Ppt genetics problems

GENETICGENETICPROBLEMSPROBLEMS

Question #1Question #1

• How many different kinds of gametesgametes could the following individuals produce?

1.1. aaBbaaBb

2.2. CCDdeeCCDdee

3.3. AABbCcDDAABbCcDD

4.4. MmNnOoPpQqMmNnOoPpQq

5.5. UUVVWWXXYYZzUUVVWWXXYYZz


• Remember the formula Remember the formula 22nn

• Where Where n = # of heterozygousn = # of heterozygous

1.1. aaBbaaBb = 2= 2

2.2. CCDdeeCCDdee = 2= 2

3.3. AABbCcDDAABbCcDD = 4= 4

4.4. MmNnOoPpQqMmNnOoPpQq = 32= 32

5.5. UUVVWWXXYYZzUUVVWWXXYYZz = 2= 2


• In dogs, wire-haired is due to a In dogs, wire-haired is due to a dominant gene (W), smooth-dominant gene (W), smooth-haired is due to its recessive haired is due to its recessive allele (w).allele (w).

• WW, Ww WW, Ww = wire haired= wire haired• wwww = smooth haired= smooth haired

Question #2AQuestion #2A

• If a homozygous wire-haired dog is If a homozygous wire-haired dog is mated with a smooth-haired dog, what mated with a smooth-haired dog, what type of offspring could be produced.type of offspring could be produced.

W WW W

ww

ww


W WW W

ww Ww WwWw Ww

fg Ffg F11 generation generation

ww Ww WwWw Ww all all heterozygousheterozygous

Question #2BQuestion #2B

• What type(s) of offspring could be What type(s) of offspring could be produced in the produced in the FF22 generation generation??

• Must breed the Must breed the FF11 generation generation to get the to get the FF22..

• Results of F1 Cross: Ww x WwResults of F1 Cross: Ww x Ww


W wW w

WW WW Ww WW Ww FF22 generation generation

ww Ww ww Ww ww

genotype: 1:2:1 ratiogenotype: 1:2:1 ratio

phenotype: 3:1 ratiophenotype: 3:1 ratio

Question #2CQuestion #2C

• Two wire-haired dogs are mated. Among Two wire-haired dogs are mated. Among the offspring of their first litter is a the offspring of their first litter is a smooth-haired pup.smooth-haired pup.

• If these, two wire-haired dogs mate If these, two wire-haired dogs mate again, what are the chances that they again, what are the chances that they will produce will produce another smooth-haired pup?another smooth-haired pup?

• What are the chances that the pup will What are the chances that the pup will wire-haired pup?wire-haired pup?


W wW w

WW WW Ww WW Ww FF22 generation generation

ww Ww ww Ww ww

- 1/4 or 25% chance for smooth-haired- 1/4 or 25% chance for smooth-haired

- 3/4 or 75% chance for wire-haired- 3/4 or 75% chance for wire-haired

Question #2DQuestion #2D

• A wire-haired male is mated with a A wire-haired male is mated with a smooth-haired female. The mother of smooth-haired female. The mother of the wire-haired male was smooth-the wire-haired male was smooth-haired.haired.

• What are the What are the phenotypesphenotypes and and genotypesgenotypes of the pups they could of the pups they could produce?produce?

• Show the results of crossing: Ww x wwShow the results of crossing: Ww x ww


W wW w

ww Ww ww Ww ww

ww Ww ww Ww ww

phenotypes:phenotypes: 1:1 ratio1:1 ratio

genotypes:genotypes: 1:1 ratio1:1 ratio

Question #3Question #3• In snapdragons, In snapdragons, red flower (R)red flower (R) color is color is

incompletely dominantincompletely dominant over over white white flower (r)flower (r) color. color.

• The heterozygous (Rr) plants have The heterozygous (Rr) plants have pinkpink flowers.flowers.

RR RR - red flowers- red flowers

Rr Rr - pink flowers- pink flowersrr rr - white flowers- white flowers


• If a If a red-floweredred-flowered plant is crossed plant is crossed with a with a white-flowered plantwhite-flowered plant, what , what are the genotypes and are the genotypes and phenotypes of the plants Fphenotypes of the plants F11 generation?generation?

• RRRR x x rr rr


R RR R

rr Rr RrRr Rr FF11 generationgeneration

r r Rr RrRr Rr

phenotypes:phenotypes: 100%100% pinkpinkgenotypes:genotypes: 100%100%

heterozygousheterozygous


• What genotypes and What genotypes and phenotypes will be produced in phenotypes will be produced in the Fthe F22 generation? generation?

• Rr Rr x x RrRr


R rR r

RR RRRR Rr Rr FF22 generationgeneration

rr RrRr rrrr

phenotypes: 1:2:1 ratiophenotypes: 1:2:1 ratio

genotypes: 1:2:1 ratiogenotypes: 1:2:1 ratio


• What kinds of offspring can be What kinds of offspring can be produced if a produced if a red-flowered plantred-flowered plant is crossed with a is crossed with a pink-flowered pink-flowered plantplant??

• RRRR x x RrRr


R R

R RR RR

r Rr Rr

50%: red flowered50%: pink flowered


• What kind of offspring is/are What kind of offspring is/are produced if a produced if a pink-flowered plantpink-flowered plant is crossed with a is crossed with a white-flowered white-flowered plantplant??

• Rr Rr x x rr rr


R r

r Rr rr

r Rr rr

50%: white flowered50%: pink flowered


• In humans, colorblindness (cc) In humans, colorblindness (cc) is a recessive sex-linked trait.is a recessive sex-linked trait.

• Remember:Remember: XX - femaleXX - female

XY - maleXY - male


• Two normal people have a colorblind son.Two normal people have a colorblind son.

• What are the What are the genotypesgenotypes of the parents? of the parents?

• XXCCXX_?_? x x X XCCYY

• What are the What are the genotypesgenotypes and and phenotypesphenotypes possible among their other children?possible among their other children?


XXCC Y Y parents parents

XXCC X XCCXXCC X XCCYY

XXcc X XCCXXcc XXccYY

50%: female (one normal, one a carrier)50%: female (one normal, one a carrier)

50%: male (one normal, one colorblind)50%: male (one normal, one colorblind)


• A couple has a A couple has a colorblindcolorblind daughter.daughter.

• What are the possible What are the possible genotypesgenotypes and and phenotypesphenotypes of the parents of the parents and the daughter?and the daughter?

Question #4BQuestion #4B XXcc Y Y

XXCC X XCCXXcc X XCCYY

XXcc XXccXXcc X XccYY

parents: parents: XXccYY and and XXCCXXcc or or XXccXXcc

father father colorblindcolorblind

mother mother carriercarrier or or colorblindcolorblind

daughter: daughter: XXccXXcc - colorblind - colorblind


• In humans, the presence of In humans, the presence of frecklesfreckles is due to a dominant gene is due to a dominant gene (F)(F) and the and the non-frecklednon-freckled condition condition is due to its recessive allele is due to its recessive allele (f).(f).

• Dimpled cheeks (D)Dimpled cheeks (D) are dominant are dominant to to non-dimpled cheeks (d).non-dimpled cheeks (d).


• Two persons with freckles and dimpled Two persons with freckles and dimpled cheeks have two children: one has cheeks have two children: one has freckles but no dimples and one has freckles but no dimples and one has dimples but no freckles.dimples but no freckles.

• What are the What are the genotypesgenotypes of the of the parents?parents? Parents:Parents: F_F___DD____ x F x F____DD____

Children: FChildren: F____dd dd x x ff ffDD____


• What are the possible phenotypes What are the possible phenotypes and genotypes of the children that and genotypes of the children that they could produce?they could produce?

• Cross: Cross: FfDd x FfDdFfDd x FfDd

• This is a This is a dihybrid crossdihybrid cross


• Possible gametes for both: FD Fd fD Possible gametes for both: FD Fd fD fdfd

FD FD Fd fD fd Fd fD fd

FD FD FFDD FFDd FfDD FfDdFFDD FFDd FfDD FfDd

Fd Fd FFDd FFdd FfDd Ffdd FFDd FFdd FfDd Ffdd

fDfD FfDD FfDd ffDD ffDd FfDD FfDd ffDD ffDd

fdfd FfDd Ffdd ffDd ffdd FfDd Ffdd ffDd ffdd


PhenotypePhenotype:: Freckles/Dimples:Freckles/Dimples: 99

Freckles/no dimples:Freckles/no dimples:33

no freckles/Dimples:no freckles/Dimples:33

no freckles/no dimples:no freckles/no dimples:11

Phenotypic ratio will always been Phenotypic ratio will always been 9:3:3:1 for all F1 dihybrid crosses.9:3:3:1 for all F1 dihybrid crosses.


Genotypic ratio:Genotypic ratio: FFDDFFDD - 1- 1

FFDdFFDd - 2- 2

FFddFFdd - 1- 1

FfDDFfDD - 2- 2

FfDdFfDd - 4- 4

FfddFfdd - 2- 2

ffDDffDD - 1- 1

ffDdffDd - 2- 2

ffddffdd - 1- 1


• What are the chances that they What are the chances that they would have a child whom lacks would have a child whom lacks both freckles and dimples?both freckles and dimples?

• This child will have a genotype This child will have a genotype of of ffddffdd

• Answer:Answer: 1/161/16


• A person with freckles and dimples A person with freckles and dimples whose mother lacked both freckles and whose mother lacked both freckles and dimples marries a person with freckles dimples marries a person with freckles but not dimples whose father did not but not dimples whose father did not have freckles or dimples.have freckles or dimples.

• Cross:Cross: FfDdFfDd x x FfddFfdd• Possible gametes:Possible gametes:

FD Fd fD fd x Fd fdFD Fd fD fd x Fd fd

Question #5DQuestion #5D• What are the chances that they would What are the chances that they would

have a child whom lacks both freckles have a child whom lacks both freckles and dimples?and dimples?

FD Fd fD fdFD Fd fD fd

FdFd FFDd FFdd FfDd Ffdd FFDd FFdd FfDd Ffdd

fdfd FfDd Ffdd ffDd FfDd Ffdd ffDd ffddffdd

Answer:Answer: 1/81/8


• Sixteen percent of the human population Sixteen percent of the human population is known to be able to wiggle their ears.is known to be able to wiggle their ears.

• This trait is determined to be a This trait is determined to be a recessive recessive gene.gene.

• These is a These is a population geneticspopulation genetics question. question.

• Use the following equation: Use the following equation: 1 = p1 = p22 + 2pq + 2pq + q+ q22

Question #6AQuestion #6A• What of the population is What of the population is homozygous homozygous

dominantdominant for this trait? for this trait?

• qq22 = 16% or .16: = 16% or .16: q2 = q2 = .16.16

q = .4q = .4• then usethen use :: 1 = p + q1 = p + q

1 = p + .41 = p + .4

1- .4 = p1- .4 = p

p = .6p = .6

• Now use pNow use p2 2 for answer: .6for answer: .622 = .36 or 36% = .36 or 36%


• What of the population is What of the population is heterozygous heterozygous for this trait?for this trait?

• We know thatWe know that q = .4q = .4 andand p = .6p = .6

• Now use 2pq for answer: 2(.6)(.4) Now use 2pq for answer: 2(.6)(.4)

= .48 or 48%= .48 or 48%


• In dogs, the inheritance of hair color In dogs, the inheritance of hair color involves a gene B for black hair and involves a gene B for black hair and gene b for brown hair b.gene b for brown hair b.

• A A dominant Cdominant C is also involved. It must is also involved. It must be present for the be present for the colorcolor to be to be synthesized.synthesized.

• If this gene is not present, a If this gene is not present, a blondblond condition results.condition results.

BB, BbBB, Bb - black hair- black hair CC, CcCC, Cc - colorcolor

bbbb - brown hair- brown hair cccc - blondblond

Question #7AQuestion #7A• A brown haired male, whose father was A brown haired male, whose father was

a blond, is mated with a black haired a blond, is mated with a black haired female, whose mother was brown female, whose mother was brown haired and her father was blond.haired and her father was blond.

Male: Male: bbCc (gametes: bC bc)bbCc (gametes: bC bc)

Female: Female: BbCc (gametes: BC Bc BbCc (gametes: BC Bc bCbC bc)bc)

• What is the expected ratios of What is the expected ratios of their offspring?their offspring?


BCBC Bc bC Bc bC bcbc

bCbC BbCC BbCc bbCC bbCc BbCC BbCc bbCC bbCc

bc bc BbCc Bbcc bbCc bbcc BbCc Bbcc bbCc bbcc

Offspring ratios:Offspring ratios:Black:Black: 3/83/8

Brown:Brown: 3/83/8

Blond:Blond: 2/8 or 1/42/8 or 1/4


• Henry Anonymous, a film star, was involved in a Henry Anonymous, a film star, was involved in a paternity case. The woman bringing suit had paternity case. The woman bringing suit had two children, on whose two children, on whose blood type blood type waswas A A and and the other whose the other whose blood type blood type waswas B B..

• Her Her blood typeblood type was was OO, the same as Henry’s!, the same as Henry’s!

• The judge in the case awarded damages to the The judge in the case awarded damages to the woman damages to the woman, saying that woman damages to the woman, saying that Henry had to be the father of at least one of the Henry had to be the father of at least one of the children.children.

Question #8AQuestion #8A• Obviously, the judge should be sentenced Obviously, the judge should be sentenced

to Biology. For Henry to have been the to Biology. For Henry to have been the father of both children, his father of both children, his blood typeblood type would have had to be what?would have had to be what?

IIAA I IBB Answer Answer

ii IIAAi i I IBBii

ii IIAAi Ii IBBii

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Ppt genetics problems