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GENETIC PROBLEMS. Question #1. How many different kinds of gametes could the following individuals produce? 1.aaBb 2.CCDdee 3.AABbCcDD 4.MmNnOoPpQq 5.UUVVWWXXYYZz. Question #1. Remember the formula 2 n Where n = # of heterozygous 1.aaBb= 2 2.CCDdee= 2

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Genetic problems

GENETICPROBLEMS


Question 1
Question #1

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

    1.aaBb

    2.CCDdee

    3.AABbCcDD

    4.MmNnOoPpQq

    5.UUVVWWXXYYZz


Question 11
Question #1

  • Remember the formula 2n

  • Where n = # of heterozygous

    1.aaBb= 2

    2.CCDdee= 2

    3.AABbCcDD= 4

    4.MmNnOoPpQq= 32

    5.UUVVWWXXYYZz= 2


Question 2
Question #2

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

  • WW, Ww = wire haired

  • ww= smooth haired


Question 2a
Question #2A

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

    W W

    w

    w


Question 2a1
Question #2A

W W

wWw Ww

fg F1 generation

wWw Wwall heterozygous


Question 2b
Question #2B

  • What type(s) of offspring could be produced in the F2 generation?

  • Must breed the F1 generation to get the F2.

  • Results of F1 Cross: Ww x Ww


Question 2b1
Question #2B

W w

W WW Ww F2 generation

w Ww ww

genotype: 1:2:1 ratio

phenotype: 3:1 ratio


Question 2c
Question #2C

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

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

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


Question 2c1
Question #2C

W w

W WW Ww F2 generation

w Ww ww

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

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


Question 2d
Question #2D

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

  • What are the phenotypes and genotypes of the pups they could produce?

  • Show the results of crossing: Ww x ww


Question 2d1
Question #2D

W w

w Ww ww

w Ww ww

phenotypes:1:1 ratio

genotypes:1:1 ratio


Question 3
Question #3

  • In snapdragons, red flower (R) color is incompletely dominant over white flower (r) color.

  • The heterozygous (Rr) plants have pink flowers.

    RR - red flowers

    Rr - pink flowers

    rr - white flowers


Question 3a
Question #3A

  • If a red-flowered plant is crossed with a white-flowered plant, what are the genotypes and phenotypes of the plants F1 generation?

  • RR x rr


Question 3a1
Question #3A

R R

rRr RrF1generation

r Rr Rr

phenotypes:100%pinkgenotypes:100%heterozygous


Question 3b
Question #3B

  • What genotypes and phenotypes will be produced in the F2 generation?

  • Rr x Rr


Question 3b1
Question #3B

R r

RRR RrF2generation

rRrrr

phenotypes: 1:2:1 ratio

genotypes: 1:2:1 ratio


Question 3c
Question #3C

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

  • RR x Rr


Question 3c1
Question #3C

R R

RRR RR

r Rr Rr

50%:red flowered

50%:pink flowered


Question 3d
Question #3D

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

  • Rr x rr


Question 3d1
Question #3D

R r

rRrrr

rRrrr

50%:white flowered

50%:pink flowered


Question 4
Question #4

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

  • Remember:XX - female

    XY - male


Question 4a
Question #4A

  • Two normal people have a colorblind son.

  • What are the genotypes of the parents?

  • XCX_? x XCY

  • What are the genotypes and phenotypes possible among their other children?


Question 4a1
Question #4A

XC Y parents

XC XCXC XCY

Xc XCXcXcY

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

50%: male (one normal, one colorblind)


Question 4b
Question #4B

  • A couple has a colorblind daughter.

  • What are the possible genotypes and phenotypes of the parents and the daughter?


Question 4b1
Question #4B

Xc Y

XC XCXc XCY

XcXcXc XcY

parents: XcY and XCXcor XcXc

father colorblind

mother carrier or colorblind

daughter: XcXc - colorblind


Question 5
Question #5

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

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


Question 5a
Question #5A

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

  • What are the genotypes of the parents?Parents:F__D__ x F__D__

    Children: F__dd x ffD__


Question 5b
Question #5B

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

  • Cross: FfDd x FfDd

  • This is a dihybrid cross


Question 5b1
Question #5B

  • Possible gametes for both: FD Fd fD fd

    FD Fd fD fd

    FD FFDD FFDd FfDD FfDd

    Fd FFDd FFdd FfDd Ffdd

    fD FfDD FfDd ffDD ffDd

    fd FfDd Ffdd ffDd ffdd


Question 5b2
Question #5B

Phenotype:Freckles/Dimples:9

Freckles/no dimples:3

no freckles/Dimples:3

no freckles/no dimples:1

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


Question 5b3
Question #5B

Genotypic ratio:FFDD- 1

FFDd- 2

FFdd- 1

FfDD- 2

FfDd- 4

Ffdd- 2

ffDD- 1

ffDd- 2

ffdd- 1


Question 5c
Question #5C

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

  • This child will have a genotype of ffdd

  • Answer:1/16


Question 5d
Question #5D

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

  • Cross:FfDd x Ffdd

  • Possible gametes:

    FD Fd fD fd x Fd fd


Question 5d1
Question #5D

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

    FD Fd fD fd

    Fd FFDd FFdd FfDd Ffdd

    fd FfDd Ffdd ffDd ffdd

    Answer:1/8


Question 6
Question #6

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

  • This trait is determined to be a recessive gene.

  • These is a population genetics question.

  • Use the following equation: 1 = p2 + 2pq + q2


Question 6a
Question #6A

  • What of the population is homozygous dominant for this trait?

  • q2 = 16% or .16:q2 = .16

    q = .4

  • then use:1 = p + q

    1 = p + .4

    1- .4 = p

    p = .6

  • Now use p2 for answer: .62 = .36 or 36%


Question 6b
Question #6B

  • What of the population is heterozygous for this trait?

  • We know thatq = .4andp = .6

  • Now use 2pq for answer: 2(.6)(.4) = .48 or 48%


Question 7
Question #7

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

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

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

    BB, Bb- black hairCC, Cc- color

    bb- brown haircc- blond


Question 7a
Question #7A

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

    Male: bbCc (gametes: bC bc)

    Female: BbCc (gametes: BC Bc bCbc)

  • What is the expected ratios of their offspring?


Question 7a1
Question #7A

BC Bc bCbc

bC BbCC BbCc bbCC bbCc

bc BbCc Bbcc bbCc bbcc

Offspring ratios:

Black:3/8

Brown:3/8

Blond:2/8 or 1/4


Question 8
Question #8

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

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

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


Question 8a
Question #8A

  • Obviously, the judge should be sentenced to Biology. For Henry to have been the father of both children, his blood type would have had to be what?

    IA IB Answer

    iIAi IBi

    iIAi IBi


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