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Genetics News • Problem Set 7 available • Summary for Exam II - Do it NOW! • Problem Set 8 available • Exam I … not available • Office hours underutilization Our answers and comments on web. Click on Course at a Glance; then Click on Exams; Answer Keys

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G e n e t i c s N e w s. Summary for Exam II - Do it NOW!. Office hours underutilization. Problem Set 7 available. Problem Set 8 available. Exam I …. not available. Our answers and comments on web. Click on Course at a Glance; then Click on Exams; Answer Keys. Topics for Today. - PowerPoint PPT Presentation

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Page 1: G e n e t i c s N e w s

Genetics News

• Problem Set 7 available

• Summary for Exam II- Do it NOW!

• Problem Set 8 available

• Exam I … not available

• Office hours underutilization

Our answers and comments on web. Click on Course at a Glance; then Click on Exams; Answer Keys

Page 2: G e n e t i c s N e w s

Topics for Today

• Sex linkage/Sex limitation

•Calculation of probabilities from pedigree

• Extension of Mendelian Genetics - Multiple alleles determining one trait - Multiple genes determining one trait - Multiple genes with additive effect

Page 3: G e n e t i c s N e w s

Study Question 6

How is the trait inherited?

Autosomal dominant

Dd

Dd Dd x

Page 4: G e n e t i c s N e w s

Study Question 6

How is the trait inherited?

Autosomal dominant

Autosomal recessivedd

Dd Dd

xdd

Dd Dd

dd

Dd Dd

Page 5: G e n e t i c s N e w s

Study Question 6

How is the trait inherited?

Autosomal dominant

Autosomal recessive

Sex-linked dominant

XDY

XDY XXD

x

Page 6: G e n e t i c s N e w s

Study Question 6

How is the trait inherited?

Autosomal dominant

Autosomal recessive

Sex-linked dominant

Sex-linked recessive

XdY

XXd

XdY

XXdXY

XdY

XXd XY

Page 7: G e n e t i c s N e w s

Study Question 6Could it be an autosomal dominant trait limited to

males?

Dd

Dd Dd

Dd

dd Dd

Dd

ddDdYes

Page 8: G e n e t i c s N e w s

Study Question 8What is the probability that IV.1 will get the disease?

P(III.2 x III.3 XdY)Specify:

Parse: P(III.2 is XXd)

AND P(III.2 passes Xd)XdY ?

AND P(III.3 is XY)

AND P(III.3 passes Y)

XXd ? XY

1/2

1/2

1

?Solve:

XXd

1/2

Probability = 1/8

Page 9: G e n e t i c s N e w s

Multiple Genes/One Trait

AB Ab aB ab

AB AABB AABb AaBB AaBb

Ab AABb AAbb AaBb Aabb

aB AaBB AaBb aaBB aaBb

ab AaBb Aabb aaBb aabb

Substrate1 Product1 Substrate2 Product2 A B

Conventional dihybrid cross

Page 10: G e n e t i c s N e w s

Multiple Genes/One Trait

AB Ab aB ab

AB AABB AABb AaBB AaBb

Ab AABb AAbb AaBb Aabb

aB AaBB AaBb aaBB aaBb

ab AaBb Aabb aaBb aabb

Substrate1 Substrate2 Product A B

AB Ab aB ab

AB AABB AABb AaBB AaBb

Ab AABb AAbb AaBb Aabb

aB AaBB AaBb aaBB aaBb

ab AaBb Aabb aaBb aabb

Dihybrid cross of interacting genes

9 : 3 : 3 : 1 = 9 : 7

Page 11: G e n e t i c s N e w s

Multiple Genes/One Trait

AB Ab aB ab

AB AABB AABb AaBB AaBb

Ab AABb AAbb AaBb Aabb

aB AaBB AaBb aaBB aaBb

ab AaBb Aabb aaBb aabb

chorismate compound X tryptophan M N

AB Ab aB ab

AB AABB AABb AaBB AaBb

Ab AABb AAbb AaBb Aabb

aB AaBB AaBb aaBB aaBb

ab AaBb Aabb aaBb aabb

Dihybrid cross of interacting genes

9 : 3 : 3 : 1 = 9 : 7

Page 12: G e n e t i c s N e w s

Study Question 17

You cross a vermilion-eyed fly with a cinnibar-eyed fly to obtain F1 progeny, then you cross these progeny to get F2 progeny. What phenotypic ratio do you expect?

Page 13: G e n e t i c s N e w s

Study Question 17

CV Cv cV cv

CV CCVV CCVv CcVV CcVv

Cv CCVv CCvv CcVv Ccvv

cV CcVV CcVv ccVV ccVv

cv CcVv Ccvv ccVv ccvv

Substrate1 Substrate2 Product V C

Dihybrid cross of interacting genes CV Cv cV cv

CV CCVV CCVv CcVV CcVv

Cv CCVv CCvv CcVv Ccvv

cV CcVV CcVv ccVV ccVv

cv CcVv Ccvv ccVv ccvv

Page 14: G e n e t i c s N e w s

Study Question 17

Substrate1 Substrate2 Product V C

Dihybrid cross of interacting genes CV Cv cV cv

CV CCVV CCVv CcVV CcVv

Cv CCVv CCvv CcVv Ccvv

cV CcVV CcVv ccVV ccVv

cv CcVv Ccvv ccVv ccvv

9 : ? : ? : ?

Page 15: G e n e t i c s N e w s

Study Question 17

Substrate1 Substrate2 Product V C

Dihybrid cross of interacting genes CV Cv cV cv

CV CCVV CCVv CcVV CcVv

Cv CCVv CCvv CcVv Ccvv

cV CcVV CcVv ccVV ccVv

cv CcVv Ccvv ccVv ccvv

9 : 3 : ? : 1

Page 16: G e n e t i c s N e w s

Multiple Genes/One Trait

Substrate1 Substrate2 Product V C

Dihybrid cross of interacting genes

9 : 3 : 3 : 1 = 9 : 3 : 4

CV Cv cV cv

CV CCVV CCVv CcVV CcVv

Cv CCVv CCvv CcVv Ccvv

cV CcVV CcVv ccVV ccVv

cv CcVv Ccvv ccVv ccvv

9 : 3 : 3 : 1

Page 17: G e n e t i c s N e w s

Study Question 19

AB Ab aB ab

AB AABB AABb AaBB AaBb

Ab AABb AAbb AaBb Aabb

aB AaBB AaBb aaBB aaBb

ab AaBb Aabb aaBb aabb

Substrate1 Product1 Substrate2 Product2 A B

Conventional dihybrid cross

Pathway where dihybrid

cross gives 15:1 ratio?

Page 18: G e n e t i c s N e w s

Study Question 19

AB Ab aB ab

AB AABB AABb AaBB AaBb

Ab AABb AAbb AaBb Aabb

aB AaBB AaBb aaBB aaBb

ab AaBb Aabb aaBb aabb

Substrate1 Product1 Substrate2 Product2 A B

What kind of genetic situation can you imagine that would lead to a 15:1 phenotypic ratio?

9 : 3 : 3 : 1 9 : 3 : 3 : 1 15 : 1

AB Ab aB ab

AB AABB AABb AaBB AaBb

Ab AABb AAbb AaBb Aabb

aB AaBB AaBb aaBB aaBb

ab AaBb Aabb aaBb aabb

Page 19: G e n e t i c s N e w s

Study Question 20

You find corn with kernels in the ratio:

Yellow 247Golden 154Sickly green 23

Hypothesis?

Hypothesis: One gene, multiple alleles

AB x CD 1:(1:1):1= 1:2:1

Page 20: G e n e t i c s N e w s

Study Question 20One gene hypothesis

You find corn with kernels in the ratio:

Yellow 247Golden 154Sickly green 23

TOTAL 424

A B

C AC BC

D AD BD

Exp Obs

Unit

= 424/4 = 106

212106106424

Page 21: G e n e t i c s N e w s

Study Question 20Two gene hypothesis

You find corn with kernels in the ratio:

Yellow 247Golden 154Sickly green 23

Hypothesis?

Hypothesis: Two genes for one trait

AaBb x AaBb 9:3:3:1

Page 22: G e n e t i c s N e w s

Study Question 20Two gene hypothesis

You find corn with kernels in the ratio:

Yellow 247Golden 154Sickly green 23

TOTAL 424

Exp Obs

Unit

= 424/16

= 26.5

AB Ab aB ab

AB AABB AABb AaBB AaBb

Ab AABb AAbb AaBb Aabb

aB AaBB AaBb aaBB aaBb

ab AaBb Aabb aaBb aabb

238.5 = 26.5 x 9159 = 26.5 x 6 26.5 = 26.5 x 1424