 1 y

Frequency: Gyy

Yy

(GYY)(Gyy)

½ Yy, ½ yy

(GYy)(Gyy)

yy

(Gyy)2

Female Parents in Population

Males

Adding up Yy Offspring

Page 16: Friday’s Class 3 Extra Credit Questions

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.

Page 17: Friday’s Class 3 Extra Credit Questions

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.

Freq. of A allele

Freq. of a allele

= { PY 2 , 2PY Py, Py2 }{GYY, GYy, Gyy}

and parents PY = offspring PY

Page 18: Friday’s Class 3 Extra Credit Questions

The Hardy – Weinberg Equilibrium is one of thePopulation 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 PY = offspring PY

Page 19: Friday’s Class 3 Extra Credit Questions

The Hardy – Weinberg Equilibrium is one of thePopulation 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!

Page 20: Friday’s Class 3 Extra Credit Questions

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, , 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.

Page 21: Friday’s Class 3 Extra Credit Questions

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.

Page 22: Friday’s Class 3 Extra Credit Questions

Offspringpopulation

= A alleles = a alleles

Parentpopulation

ReproductionWith Mutation

Page 23: Friday’s Class 3 Extra Credit Questions

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.

A aμMechanism of

MutationAllele in the

Parent

Mutant Allele in the Gamete and then

In the Offspring

Aa

Allele in the Parent

Mutant Allele in the Gamete and then

In the Offspring

Page 24: Friday’s Class 3 Extra Credit Questions

Change in allele frequency, Pa, as a result of mutation

A aμMechanism of

Mutation

Parent Frequencies:{PA, Pa}

Offspring Frequencies:{PA’, Pa’}

ReproductionWith Mutation

How similar are PA’ and PA?

Page 25: Friday’s Class 3 Extra Credit Questions

Pa’ = (1- v) Pa + μPA

The change in allele frequency, Pa, caused by mutation

Freq of a allele in offspring

after mutation

Non-Mutation rate times the

Freq of a before mutation

Mutation rate from A to a times

the Freq of A before mutation

ΔPa = Pa’– Pa = μ – ( + )Pa

Parent Frequencies:{PA, Pa}

Offspring Frequencies:{PA’, Pa’}

ReproductionWith Mutation

Page 26: Friday’s Class 3 Extra Credit Questions

Change in allele frequency, Pa, as a result of mutation

ΔPa = Pa’– Pa = μ – ( + )Pa

At the Mutation Equilibrium, ΔPa = 0. 0 = μ – ( + )P*a

P*a = μ/( + ) =The Equilibrium Allele Frequency =

Rate at which A is wrongly copied as a,Relative to all errors at that gene.

Parent Frequencies:{PA, Pa}

Offspring Frequencies:{PA’, Pa’}

ReproductionWith Mutation