# 2014 - Mendelian Genetics Quiz 2 - answer key

```SBI3U Biology – Genetics
Name_____________________________________
Mendelian Genetics Quiz #2
December 25, 2014
1. A rooster with gray feathers is mated with a hen of the same phenotype. Among their offspring,
15 chicks are gray, 6 are black, and 8 are white.
•
What is the simplest explanation for the inheritance of these colors in chickens?
Since we have 3 phenotypes, one of which is a blend of the other two, the simplest
hypothesis would be that this is incomplete dominance. See Punnett square for further
demonstration.
Black allele: FB
White allele: FW
Therefore:
• black chicken = FBFB
• grey chicken = FBFW
• white chicken = FWFW
Cross: Grey (FBFW) x Grey (FBFW)
FB
FW
FB
FBFB
FBFW
FW
FBFW
FWFW
And our phenotypic ratio will be black : grey : white = 1 : 2 : 1
which fits in with observed numbers of offspring (6 : 15 : 8).
•
What offspring would you predict from the mating of a gray rooster and a black hen?
Cross: Grey (FBFW) x Black (FBFB)
FB
FB
FB
FBFB
FBFB
FW
FBFW
FBFW
And our phenotypic ratio will be black : grey = 2 : 2 or 50% black and 50% grey.
2. A man has six fingers on each hand and six toes on each foot. His wife and their daughter have
the normal number of digits (5). Extra digits is a dominant trait. What fraction of this couple's
children would be expected to have extra digits?
This seems to be a simple Mendelian genetics problem with one dominant and one
recessive
trait so let’s try the Punnett square to verify.
Multidactyly (multiple fingers): M
Normal number of fingers : m
Therefore:
• Six fingers = MM or Mm
• Normal fingers = mm
Since the question says a six-fingered man has a child with normal fingers, we will
assume that he is heterozygous (Mm) since he will have had to give a little m to his
daughter.
Cross: Mm x mm
M
m
m
Mm
mm
m
Mm
mm
And our phenotypic ratio will be six fingered : normal = 2 : 2 or 50% chance of either.
3. A black guinea pig crossed with an albino guinea pig produced 12 black offspring. When the
albino was crossed with a second one, 7 blacks and 5 albinos were obtained.
•
What is the best explanation for this genetic situation?
This also seems to be a simple Mendelian genetics problem with one dominant and one
recessive
trait so let’s try the Punnett squares to verify.
Black : B
Albino : b
Therefore:
• Black = BB or Bb
• Albino = bb
Therefore, in the first cross, since there only were black offspring, we can assume
that it was a BB crosses with a bb
Cross: BB x bb
B
B
b
Bb
Bb
b
Bb
Bb
And our phenotypic ratio will be 100% black, just like the observed.
was a Bb
Next, since the next mating produced both black and white, we can assume that it
with the bb.
Cross: Bb x bb
probable
•
BEST
B
B
b
Bb
bb
b
Bb
bb
And our phenotypic ratio will be 50% black and 50% albino, which makes it statistically
to get offspring of 7 black and 5 albino.
Write genotypes for the parents, gametes, and offspring.
These are all present in the answer above. Note that what is presented above is the
4. A man with group A blood marries a woman with group B blood. Their child has group O blood.
•
What are the genotypes of these individuals?
Since this is an ABO blood type question, we use the symbols presented in class
The man with type A has to be either IAIA or IAi.
The woman with type B has to be either IBIB or IBi.
The child with type O has to be ii.
Therefore, in order to get that child, each parent would have had to donate a little i
so the cross would be:
Cross: Type A (IAi) x Type B (IBi)
•
What other genotypes and in what frequencies, would you expect in offspring from this
marriage?
Cross: Type A (IAi) x Type B (IBi)
IA
i
IB
IAIB
IBi
i
IAi
ii
And the ratios are IAIB : IAi : IBi : ii = AB : A : B : O = 1 : 1 : 1 : 1
5. Color pattern in a species of duck is determined by one gene with three alleles. Alleles H and I
are codominant, and allele i is recessive to both. How many phenotypes are possible in a flock of
ducks that contains all the possible combinations of these three alleles?
This, while using different letters, will work exactly like the ABO blood-typing.
Phenotypes:
• “H” phenotype = HH or Hi
• “I” phenotype = II or Ii
•
“HI” phenotype = HI
•
“i” phenotype = ii
6. Phenylketonuria (PKU) is an inherited disease caused by a recessive allele. If a woman and her
husband are both carriers, what is the probability of each of the following?
Alleles:
• Normal, dominant allele
• Disease, recessive allele
=P
=p
Cross: Pp x Pp
P
p
P
PP
Pp
p
Pp
pp
And our ratios will be:
• PP : Pp : pp = 1 : 2 : 1 = 25% : 50% : 25%
• Normal : diseased = 3 : 1 = 75% : 25%
a) all three of their children will be of normal phenotype
Product rule = 1st child normal x 2nd child normal x 3rd child normal = 75% x 75% x 75%
= 42.2%
b) one or more of the three children will have the disease
This question and “d” are beyond what we are doing in class but:
all normal
If the probability of them all being normal is 42.2%, the probability that they are NOT
is 100%-42.2% = 57.8%.
Saying that one or more will have the disease is another way of writing not all normal.
c) all three children will have the disease
Product rule = 1st child disease x 2nd child disease x 3rd child disease = 25% x 25% x 25%
= 1.56%
d) at least one child out of three will be phenotypically normal
100%-1.56% = 98.44%
(Note: Remember that the probabilities of all possible outcomes always add up to 1)
```