Genetic problems
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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

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GENETICPROBLEMS


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 2n

  • Where n = # of heterozygous

    1.aaBb= 2

    2.CCDdee= 2

    3.AABbCcDD= 4

    4.MmNnOoPpQq= 32

    5.UUVVWWXXYYZz= 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

  • 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 #2A

W W

wWw Ww

fg F1 generation

wWw Wwall heterozygous


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 #2B

W w

W WW Ww F2 generation

w Ww ww

genotype: 1:2:1 ratio

phenotype: 3:1 ratio


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 #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

  • 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 #2D

W w

w Ww ww

w Ww ww

phenotypes:1:1 ratio

genotypes:1:1 ratio


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

  • 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 #3A

R R

rRr RrF1generation

r Rr Rr

phenotypes:100%pinkgenotypes:100%heterozygous


Question #3B

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

  • Rr x Rr


Question #3B

R r

RRR RrF2generation

rRrrr

phenotypes: 1:2:1 ratio

genotypes: 1:2:1 ratio


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 #3C

R R

RRR RR

r Rr Rr

50%:red flowered

50%:pink flowered


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 #3D

R r

rRrrr

rRrrr

50%:white flowered

50%:pink flowered


Question #4

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

  • Remember:XX - female

    XY - male


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 #4A

XC Y parents

XC XCXC XCY

Xc XCXcXcY

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

50%: male (one normal, one colorblind)


Question #4B

  • A couple has a colorblind daughter.

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


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

  • 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

  • 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

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

  • Cross: FfDd x FfDd

  • This is a dihybrid cross


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 #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 #5B

Genotypic ratio:FFDD- 1

FFDd- 2

FFdd- 1

FfDD- 2

FfDd- 4

Ffdd- 2

ffDD- 1

ffDd- 2

ffdd- 1


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

  • 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 #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

  • 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

  • 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

  • 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

  • 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

  • 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 #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

  • 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

  • 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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