Practice Problems in Mendelian Genetics Answers
I Problems Involving One Gene
1. S: short hair; s: long hair
SS X ss
2. ss X _ _
should yield 100% heterozygous, short haired cats
yielded two S_ and 3 ss
Unknown parent is most likely heterozygous to produce 1:1 ratio in offspring.
3. P: Widow’s peak;
p: straight hairline
Mr Smith Mrs 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
4. F: Free earlobes; f: attached
Mr Jones: Mrs Jones
ff F_
Three children are ff and three are F_.
heterozygous.
5. HH: tightly curled; H1H1: straight
Mrs Jones and children with dominant character must be
HH1: wavy
Mr. Anderson Mrs. Anderson
HH
HH
Child: HH1 does not appear possible from these parents.
6. 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.
7. 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).
8. 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.
II
Problems Involving Two Genes
1. D: dark;
d: light;
CC: curly; Cc: wavy cc: straight
Father
Mother
D_CC -----ddcc
Daughter-----Husband
DdCc
ddCc
a.
dC dc
DC DdCC DdCc
Dc DdCc Ddcc
dC ccCC ccCc
dc ddCc ddcc
b. Chance of child with hair type like father’s: 1/8
2. D: dark;
d: Siamese S: short hair; s: long hair
S_dd female 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.
3. C: convex nose line; c: straight R: tongue roller; r: non roller
Elizabeth John
ccR_
C_R_
Children: Ellen: C_R_
Dan: ccR_
Anne: C_rr
Peter: ccrr
a. Must be heterozygous for dominant traits to have homozygous recessive children.
b. 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.
c. 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.
4. ssDD
X
SSdd
Expected F1: 100% heterozygous.
If F1 crossed, chance of S_dd offspring:
3/4
x 1/4 = 3/16.
III. Problems Involving Sex Linkage
1. Cross 1 – offspring all females red eyed hybrid XWXR; all males white eyed hemizygous XWY
Cross 2 – offspring all females red eyed hybrid XWXR; all males red eyed hemizygous XRY
2. all of their daughters will be carriers; and all of their sons will have color blindness.
3. all of their daughters will be carriers; and all of their sons will be normal vision; predict 4 normal vision
carrier girls and 4 colorblind boys.
4.
a. 50% chance Mickey is colorblind
b. Marian’s father; Martin gave him his Y; Marian gave him his X which she inherited from her father.
c. do this…..
d. wouldn’t matter. Boys don’t get any X information from their fathers.
IV Problems Involving Genes With Multiple Alleles
1.
Parent 1 Parent 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.
2. 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.
3. Alleged father:
ii
Mother: IA
Child: type IAIB
a. Mother could be IAIA or heterozygous IAi
b. Genetics do not support court’s decision.
4. 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.
5. Man:
ii
Woman: 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.
7. Mother:
IA__ Man #1:
IAIB Man #2:
IB__
Man #3:
Daughter:
ii
a. In terms of blood type, either #2 or #3 could be the father.
b. Sex-linked: not this unit. #2’s the man.
ii