Practice Problems in Mendelian Genetics
I
Answers
Problems Involving One Gene
1. S: short hair;
SS X ss
s: long hair
should yield 100% heterozygous, short haired cats
2. ss X _ _
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_. Mrs Jones and children with dominant character must
be heterozygous.
5. HH: tightly curled;
H1H1: straight
HH1: wavy
Mr. Anderson
Mrs. Anderson
HH
HH
1
Child: HH 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.
9. Very little. Dominant does not mean frequent; recessive does not mean rare. One allele required
to show dominant and two required to show recessive.
10. 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,
a. Expect 75 living offspring
b. Expect 25 normal wings
c. Expect 50 curly wings.
11. DD: normal
Dd: dwarf;
dd: lethal
Dd XDd: 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: wavy
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;
cc: straight
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;
Elizabeth
ccR_
c: straight
John
C_R_
R: tongue roller;
r: non roller
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.
5. D: Dry,
d: water
G: trotter
Stallion
Mare (one parent was ddgg)
ddgg
ddG_
Chance of ddG_ foal: 1 x 1/2 = 0.5
g: pacer
6. 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.
7. L: long stems;
l: short
Y: yellow seeds;
y: green
100 LlYy (heterozygous because of homozygous recessive parents) produced 1600 offspring
(progeny)
a. Chance of L_yy from this cross: 3/4 x 1/4 = 3/16
b. Expected ratio of yellow to green: 3:1
c. Overall ratio in equivalent of F2 of a dihybrid cross: 9:3:3:1
IV Problems Involving Genes With Multiple Alleles
1. Parent 1
Parent 2
A
I __
IB__
Children:
1. IA__,
2. IB__,
3. IAIB,
Parents must be heterozygous with recessive allele.
First and second child also heterozygous with recessive allele.
4. ii
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.
6. 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
Rr
RR1
R1R1
Rr
rr
red
red
purple
blue
blue
white
Much like human blood types: three alleles, in this case the two
alleles are dominant over a third, but incompletely dominant
with respect to each other, as opposed to co-dominant as IA and IB
alleles.
7. Mother:
IA__
Man #1: IAIB Man #2: IB__
Man #3: ii
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.