Practice Problems in Mendelian Genetics
Important Note: Get in the habit right from the first of writing down the work necessary to solve the
problems you do. You will be required to show work on any assignment or exam problem.
II.
Problems Involving Two Genes
1.
A man with dark (dominant), curly (see problem 5 from previous homework) hair marries a woman with
light, straight hair. Their daughter, who happens to have dark hair, marries a man with light, wavy hair.
Answer the following questions about this dark-haired daughter and her family.
a. Draw a Punnett’s square for this marriage, and predict the phenotypic ratio among the offspring
of the daughter and her husband.
Curly
: HC
Straight allele : HS
Genotypes:
 Curly haired person
 Wavy
 Straight
: HC HC
: HC HS
: HSHS
Dark hair color: D
Light hair color: d
To find the genotype of the daughter:
Cross: Dd HCHC x dd HSHS
D HC
d HS
C
Dd H H
d HC
S
dd HCHS
Therefore, since the question says that the daughter has dark hair, she must be Dd H CHS.
Next, the cross between the daughter and her husband:
Cross: Dd HCHS x dd HCHS
d HC
HC
D
d HC
D HS
d HS
C
d HS
C
Dd HCHS
dd HCHC
dd HCHS
Dd HCHS
Dd HSHS
dd HCHS
dd HSHS
Dd H H
Offspring ratio:
Dark curly : Dark wavy : Dark straight : Light curly : Light wavy : Light Straight = 1 : 2 : 1 : 1 : 2 : 1
b.
What is the chance that they will have a child with hair just like his or her father’s?
A light wavy child is 2 out of 8 or 1/4th or 25%.
2.
In cats, again, black color is dominant to a special, temperature-sensitive albino gene which produces
cats with dark legs, faces and tails (Siamese cats, in case you don’t recognize it). A short haired
(dominant) Siamese colored female is bred to a long-haired black male. They have eight kittens: 2
black, short-haired; 2 black, long-haired; 2 Siamese, short-haired; and 2 Siamese, long-haired. What
were the genotypes of the two parents?
Black
:B
Siamese : b
Short
Long
:S
:s
Parents:

The female could be SS bb or Ss bb but since we know she has kittens that are long haired,
she needs to have a “s” so she is Ss bb

Same with the male. He could be ss BB or ss Bb. Since he has kittens that are Siamese, he
needs to have a “b” so he is ss Bb
To verify:
Cross: Ss bb x ss Bb
sB
sb
Sb
sb
Ss Bb
ss Bb
Ss bb
ss bb
Short black : Long black : Short Siamese : Long Siamese = 1 : 1 : 1 : 1
3.
Elizabeth is married to John, and they have four children. Elizabeth has a straight nose (recessive)
and is able to roll her tongue (dominant). John is also able to roll his tongue, but he has a convex
(Roman) nose (dominant). Of their four children, Ellen is just like her father, and Dan is just like his
mother. The other children—Anne, who has a convex nose, and Peter, who has a straight nose—are
unable to roll their tongues. Please answer the following questions about this family.
a. What are the genotypes of Elizabeth and John?
Convex nose
Straight nose
:N
:n
Roller
Non-roller
:R
:r
Elizabeth has a straight nose and is a roller. She could be nn RR or nn Rr. Since she has children that
are non-rollers, she must have given them a “r”, so therefore she has to be nn Rr.
John has a convex nose and is a roller. We can use the same logic as Elizabeth and therefore say he
has to be Rr so he could be NN Rr or Nn Rr. And, since he has straight nosed children, he needs to be
Nn Rr.
NR
Nr
nR
nr
nR
nr
Nn RR
Nn Rr
Nn Rr
Nn rr
nn RR
nn Rr
nn Rr
nn rr
Convex nose roller : Convex nose non-roller : Straight nose roller : Straight nose non-roller
3
:
1
:
3
:
1
b.
Elizabeth’s father was a straight-nosed roller, while her mother was a convex-nosed non-roller.
What can you figure out about their genotypes?
Elizabeth: nn Rr
Elizabeth’s father: nn RR or nn Rr
Elizabeth’s mother: NN rr or Nn rr
Since Elizabeth has two little n’s she must have gotten one from each parent and therefore her
mother needs to be Nn rr.
In terms of Elizabeth’s father, nothing further can be figured out.
c.
John’s father was a straight-nosed roller, while his mother was a convex-nosed roller. What can
you determine about their genotypes?
John: Nn Rr
John’s father: nn RR or nn Rr
John’s mother: NN RR or Nn RR or NN Rr or Nn Rr
There doesn’t seem to be anyway to logical to narrow down the choices.
d.
4.
Diagram the three described generations of this family in accepted pedigree form, including the
phenotypes for these two traits.
If a pure-breeding (homozygous) black (dominant), long-haired (recessive) cat is mated to a purebreeding Siamese, short-haired cat, and one of their male offspring is mated to one of their female
offspring, what is the chance of producing a Siamese colored, short-haired kitten?
Black: B
Siamese: b
Short: S
Long: s
Parental cross: BB ss x bb SS
All offspring with be Bb Ss.
F1 cross: Bb Ss x Bb Ss
This is the big, long 4 by 4 punnett square. Feel free to try it if you need the practice.
The phenotypic ratio will be:
Black short : Black long : Siamese short : Siamese long = 9 : 3 : 3 : 1
Therefore, the chance of a Siamese short haired cat will be 3 out of 16 or 3 / 16 = 18.75%
5.
In horses, one which runs best in water (or in wet conditions) is called (WATER), and one which runs
best in dry conditions is called (DRY). (WATER) is recessive to (DRY). A horse can also be either a
trotter, which we will designate (GAIT) or a pacer, which we will designate (PACE). (PACE) is recessive
to (GATE). We have mated two horses, a stallion named Halter-Man and a mare named Erlich-Mane.
Halter-Man is a (WATER)(PACER), while Erlich-mane is a (DRY)(GAIT). One of Erlich-Mane’s parents
was a (WATER)(PACER). What are the chances of Erlich-Mane and Halter-Man producing a
(WATER)(GAIT) foal (that’s a baby horse, in case you didn’t know)?
Oh boy…trying to use their conventions is a recipe for disaster so putting it into letters
Best dry (DRY) : B
Best wet (WET) : b
Runs trot (GAIT)
Runs pace (PACE)
Cross:
:R
:r
Halter-Man
bb rr
x
x
Erlich-Mane (who had a bb rr parent, therefore who is heterozygous)
Bb Rr
br
BR
Br
bR
br
Bb Rr
Bb rr
bb Rr
bb rr
DRY GAIT : DRY PACE : WATER GAIT : WATER PACE = 1 : 1: 1 : 1
Therefore, the chance of producing a WATER GAIT foal is 25%
6.
When a male pig from a line of true-breeding (homozygous) black, solid-hooved pigs was crossed to a
female from a breed (homozygous) of red, cloven-hooved pigs, their several progeny all looked alike
with regard to color and hooves. These progeny were all mated to members of the same breed as their
red, cloven-hooved mother pig. The offspring from this final cross were: 11 black, cloven-hooved; 8
black, solid-hooved; 14 red, cloven-hooved; and 10 red, solid-hooved. For each of these two genes (coat
color and hoof type) determine which allele is the dominant one. Explain your reasoning. What were
the phenotypes of the progeny produced by the first mating in this problem?
Having just done question 5, that ratio at the end of 1:1:1:1 looks awfully similar to the offspring here
so it’s basically the same question/crosses…but if we didn’t have that question…
Cross: Black, solid hooved x Red, cloven hooved
The question doesn’t tell us the genotype or phenotype of the offspring but does say that 100% of
them look the same. Since the question also states that both of the original pigs are homozygous we
can say that whatever these offspring are, they are completely heterozygous. Think of Mendel’s F1
generation.
Next cross: Unknown heterozygous offspring x Red, cloven hooved
Offspring:

11 black, cloven

8 black, solid

14 red, cloven

10 red, solid
Ok, crazy logic time. The Red, cloven pig could be (using simple letters):

RR CC

RR cc

rr CC

rr cc
If the first one was true, crossing a RR CC with a heterozygote (Rr Cc) would give only one type of
offspring, red cloven, so that one is out…(check the Punnett square if you can’t visualize)
If the second one was true, crossing a RR cc with the Rr Cc would give two types of offspring red
solid and red cloven, so that one is out…
If the third one was true, crossing a rr CC with the Rr Cc would give two types of offspring black solid
and black cloven, so that one is out…
Finally the fourth one:
Cross: rr cc x Rr Cc
rc
RC
Rc
rC
rc
Rr Cc
Rr cc
rr Cc
rr cc
Ratio:
Black solid : Black cloven : Red solid : Red cloven = 1 : 1 : 1 : 1
Just like the offspring!
Therefore:
Black : R
Red : r
Solid : C
Cloven : c
And the original parents were:
Black solid x Red cloven
RR CC
x rr cc
7.
In garden peas, long stems are dominant to short stems, and yellow seeds are dominant to green
seeds. 100 long/yellow pea plants, all of which had one short/green parent, are interbred (bred to
each other). 1600 progeny result. Please answer the following questions about these progeny.
a. Assuming that these two genes are unlinked, about how many long/green pea plants would you
expect to find among the offspring?
b. What ratio of yellow to green seed color would you expect among the offspring?
c. What would you expect the overall phenotypic ratio among the 1600 offspring to be (taking into
consideration both traits)?