Download Genetics Probability Practice Problems

Practice Genetics Problems – Calculating Probabilities
1. The genotype of F1 individuals in a tetrahybrid cross is AaBbCcDd
(AaBbCcDd x AaBbCcDd). Assuming independent assortment of these four genes, what
are the probabilities that F2 offspring will have the following genotypes?
A) aabbccdd
B) AaBbCcDd
C) AABBCCDD
D) AaBBccDd
E) AaBBCCdd
2. Karen and Steve each have a sibling with sickle-cell disease (recessive disorder).
Neither Karen nor Steve nor any of their parents have the disease, and none of them have
been tested to reveal sickle-cell trait. Based on this incomplete information, calculate the
probability that if this couple has a child, the child will have sickle-cell disease.
3. Imagine that you are a genetic counselor, and a couple planning to start a family comes
to you for information. Charles was married once before, and he and his first wife had a
child with cystic fibrosis (a recessive disorder). The brother of his current wife, Elaine,
died of cystic fibrosis. What is the probability that Charles and Elaine will have a baby
with cystic fibrosis? (Neither Charles nor Elaine has cystic fibrosis.)
Problem #1 Soultion: First, create Punnett Squares for the 4 individual Monohybrid
crosses:
A
a
B
b
A
AA
Aa
B
BB
Bb
a
aA
aa
b
bB
bb
C
c
D
d
C
CC
Cc
D
DD
Dd
c
cC
cc
d
dD
dd
Each Monohybrid cross can now have individual probabilities calculated for each
genotype as follows:
Monohybrid Cross #1, Aa x Aa  AA = ¼; Aa = ½; aa = ¼
Monohybrid Cross #2, Bb x Bb  BB = ¼; Bb = ½; bb = ¼
Monohybrid Cross #3, Cc x Cc  CC = ¼; Cc = ½; cc = ¼
Monohybrid Cross #4, Dd x Dd  DD = ¼; Dd = ½; dd = ¼
Part A) now wants us to consider the probability of having an offspring with the genotype
of “aabbccdd,” so we must consider the four individual probabilities linked (“aa” and
“bb” and “cc” and “dd”), leading us to multiplication of the individual probabilities as
follows:
aa x bb x cc x dd = ?
¼ x ¼ x ¼ x ¼ = 1/256
Part B) now wants us to consider the probability of having an offspring with the genotype
of “AaBbCcDd,” so we must consider the four individual probabilities linked (“Aa” and
“Bb” and “Cc” and “Dd”), leading us to multiplication of the individual probabilities as
follows:
Aa x Bb x Cc x Dd = ?
½ x ½ x ½ x ½ = 1/16
Part C) now wants us to consider the probability of having an offspring with the genotype
of “AABBCCDD,” so we must consider the four individual probabilities linked (“AA”
and “BB” and “CC” and “DD”), leading us to multiplication of the individual
probabilities as follows:
AA x BB x CC x DD = ?
¼ x ¼ x ¼ x ¼ =
1/256
Part D) now wants us to consider the probability of having an offspring with the genotype
of “AaBBccDd,” so we must consider the four individual probabilities linked (“Aa” and
“BB” and “cc” and “Dd”), leading us to multiplication of the individual probabilities as
follows:
Aa x BB x cc x Dd = ?
½ x ¼ x ¼ x ½ = 1/64
Part E) now wants us to consider the probability of having an offspring with the genotype
of “AaBBCCdd,” so we must consider the four individual probabilities linked (“Aa” and
“BB” and “CC” and “dd”), leading us to multiplication of the individual probabilities as
follows:
Aa x BB x CC x dd = ?
½ x ¼ x ¼ x ¼ = 1/128
Problem #2 Solution:
First, create a list to summarize the information provided above:
Karen – No Disease
Karen’s Sibling – Disease
Karen’s Parents – Neither has Disease
Steve – No Disease
Steve’s Sibling – Disease
Steve’s Parents – Neither has Disease
Second, reason through what each individual above has for a genotype based on the
phenotypes provided – we will assign a “S” for the dominant allele and a “s” for the
recessive allele:
Karen = SS or Ss
Steve = SS or Ss
Karen’s Sibling = ss
Steve’s Sibling = ss
Karen’s Parents = both are Ss
Steve’s Parents = both are Ss
We know that Karen’s sibling and Steve’s sibling must be “ss” because each sibling has
the disease (this only manifests when there are 2 copies of the “s” allele). Where did each
sibling get those alleles? – from their parents. Therefore, Karen’s and Steve’s parents
must have at least one copy of the “s” allele; however, the parents do not have the
disease, therefore they must be heterozygotes (carriers with genotype of “Ss”). Still we
do not know what Karen’s or Steve’s genotype is specifically – it could be either “SS” or
“Ss” since either of these is a possibility that yields no disease, resulting from having
heterozygote parents (Ss x Ss) – we know for sure that neither Karen nor Steve can be
“ss” because they do not have the disease. See below:
Karen’s Parental Cross
S
s
Possible genotypes
S
SS
Ss
for Karen
s
sS
ss
Not a possible
genotype for
Karen
Steve’s Parental Cross
S
s
Possible
S
SS
Ss
genotypes
for Steve
s
sS
ss
Not a possible
genotype for
Steve
Third, we now need to know the probability of Karen and Steve each being heterozygotes
(“Ss”) since this is the only way there is the possibility of having a child with Sickle-cell
Disease. As we can see from above, Karen has a 2 in 3 chance of being “Ss” and so does
Steve. Therefore the probability that both are heterozygous (Karen = Ss and Steve = Ss),
we take 2/3 x 2/3 = 4/9.
Fourth, we must now calculate the final individual probability of Karen and Steve’s child
having the disease:
Karen and Steve Cross
S
s
S
SS
Ss
s
sS
ss
Child with disease = ¼ probability
Lastly, we put all of this together in determining the combined probability that
Karen is heterozygous and Steve is heterozygous and that their child has the disease 
2/3
x
2/3
x
¼
= 4/36 = 1/9
Problem #3 Solution: Use problem #2 as a guide to reason your way through to the
realization that the probability of Charles being heterozygous is 1, and the probability of
Elaine being heterozygous is 2/3, and the probability that their child has the disease is ¼
 therefore,
1 x 2/3 x ¼ = 2/12 = 1/6