Transcript
Page 1: Inheritance Patterns and Probability July 2008. Pedigrees

Inheritance Patterns and Probability

July 2008

Page 2: Inheritance Patterns and Probability July 2008. Pedigrees

Pedigrees

Page 3: Inheritance Patterns and Probability July 2008. Pedigrees

1 2

21 3

1

I.

II.

III.

This pedigree shows a family with a form of deafness that is inherited in a recessive manner. Members of the family with filled symbols are deaf. Which members of this family are definitely heterozygous (Dd)?

A. I-1 and I-2

B. I-1, I-2, and II-1

C. I-1, II-1, and II-3

D. I-1, I-2, and II-3

E. I-1, I-2, II-1, and II-3

Dd, DD = normaldd = deaf

Page 4: Inheritance Patterns and Probability July 2008. Pedigrees

1 2

21 3

1

I.

II.

III.

If II-2 and II-3 just had another baby boy. What is the chance that he is deaf?

A. 1/8

B. 1/4

C. 1/2

D. 3/4

E. 1

Dddd

dd

Page 5: Inheritance Patterns and Probability July 2008. Pedigrees

1 2

21 3

1

I.

II.

III.

dd

dd

1 2

21 3

1

I.

II.

III.

dd

dd

Dd Dd

Dd or DD

Dd Dd

Dd or DDDd Dd

family 1family 2

What are the chances that II-1 from family 1 and II-1 from family 2 will have a deaf child together?

A. 1/4

B. 1/9

C. 4/9

D. 1/16

Page 6: Inheritance Patterns and Probability July 2008. Pedigrees

Based on the pedigree above, which inheritance pattern can be ruled out?

A. Autosomal dominant

B. Autosomal recessive

C. X-linked dominant

D. X-linked recessive

E. None of the above

Page 7: Inheritance Patterns and Probability July 2008. Pedigrees

Based on the pedigree above, which inheritance pattern can be ruled out?

A. Autosomal dominantB. Autosomal recessiveC. X-linked dominantD. X-linked recessiveE. None of the above

Page 8: Inheritance Patterns and Probability July 2008. Pedigrees

Based on the pedigree above, which inheritance pattern can be ruled out?

A. X-linked dominant

B. X-linked recessive

C. neither of the above

Page 9: Inheritance Patterns and Probability July 2008. Pedigrees

?

Phenylketonuria (PKU) is an inherited disorder that can lead to mental retardation if left untreated. PKU is inherited in an recessive manner. What is the chance that the boy marked with a “?” in the pedigree will have PKU?

A. 1/3

B. 1/4

C. 1/6

D. 1/8

Page 10: Inheritance Patterns and Probability July 2008. Pedigrees

I.

II.

III.

1 2

1

3 4

12 3 4 5

?

You would like to use mitochondrial DNA to try to determine if III-1 is a member of the family shown in this pedigree. II-2 and II-3 are dead as indicated with a slash and you are unable to collect mitochondrial DNA from them.If III-1 is a member of this family his mitochondrial DNA should match:

A) I-1 and II-1 only B) I-1, I-2 and II-1 onlyC) I-1, I-3, II-1, and II-4 onlyD) I-3 and II-4 only E) I-3, II-4, and II-5 only

Page 11: Inheritance Patterns and Probability July 2008. Pedigrees

Diastrophic dysplasiaAutosomal recessive

“D” normal allele

“d” mutant allele

Matt Amy

Ron Peggy Gordon Pat

What is Matt’s genotype (Matt has diastrophic dysplasia)?

A. ddaa

B. ddAa

C. Ddaa

D. DdAa

E. ddAA

Achondroplasia Autosomal dominant “A” mutant allele

“a” normal allele

Page 12: Inheritance Patterns and Probability July 2008. Pedigrees

Diastrophic dysplasiaAutosomal recessive SLC26A2 gene

“D” normal allele

“d” mutant allele

Matt Amy

Ron Peggy Gordon Pat

ddaa

What is Ron’s genotype?

A. ddaa

B. Ddaa

C. DDaa

Achondroplasia Autosomal dominant FGFR3 gene

“A” mutant allele

“a” normal allele

DDAa (note AA embryos are not viable)

Ddaa Ddaa

Page 13: Inheritance Patterns and Probability July 2008. Pedigrees

Diastrophic dysplasiaAutosomal recessive SLC26A2 gene

“D” normal allele

“d” mutant allele

Matt Amy

Ron Peggy Gordon Pat

ddaa

What is Pat’s genotype?

A. DDAa

B. DDAA

C. Ddaa

D. None of the above

Achondroplasia Autosomal dominant FGFR3 gene

“A” mutant allele

“a” normal allele

DDAa (note AA embryos are not viable)

Ddaa Ddaa DDaa DDaa

Page 14: Inheritance Patterns and Probability July 2008. Pedigrees

Diastrophic dysplasiaAutosomal recessive SLC26A2 gene

“D” normal allele

“d” mutant alleleMatt Amy

Jeremy Zach Molly Jacob

ddaa

What is Zach’s genotype?

A. Ddaa

B. DdAa

C. DdAA

Achondroplasia Autosomal dominant FGFR3 gene

“A” mutant allele

“a” normal allele

DDAa (note AA embryos are not viable)

Page 15: Inheritance Patterns and Probability July 2008. Pedigrees

Blue cootie disease is an autosomal recessive disorder that causes cooties to be blue. Wild type cooties are red. R is the dominant color allele and r is the recessive color allele.

What is the phenotype of the twins’ father?A) RRB) RrC) rrD) red

? ?

Page 16: Inheritance Patterns and Probability July 2008. Pedigrees

Blue cootie disease is an autosomal recessive disorder that causes cooties to be blue. Wild type cooties are red. R is the dominant color allele and r is the recessive color allele.

What is the genotype of the twins’ father?A) RRB) RrC) rrD) 1/2 Rr, 1/2 RR

? ?

Page 17: Inheritance Patterns and Probability July 2008. Pedigrees

Blue cootie disease is an autosomal recessive disorder that causes cooties to be blue. Wild type cooties are red. R is the dominant color allele and r is the recessive color allele.

What is the genotype of the twins' mother?A) RRB) RrC) ½ Rr, ¼ RRD) 2/3 Rr, 1/3 RR

? ?

Page 18: Inheritance Patterns and Probability July 2008. Pedigrees

Blue cootie disease is an autosomal recessive disorder that causes cooties to be blue. Wild type cooties are red. R is the dominant color allele and r is the recessive color allele.

What is the probability that the first twin bornwill have blue cootie disease?A) 1/4B) 1/3C) 1/6D) 0

? ?

Page 19: Inheritance Patterns and Probability July 2008. Pedigrees

The next few questions are not about pedigrees, but follow the cootie example

Page 20: Inheritance Patterns and Probability July 2008. Pedigrees

Antennaless is an autosomal recessive disorder that leads to cooties without antenna. A is the dominant WT allele and a is the mutant recessive allele.A true-breeding WT red cootie mates with a true-breeding blue antennaless cootie.

What is the phenotype of the F1 generation?A) All red with antennae B) All red but half with antennae and half withoutC) 9 red antennae: 3 red no antennae: 3 blue antennae: 1 blue no antennaeD) ¼ red with antennae, ¼ red without antennae, ¼ blue with antennae, ¼ blue without antennae

XP

Page 21: Inheritance Patterns and Probability July 2008. Pedigrees

You allow the F1 generation to mate and produce offspring (F2 generation).

What is the probability that an F2 cootie will bered? A) 1/4B) 1/2C) 3/4D) 1

XP

RrAa

F1

Page 22: Inheritance Patterns and Probability July 2008. Pedigrees

Here is the F2 generation (RrAa X RrAa)

What is expected number of red cooties with antennae?A) 963B) 1700C) 1760D) 2063

1767

543

598

221

observed expected (O-E)2/E

Do the red and antenna gene follow rules of independent assortment?

Page 23: Inheritance Patterns and Probability July 2008. Pedigrees

Here is the F2 generation (RrAa X RrAa)

What is (O-E)2/E for the blue antennaless group?A) 625B) 25C) 3.2D) 0.13

1767 ---

543 586.7

598 586.7

221 196

observed expected (O-E)2/E

Do the red and antenna gene follow rules of independent assortment?

Page 24: Inheritance Patterns and Probability July 2008. Pedigrees

Here is the F2 generation (RrAa X RrAa) 3138 total

Do the red and antenna gene follow rules of independent assortment?A) Yes, accept hypothesis – differences are likely due to chanceB) Yes, accept hypothesis – differences are not likely due to chanceC) No, reject hypothesis – differences are likely due to chanceD) No, reject hypothesis – differences are not likely due to chance

1767 --- 0.02543 586.7 3.3598 586.7 0.21221 196 ---

observed expected (O-E)2/E

Page 25: Inheritance Patterns and Probability July 2008. Pedigrees

Calculating probability of inheritance (monhybrid, dihybrid crosses)

Page 26: Inheritance Patterns and Probability July 2008. Pedigrees

Results of the F1 cross Yy X Yy

What is the phenotype of the circled green pea?A) YYB) YyC) yyD) greenE) need more information

Page 27: Inheritance Patterns and Probability July 2008. Pedigrees

Results of the F1 cross Yy X Yy

What is the genotype of the circled yellow pea?A) YYB) YyC) yyD) yellowE) need more information

Page 28: Inheritance Patterns and Probability July 2008. Pedigrees

What is the genotype of the yellow, round parent?Y - Yellowy - Green

R - Roundr - wrinkled A: YYRR

B: YyRRC: YYRrD: YyRrE: Cannot be determined

Plant 1: Yellow, round peas

Plant 2: Green, wrinkled peasX

F1:

1/2 Yellow, round peas

1/2 Yellow, wrinkled peas

P:

Page 29: Inheritance Patterns and Probability July 2008. Pedigrees

Use Mendel’s Dihybrid cross results:

XP

F1

F2

315 101 108 32

Given this data, what do you think the ratio of offspring is?

A: 3:1B: 1:2:1C: 9:3:3:1D: 2:1

Page 30: Inheritance Patterns and Probability July 2008. Pedigrees

Results of the F1 cross Yy X Yy

The test cross that would most clearly distinguish the genotype of the circled yellow pea is:A) Yellow pea 1 X Yellow pea 2B) Yellow pea 2 X Yellow pea 3C) Yellow pea 2 X Green pea 4D) You would need to do all of the above crosses

1 2 3 4

Page 31: Inheritance Patterns and Probability July 2008. Pedigrees

Genotype and phenotype

PhenotypesGenotypes

You cross a yellow with a green and see a 50:50 ratio of greenand yellow progeny. What is the genotype of theoriginal yellow pea?

A) YYB) YyC) yyD) Need more information

Page 32: Inheritance Patterns and Probability July 2008. Pedigrees

Dihybrid cross

Mating between individuals that differ in two traits

X

Round, Yellow Wrinkled, Green

RRYY rryy

What are the possible gametes produced by the F1 peas?A) rryy, RrYy, RRYYB) R, r, Y, yC) Rr, Yy, RR, rr, YY, yyD) RY, Ry, rY, ry

P

F1

RrYy

100% Round, Yellow

Page 33: Inheritance Patterns and Probability July 2008. Pedigrees

Dihybrid cross

F1

X

RrYy

RrYy

RY Ry rY ry

RY

Ry

rY

ry

F2 Generation

RRYY RYRy RrYY RrYy

RRYy RRyy RrYy Rryy

RrYY RrYy rrYY rrYy

RrYy Rryy rrYy rryy

Question 6: What fraction of theF2 generation is green?

A) 1/16B) 1/2C) 1/9D) 1/4

Page 34: Inheritance Patterns and Probability July 2008. Pedigrees

What is the phenotype ratio of progeny in F1 generation of the following cross?

RrYy rryy

X

Round, yellow Wrinkled, green

Round, Yellow

Wrinkled, Yellow

Round, Green

Wrinkled, Green

9

3

3

1

3

1

3

1

1

1

1

1

1

3

3

9

A B C D

Page 35: Inheritance Patterns and Probability July 2008. Pedigrees

Can you use the outcome todeduce the parental genotype?

• Suppose you cross a yellow and green and get 50% yellow and 50% green?

What are the parental genotypes?A) YY X yyB) Yy x yyC) yy x yyD) Yy x Yy

Page 36: Inheritance Patterns and Probability July 2008. Pedigrees

Monohybrid cross probability

• Consider Yy X Yy cross

What is the probability ofgetting a Y from parent 1?

A) 1/4B) 1/2C) 1D) 1/16

Page 37: Inheritance Patterns and Probability July 2008. Pedigrees

Monohybrid cross probability

• Consider Yy X Yy cross

What is the probability ofgetting a Y from oneparent *AND* Y from theother parent (i.e. YY)?

A) 1/4B) 1/2C) 1D) 1/16

Page 38: Inheritance Patterns and Probability July 2008. Pedigrees

Monohybrid cross probability

• Consider Yy X Yy cross

What is the probability ofbeing Yellow (i.e. YY ORYy)?

A) 1/4B) 1/2C) 3/4D) 1

Page 39: Inheritance Patterns and Probability July 2008. Pedigrees

Consider the following cross:AaBBCcddEe X aabbCCDdEeWhat is the probability their first offspring will be aaBbCCDdee?

A)1/8B)1/16C)1/32D)1/64E)Cannot be determined

Page 40: Inheritance Patterns and Probability July 2008. Pedigrees

What is the probability of rolling a two OR a three with one role of a six-sided die?

A)1/3B)1/2C)1/6D)1/36E)1/64

Page 41: Inheritance Patterns and Probability July 2008. Pedigrees

A male smurf has an dominant X-linked disorderthat causes red skin. He marries smurfette (who is normal blue).

What are the possible phenotypes of their male children?A) all blue skinB) all red skinC) patches of red and blue skinD) more than one of the above

Page 42: Inheritance Patterns and Probability July 2008. Pedigrees

A male smurf has an dominant X-linked disorder that causes red skin. He marries smurfette (who is normal blue).

What are the possible phenotypes of their female children?A) all blue skinB) all red skinC) patches of red and blue skinD) more than one of the above

Page 43: Inheritance Patterns and Probability July 2008. Pedigrees

Statistical Analysis of Crosses

Page 44: Inheritance Patterns and Probability July 2008. Pedigrees

Use Mendel’s Dihybrid cross results:

XP

F1

F2

315 101 108 32

3. Calculate Expected (e) numbers for each class if hypothesis correctHow many seeds would you expect to be green and round (to the nearest whole

number)?A: 100B: 104C: 105D: 108

Total seeds observed = 556

Page 45: Inheritance Patterns and Probability July 2008. Pedigrees

Observed expected (o-e)2/e

5. Calculate Degree of Freedom

3. Calculate Expected (e) numbers for each class if hypothesis correct

4. Calculate X2 = ∑ (o -e)2/eAlways use real numbers, not % or fraction

∑ means ’Sum of all classes’

Use a table:

X2 =

315 312 (315-312)2/312=0.029101 104 0.087108 104 0.15432 35 0.257

0.527

A: 1B: 4C: 3

Page 46: Inheritance Patterns and Probability July 2008. Pedigrees

6. Look up probability (p) for X2 at a given df in the table

A: Accept the hypothesis B: Reject the hypothesis

Page 47: Inheritance Patterns and Probability July 2008. Pedigrees

XP

F1

F2

315 101 108 32

How many degrees of freedom are there in the F2 generation of the following cross?

A) 1B) 2C) 3D) 4E) 5

Page 48: Inheritance Patterns and Probability July 2008. Pedigrees

X2 = 535

A) Accept the hypothesisB) Reject the hypothesis

What if his results had been 5120 yellow and 2903 green?Could Mendel still accept his hypothesis?

Page 49: Inheritance Patterns and Probability July 2008. Pedigrees

What does P = 0.005mean for the 28:20 ratio?

A) 28:20 is likely to be 3:1B) 28:20 is NOT likely to be 3:1C) 28:20 is not statistically “significant” and so

cannot be used to assess 3:1 ratioD) This experiment is totally flawed and cannot

be interpretted

Page 50: Inheritance Patterns and Probability July 2008. Pedigrees

Exceptions to Mendel’s Laws (maternal,

cytoplasmic/mitochondrial, sex-limited, co-dominance,

incomplete dominance, lethal, epistatsis, heterozygous advantage, imprinting)

Page 51: Inheritance Patterns and Probability July 2008. Pedigrees

Maternal

Page 52: Inheritance Patterns and Probability July 2008. Pedigrees

A maternal effect gene exists in a dominant N (normal) allele and a recessive n (mutant) allele. What would be the ratios of genotypes and phenotypesfor the following cross?

nn female X NN male

A) all Nn, all normalB) all Nn, all mutantC) all nn, all mutantD) all NN, all normal

Page 53: Inheritance Patterns and Probability July 2008. Pedigrees

A maternal effect gene exists in a dominant N (normal) allele and a recessive n (mutant) allele. What would be the ratios of genotypes and phenotypesfor the following cross?

NN female X nn male

A) all Nn, all normalB) all Nn, all mutantC) all nn, all mutantD) all NN, all normal

Page 54: Inheritance Patterns and Probability July 2008. Pedigrees

Worm Mel2 gene products are deposited into the egg by the mother and are required from embryonic development. Mutations in the mel2 gene are recessive and cause maternal effect embryonic lethality.

In a cross between mel2 heterozygotes, what percent of embryos will die?

A) 100%B) 50%C) 25%D) 0%

Page 55: Inheritance Patterns and Probability July 2008. Pedigrees

Zebrafish Ack15 gene products are deposited into the egg by the mother and are required from embryonic development. Mutations in the ack15 gene are recessive and cause maternal effect embryonic lethality.

In a cross between ack15 homozygous mutant female and a heterozygous male, what percent of embryos will die?

A) 100%B) 50%C) 25%D) 0%

Page 56: Inheritance Patterns and Probability July 2008. Pedigrees

You are studying Drosophila mulleri and you discover a maternal effect gene you call nanu. nanu mRNA is localized to the anterior end of the embryo and promotes the formation of anterior structures. Mutations in the nanu gene are recessive.

In a cross between two nanu heterozygotes, how many of the embryos will have defects in their anterior structures?

A. 100%

B. 50%

C. 25%

D. 0%

Page 57: Inheritance Patterns and Probability July 2008. Pedigrees

You are studying Drosophila mulleri and you discover a maternal effect gene you call nanu. nanu mRNA is localized to the anterior end of the embryo and promotes the formation of anterior structures. Mutations in the nanu gene are recessive.

In a cross between a nanu/nanu mutant female and a +/+ male, how many of the embryos will have defects in their anterior structures?

A. 100%

B. 50%

C. 25%

D. 0%

Page 58: Inheritance Patterns and Probability July 2008. Pedigrees

Cytoplasmic/mitochondrial

Page 59: Inheritance Patterns and Probability July 2008. Pedigrees

Sex Limited

Page 60: Inheritance Patterns and Probability July 2008. Pedigrees

Co-dominance/incomplete dominance

Variable ExpressionConditional

Page 61: Inheritance Patterns and Probability July 2008. Pedigrees

You cross a true-breeding white buffalo to a true breeding black buffalo. All the F1 are brown.

An F1 brown buffalo is crossed to the white parent. If they have 4 offspring, how many do you predict will be white?

A. 0B. 1C. 2D. 4E. Not enough information

P

F1

Page 62: Inheritance Patterns and Probability July 2008. Pedigrees

A true-breeding albino buffalo is crossed to a true-breeding black buffalo and all of the progeny are brown. Crossing the brown buffalos to each other yields an approximate ratio of 1 albino: 2 brown: 1 black.

The alleles for buffalo color show:A) complete dominanceB) partial dominanceC) co-dominance

Page 63: Inheritance Patterns and Probability July 2008. Pedigrees

CU has asked the MCDB2150 class to help it with the breeding of Ralphie buffalo. You do the following cross:

You try to establish a true breeding herd of Ralphie buffalo with a mix of short and long hair using the F1 Ralphies but you are unsuccessful. Which mode of interaction between alleles is a possible reason for your lack of success?A. CodominanceB. Incomplete dominanceC. Complete dominance D. Recessive epistasis

X

Long haired Ralphie buffalo

Short haired Ralphie buffalo

Ralphie buffalo with a mix of short and long hairs

Page 64: Inheritance Patterns and Probability July 2008. Pedigrees

The blood type alleles in humans show example(s) of: A) co-dominanceB) complete dominanceC) multiple allelesD) two of the aboveE) all of the above

Page 65: Inheritance Patterns and Probability July 2008. Pedigrees

A man with blood type A and a woman with blood type B have a child with blood type O. This couple can also have children with which blood types?

A: O onlyB: AB and OC: A, B and OD: A, B, AB and O

Page 66: Inheritance Patterns and Probability July 2008. Pedigrees

Charlie Chaplin (multiple alleles)

• Charlie was blood type O• Girlfriend was blood type A• Her (out-of-wedlock child) BFACTS:I locus (blood group) has 3 allelesA = genotype IAIA or Iai -> Ab to BB = genotype IBIB or Ibi -> Ab to AAB = genotype IAIB -> No AbO = genotype ii -> Ab to both A and B

Page 67: Inheritance Patterns and Probability July 2008. Pedigrees

Mr. Chaplin’s case

• Given IA and IB are dominant to I• His girlfriend sued for paternity who won?A) Girlfriend won - baby COULD be hisB) Chaplin won - baby COULD NOT be hisC) Hung jury, can’t tell from the facts

Page 68: Inheritance Patterns and Probability July 2008. Pedigrees

Only 66% of women with a heterozygous BRCA1 mutation get breast cancer by age 55 and most do so in only one breast.

BRCA1 mutant allele…

A: shows incomplete penetranceB: shows variable expressivityC: both

Page 69: Inheritance Patterns and Probability July 2008. Pedigrees

Lethal

Page 70: Inheritance Patterns and Probability July 2008. Pedigrees

Lethal AllelesAll the sneetches want their children to havestars on their bellies. The combination of alleles that makes a black, starless sneetch is lethal.

If a true breeding black star bellied sneetch mates with a true breading yellow sneetch, what is the probability that their first child will have a star?

YYss X yySSA) 1B) 1/2C) 1/4D) 3/16

YY, Yy = yellowyy = blackSS, Ss = starss = no star

Page 71: Inheritance Patterns and Probability July 2008. Pedigrees

Lethal AllelesAll the sneetches want their children to havestars on their bellies. The combination of alleles that makes a black, starless sneetch is lethal.

If two heterozygous yellow star-bellied sneetches mate, what is the likelihood that their first child will not have a star? YySs X YySs

A) 1B) 1/4C) 1/5D) 3/16

YS Ys yS ysYS YYSS YYSs YySs YySsYs YYSs YYss YySs YyssyS YySS YySs yySS yySsys YySs Yyss yySs yyss

Page 72: Inheritance Patterns and Probability July 2008. Pedigrees

DdAa

ttp://images.icnetwork.co.uk/upl/icnewcastle/aug2005/4/8/00015218-E719-12F1-88F40C01AC1BF814.jpg

Zach

Sue

DdAa

DA

Da

dA

da

DA Da dA da

DDAA

DDAa

DdAA

DdAa

DDAa

DDaa

DdAa

Ddaa

DdAA

DdAa

ddAA

ddAa

DdAa

Ddaa

ddAa

ddaa

What is the chance that Zach and Sue will have a child with diastrophic dysplasia and achondroplasia?

A. 1/6

B. 1/8

C. 3/16

D. 9/16

E. None of the above

SLC26A gene

FGFR3 gene

SLC26A gene

FGFR3 gene

Page 73: Inheritance Patterns and Probability July 2008. Pedigrees

What is the chance that Zach and Sue will have a child with only diastrophic dysplasia?

A. 1/3

B. 1/6

C. 1/12

D. 1/16

DdAa

ttp://images.icnetwork.co.uk/upl/icnewcastle/aug2005/4/8/00015218-E719-12F1-88F40C01AC1BF814.jpg

Zach

Sue

DdAa

DA

Da

dA

da

DA Da dA da

DDAA

DDAa

DdAA

DdAa

DDAa

DDaa

DdAa

Ddaa

DdAA

DdAa

ddAA

ddAa

DdAa

Ddaa

ddAa

ddaaSLC26A gene

FGFR3 gene

SLC26A gene

FGFR3 gene

Page 74: Inheritance Patterns and Probability July 2008. Pedigrees

Epistasis

Page 75: Inheritance Patterns and Probability July 2008. Pedigrees

Crossing F2 yellows to brown parent gave a mix of brown, yellow and black. Which model does this support?

A: partial dominanceB: recessive epistasis

Partial Dominance Model1: BB -> Brown2: Bb -> Black1: bb -> Yellow

Recessive epistasis Model9 B_E_ -> Black3 bbE_ -> Brown4 _ _ee -> Yellow

P Brown X Yellow

F1 Black

F2 Brown: Black: Yellow

Page 76: Inheritance Patterns and Probability July 2008. Pedigrees

Epistasis and Labrador retriever coat colorThe B locus determines if pigment can be produced

“B” codes for black pigment“b” codes for brown pigment

The E locus determines if the pigment can be deposited in the hair shaft

“E” allows dark pigment (black or brown) to be deposited“e” prevents dark pigment from being deposited

(the dogs are yellow)

What is the phenotype of a BbEe lab?

A. Black

B. Brown

C. Yellow

Page 77: Inheritance Patterns and Probability July 2008. Pedigrees

The B locus determines if pigment can be produced“B” codes for black pigment“b” codes for brown pigment

The E locus determines if the pigment can be deposited in the hair shaft

“E” allows dark pigment (black or brown) to be deposited“e” prevents dark pigment from being deposited

(the dogs are yellow)

What is the phenotype of a bbEe lab?

A. Black

B. Brown

C. Yellow

Page 78: Inheritance Patterns and Probability July 2008. Pedigrees

The B locus determines if pigment can be produced“B” codes for black pigment“b” codes for brown pigment

The E locus determines if the pigment can be deposited in the hair shaft

“E” allows dark pigment (black or brown) to be deposited“e” prevents dark pigment from being deposited

(the dogs are yellow)

What is the phenotype of a bbee lab?

A. Black

B. Brown

C. Yellow

Page 79: Inheritance Patterns and Probability July 2008. Pedigrees

The B locus determines if pigment can be produced“B” codes for black pigment“b” codes for brown pigment

The E locus determines if the pigment can be deposited in the hair shaft

“E” allows dark pigment (black or brown) to be deposited“e” prevents dark pigment from being deposited

(the dogs are yellow)

What is the phenotype of a BBee lab?

A. Black

B. Brown

C. Yellow

Page 80: Inheritance Patterns and Probability July 2008. Pedigrees

You cross a hairless mouse aaBB to a mouse with curly hair AAbb. All of the F1s have straight hair. You cross two of the F1 mice together. In the F2 generation: 18 mice have straight hair, 8 mice are hairless, and 6 have curly hair.

What is the phenotype of the aabb mice in the F2 generation?

A. hairlessB. curly hairC. straight hair

Page 81: Inheritance Patterns and Probability July 2008. Pedigrees

hairless mouse aaBB X curly hair AAbb

F1s have straight hair AaBb

F2 9 A- B-4 aa -- 3 A- bb

What is the order of function?

A. A, then BB. B, then AC. A and B act simultaneouslyD. Not enough data

18 straight hair mice A-B-

8 hairless mice aaB- & aabb

6 curly hair mice A-bb

Page 82: Inheritance Patterns and Probability July 2008. Pedigrees

What is the order of function?

A. B, then EB. E, then BC. E and B act simultaneouslyD. Not enough data

Recessive epistasis Model9 B_E_ -> Black3 bbE_ -> Brown4 _ _ee -> Yellow

Page 83: Inheritance Patterns and Probability July 2008. Pedigrees

If gene A is epistatic to gene B which intermediate will build upin a AB double mutant?

A) intermediate 1B) intermediate 2C) intermediate 3

Mutant strain A: intermediate 2 builds up

Mutant strain B: intermediate 3 builds up

Mutant strain C: intermediate 1 builds up

Page 84: Inheritance Patterns and Probability July 2008. Pedigrees

Epistasis AA – fluffy hairaa – baldBB – red hair pigmentbb – no red pigment (blue hair)

X

AaBb AaBb

What are the possible gametes produced by each chuzzel?A) Aa, BbB) AaBbC) A, a, B, bD) AB, Ab, aB, ab

Page 85: Inheritance Patterns and Probability July 2008. Pedigrees

1.You cross a true-breeding white buffalo to a true-breeding tan buffalo in the P generation

2. You find out that all of the F1 buffalos are gold

91 47 33

3. You cross two of the F1 gold buffalo together and in the F2 generation get:

The T locus determines if pigment can be produced:“T” codes for gold pigment“t” codes for tan pigment

The A locus determines if pigment can be deposited into the hair shaft:“A” allows pigment (gold or tan) to be deposited into the hair shaft“a” prevents pigment from being deposited into the hair shaft (the buffalo are white)

What is the genotype of the tan buffalo in the P generation?

A. AAtt

B. Aatt

C. Either AAtt or Aatt

Page 86: Inheritance Patterns and Probability July 2008. Pedigrees

1.You cross a true-breeding white buffalo to a true-breeding tan buffalo in the P generation

2. You find out that all of the F1 buffalos are gold

91 47 33

3. You cross two of the F1 gold buffalo together and in the F2 generation get:

The T locus determines if pigment can be produced:“T” codes for gold pigment“t” codes for tan pigment

The A locus determines if pigment can be deposited into the hair shaft:“A” allows pigment (gold or tan) to be deposited into the hair shaft“a” prevents pigment from being deposited into the hair shaft (the buffalo are white)

What is the genotype of the white buffalo in the P generation?

A. aatt

B. aaTT

C. aaTt

AAttaaTT

AaTt

Page 87: Inheritance Patterns and Probability July 2008. Pedigrees

2. You find out that all of the F1 buffalos are gold

3. You cross two of the F1 gold buffalo together and in the F2 generation get:AaTt

91 47 33

AT At aT at

AT AATT AATt AaTT AaTt

At AATt AAtt AaTt Aatt

aT AaTT AaTt aaTT aaTt

at AaTt Aatt aaTt aatt

How many of the genotypes in the Punnett square will result in a gold buffalo?

A. 9

B. 4

C. 3

D. 2

E. 1

AaTt

X

Punnett square for cross of two gold buffalo AaTt x AaTt:

Page 88: Inheritance Patterns and Probability July 2008. Pedigrees

2. You find out that all of the F1 buffalos are gold

3. You cross two of the F1 gold buffalo together and in the F2 generation get:AaTt

91 47 33

AT At aT at

AT AATT AATt AaTT AaTt

At AATt AAtt AaTt Aatt

aT AaTT AaTt aaTT aaTt

at AaTt Aatt aaTt aatt

How many of the genotypes in the Punnett square will result in a white buffalo?

A. 9

B. 4

C. 3

D. 2

E. 1

AaTt

X

Page 89: Inheritance Patterns and Probability July 2008. Pedigrees

2. You find out that all of the F1 buffalos are gold

3. You cross two of the F1 gold buffalo together and in the F2 generation get:AaTt

91 47 33

AT At aT at

AT AATT AATt AaTT AaTt

At AATt AAtt AaTt Aatt

aT AaTT AaTt aaTT aaTt

at AaTt Aatt aaTt aatt

How many of the genotypes in the Punnett square will result in a tan buffalo?

A. 9

B. 4

C. 3

D. 2

E. 1

AaTt

X

Page 90: Inheritance Patterns and Probability July 2008. Pedigrees

3. You cross two of the F1 gold buffalo together and in the F2 generation get:

91 47 33

AT At aT at

AT AATT AATt AaTT AaTt

At AATt AAtt AaTt Aatt

aT AaTT AaTt aaTT aaTt

at AaTt Aatt aaTt aatt

Approximately how many true-breeding tan buffalo are in the F2 generation in your herd?

A. 1

B. 3

C. 11

D. 22

E. 33

Page 91: Inheritance Patterns and Probability July 2008. Pedigrees

Heterozygous Advantage

Page 92: Inheritance Patterns and Probability July 2008. Pedigrees

A man who does not have sickle cell anemia and has no history of it in his family (assume he is not a carrier) marries a woman who has sickle cell anemia. They have a son.

This family is planning to travel to the Solomon Islands. Which family member(s) should take Lariam, a very expensive drug that prevents malaria?

A. FatherB. MotherC. SonD. Father and the sonE. Everyone

Page 93: Inheritance Patterns and Probability July 2008. Pedigrees

Imprinting

Page 94: Inheritance Patterns and Probability July 2008. Pedigrees

A mutation (a) occurs on an imprinted gene (A). The maternal copy of the gene is methylated and not expressed. ‘ denotes the alleles inherited from the father.

Aa A’a’

A’a Aa’ aa’ AA’

1 2 3 4

Which of the offspring will be affected?A) 1 and 3B) 2 and 3C) 3 onlyD) none of the offspring will be affected

Page 95: Inheritance Patterns and Probability July 2008. Pedigrees

Prader-Willi if the deletion is on the chromosome inherited from mom.Angelman syndrome if the deletion is on the chromosome inherited from dad. A is the normal chromosome 15 a contains the deletion

A is from mom A’ is from dad

Aa A’a’

A’a AA’ aa’

Which disorder does the mother have?A) NoneB) PWSC) ASD) need more information

1 2 3

Page 96: Inheritance Patterns and Probability July 2008. Pedigrees

Prader-Willi if the deletion is on the chromosome inherited from mom.Angelman syndrome if the deletion is on the chromosome inherited from dad. A is the normal chromosome 15 a contains the deletion

A is from mom A’ is from dad

Aa A’a’

A’a AA’ aa’

Which disorder does the offspring 1 have?A) NoneB) PWSC) ASD) need more information

1 2 3

Page 97: Inheritance Patterns and Probability July 2008. Pedigrees

Prader-Willi if the deletion is on the chromosome inherited from mom.Angelman syndrome if the deletion is on the chromosome inherited from dad. A is the normal chromosome 15 a contains the deletion

A is from mom A’ is from dad

Aa A’a’

A’a AA’ aa’

Which disorder does the offspring 3 have?A) both syndromesB) PWSC) ASD) need more information

1 2 3

Page 98: Inheritance Patterns and Probability July 2008. Pedigrees

An individual with AS married a normal individual and produced anoffspring with PWS. What is the gender of the parent with AS?

A) maleB) femaleC) need more information

Prader-Willi if the deletion is on the chromosome inherited from mom.Angelman syndrome if the deletion is on the chromosome inherited from dad.

What is the gender of the child with PWS?A) maleB) femaleC) need more information

Page 99: Inheritance Patterns and Probability July 2008. Pedigrees

Dominance vs. recessive

Page 100: Inheritance Patterns and Probability July 2008. Pedigrees

•A chuzzle populations contains 640 red chuzzels and 320 green chuzzles. •Chuzzles are not choosy about their mates. Either color will mate with the other at equal frequencies. •When red chuzzles mate all the pups are red. When red and greenchuzzles mate some pups are red and some are green.•There is no advantage (for mating or survival) based on color.

Which trait is dominant?A) redB) green

Chuzzle Population


Recommended