Practice Problems in Mendelian Geneticsanswers

Practice Problems in Mendelian Geneticsanswers

Practice Problems in Mendelian GeneticsAnswers

IProblems Involving One Gene

  1. S: short hair; s: long hair

SS X ssshould yield 100% heterozygous, short haired cats

  1. ss X _ _yielded two S_ and 3 ss

Unknown parent is most likely heterozygous to produce 1:1 ratio in offspring.

  1. P:Widow’s peak; p:straight hairline

Mr SmithMrs Smith

S_S_

First child is S_, second is ss. Second child could belong to Mr Smith on basis of hairline. If Mr and Mrs are both heterozygous, expected ratio of dominant to recessive among their children would be 3:1

  1. F:Free earlobes;f: attached

Mr Jones:Mrs Jones

ffF_

Three children are ff and three are F_. Mrs Jones and children with dominant character must be heterozygous.

  1. HH: tightly curled;H1H1: straightHH1: wavy

Mr. AndersonMrs. Anderson

HHHH

Child: HH1 does not appear possible from these parents.

  1. HH1 ----HH1 expect 1:2:1 genotypic and phenotypic ratios

Therefore, of 8 kids: two straight, four wavy and two curly haired children.

If three of the children have curly hair, would not expect that anything went “wrong”. 1:2:1 is and expected ratio, like 4 heads and 4 tails in 8 coin tosses.

  1. DD: chestnut (dark brown); D1D1: pale cream;DD1: palomino

Since only heterozygotes are palominos, it would not be possible to produce a pure-breeding herd of palomino horses. Expected ratio among offspring of two palominos: 1:2:1 (chestnut: palomino: pale cream).

  1. Carriers (heterozygous individuals) do not show symptoms, can have children with homozygous dominant individuals with no Tay Sachs showing up, but the allele will still be in the population.
  1. Very little. Dominant does not mean frequent; recessive does not mean rare. One allele required to show dominant and two required to show recessive.
  1. CC: lethal; Cc: curly wings; cc: normal (straight) wings. (Choosing to use one letter instead of two)

Cc X Cc: expect 1:2:1 ratio homozygous: heterozygous: homozygous

Of 100 eggs,

  1. Expect 75 living offspring
  2. Expect 25 normal wings
  3. Expect 50 curly wings.
  1. DD: normalDd: dwarf;dd: lethal

DdXDd: expect 2:1 ratio of dwarf to normal. 1/4 of offspring do not show up.

II Problems Involving Two Genes

  1. D: dark;d: light;CC: curly;Cc: wavycc: straight

Father Mother

D_CC -----ddcc

Daughter-----Husband

DdCc ddCc

  1. dCdc

DCDdCCDdCc

DcDdCcDdcc

dCccCCccCc

dcddCcddcc

  1. Chance of child with hair type like father’s: 1/8
  1. D: dark;d: SiameseS: short hair;s: long hair

S_ddfemale X ssD__ male

Kittens:

2: S_D_,2: ssD_,2: S_dd,2: ssdd

1:1:1:1 ratio expected if parents heterozygous for the dominant traits.

  1. C: convex nose line;c: straightR: tongue roller;r: non roller

ElizabethJohn

ccR_C_R_

Children:Ellen: C_R_

Dan:ccR_

Anne: C_rr

Peter:ccrr

  1. Must be heterozygous for dominant traits to have homozygous recessive children.
  2. Elizabeth’s father: ccR_, mother C_rr

Mom must have been heterozygous for C, can’t determine dad’s genotype for nose line from this information alone.

  1. John’s father: ccR_, mother C_R_

Cannot determine mom’s genotype for nose line from this info alone; one parent would have had to be heterozygous for tongue rolling.

  1. ssDD X SSdd

Expected F1: 100% heterozygous.

If F1 crossed, chance of S_dd offspring: 3/4 x1/4= 3/16.

  1. D: Dry,d: waterG: trotterg: pacer

StallionMare (one parent was ddgg)

ddggddG_

Chance of ddG_ foal: 1 x 1/2 = 0.5

  1. F1 100% heterozygous. If heterozygous crossed with homozygous recessive for two traits, we expect the 1:1:1:1 ratio observed. If heterozygous crossed with homozygous dominant, only dominant offspring result. Therefore original red, cloven-hooved mother must have been homozygous recessive. Black and solid are dominant. F1 would have been dominant for both.
  1. L: long stems;l: shortY: yellow seeds;y: green

100 LlYy (heterozygous because of homozygous recessive parents) produced 1600 offspring (progeny)

  1. Chance of L_yy from this cross: 3/4 x 1/4 = 3/16
  2. Expected ratio of yellow to green: 3:1
  3. Overall ratio in equivalent of F2 of a dihybrid cross: 9:3:3:1

IV Problems Involving Genes With Multiple Alleles

  1. Parent 1Parent 2

IA__IB__

Children:1. IA__,2. IB__,3. IAIB,4. ii

Parents must be heterozygous with recessive allele.

First and second child also heterozygous with recessive allele.

  1. Mrs Smith: IA__, Mr Smith: IB__

First child: IAIB, second child ii.

Both are possibly children of these parents. They would each be heterozygous with recessive allele.

  1. Alleged father: ii

Mother: IA

Child: type IAIB

  1. Mother could be IAIA or heterozygous IAi
  2. Genetics do not support court’s decision.
  1. Baby #1 could belong to either set of parents based on blood type.

Baby #2 could not belong to Mr and Mrs Simon – would have to inherit one dominant allele from Mr Simon, and be either type A or B.

  1. Man: iiWoman: IAIB

None of their children could be expected to have the blood type of either parent.

Expect half of children to be IA, half to be IB.

  1. Red X Red: Red is dominant – could produce either all reds, or three quarters red, one quarter white if both heterozygous.

Blue X Blue: Blue is dominant – as for red.

Purple X Purple: Red and Blue incompletely dominant. Heterozygous produces purple. Cross of two heterozygous individuals with incomplete dominance yields 1:2:1 genotypic and phenotypic ratios.

White X Red: White recessive to red. If all offspring are red, dominant parent was homozygous.\

If half of offspring are white, dominant parent was heterozygous.

White X Blue: White recessive to blue. As above for red X white.

White X Purple: Homozygous recessive crossed with heterozygous for blue and red alleles will produce 1:1 ratio of blue and red.

Red X Blue: Some offspring will always be heterozygous with red and blue alleles – purple. If parents are heterozygous with recessive allele, blue, red and white offspring will result.

Red X Purple: R: red, R1: blue. R_ X RR1 could produce homozygous for red, heterozygous for red, heterozygous for blue, or heterozygous with red and blue allele (purple)

Blue X Purple: as for the cross above – substitute blue for red.

RR redMuch like human blood types: three alleles, in this case the two

Rr redalleles are dominant over a third, but incompletely dominant

RR1purplewith respect to each other, as opposed to co-dominant as IA and IB

R1R1bluealleles.

Rrblue

rrwhite

  1. Mother: IA__Man #1: IAIBMan #2: IB__ Man #3: ii

Daughter: ii

  1. In terms of blood type, either #2 or #3 could be the father.
  2. Sex-linked: not this unit. #2’s the man.
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