139 SOLVEDTUTORIALEXAMPLES
139 SOLVED TUTORIAL EXAMPLES Tutorial 1 –
Example problems
In cucumbers, dull fruit (D) is dominant over glossy fruit (d), orange fruit (R) is dominant over
cream fruit (r), and bitter cotyledons (B) are dominant over nonbitter cotyledons (b). The
three characters are encoded by genes located on different pairs of chromosomes. A plant
homozygous for dull, orange fruit and bitter cotyledons is crossed with a plant that has
glossy, cream fruit and nonbitter cotyledons.
All of the F1 plants have dull, orange fruit and bitter cotyledons (DdRrBb). By
intercrossing the F1, the F2 are produced. Give the phenotypes and their expected
proportions in the F2.
To answer this question, the expected phenotypic ratios in the F2 can be calculated more easily
by examining the phenotypic ratios produced by the individual crosses of each gene locus.
F1 are intercrossed: DdRrBb x DdRrBb
Locus (gene) 1: Dd x Dd = 3/4 dull (DD and Dd); 1/4 glossy (dd)
Locus (gene) 2: Rr x Rr = 3/4 orange (RR and Rr); 1/4 cream (rr)
Locus (gene) 3:Bb x Bb = 3/4 bitter (BB and Bb); 1/4 nonbitter (bb)
dull, orange, bitter: 3/4 dull x 3/4 orange x 3/4 bitter = 27/64 dull,
orange, nonbitter:3/4 dull x 3/4 orange x 1/4 nonbitter = 9/64 dull,
cream, bitter: 3/4 dull x 1/4 cream x 3/4 bitter = 9/64 dull, cream,
nonbitter: 3/4 dull x 1/4 cream x 1/4 nonbitter = 3/64 glossy,
orange, bitter: 1/4 glossy x 3/4 orange x 3/4 bitter = 9/64 glossy,
orange, nonbitter: 1/4 glossy x 3/4 orange x 1/4 nonbitter = 3/64
glossy, cream, bitter: 1/4 glossy x 1/4 cream x 3/4 bitter = 3/64
glossy, cream, nonbitter: 1/4 glossy x 1/4 cream x 1/4 nonbitter = 1/64
In chickens, comb shape is determined by alleles at two loci (R, r and P, p). A walnut comb is
produced when at least one dominant allele R is present at one locus and at least one
dominant allele P is present at a second locus (genotype R_P_). A rose comb is produced
when at least one dominant allele is present at the first locus and two recessive alleles are
present at the second locus (genotype R_pp). A pea comb is produced when two recessive
alleles are present at the first locus and at least one dominant allele is present at the second
(genotype rrP_). If two recessive alleles are present at the first and at the second locus
pg. 1
, 139 SOLVEDTUTORIALEXAMPLES
(rrpp), a single comb is produced. Progeny with what types of combs and in what
proportions will result from the following crosses?
(a) RRPP x rrpp
All walnut (RrPp)
(b) RrPp x rrpp
1/4 walnut (RrPp), 1/4 rose (Rrpp), 1/4 pea (rrPp), 1/4 single (rrpp) (c)
RrPp x RrPp
9/16 walnut (R_P_), 3/16 rose (R_pp), 3/16 pea (rrP_), 1/16 single (rrpp)
(d) Rrpp x Rrpp
3/4 rose (R_pp), 1/4 single (rrpp)
(e) Rrpp x rrPp
1/4 walnut (RrPp), 1/4 rose (Rrpp), 1/4 pea (rrPp), 1/4 single (rrpp)
(f) Rrpp x rrpp
1/2 rose (Rrpp), 1/2 single (rrpp)
pg. 2
, 139 SOLVEDTUTORIALEXAMPLES
Tutorial 2 – Example problems
There is often more than one way to approach probability questions. Several of the following
examples will give alternative approaches to solving the same problem. The binomial
probability using Pascal’s triangle is often a fast way to calculate probabilities involving the
product rule. Below is the derivation of Pascals’ triangle. You may find it useful when
answering the tutorial questions.
p+q
p+q
2
pq+q
p2 + pq
n=2 p2 + 2pq +q2
p+q
p2q + 2pq2 + q3
p3 + 2p2q + pq2
n=3 p3 + 3p2q + 3pq2 + q3
p+q
3 2 2 3 4
p q + 3p q + 3pq +q
p4 + 3p3q + 3p2q2 + pq3
n=4 p + 4p q + 6p2q2 + 4pq3 +q4
4 3
p+q p4q +
3 2 2 3
4p q + 6p q + 4pq + q4 5 p + 4p4q +
5
6p3q2 + 4p2q3 + pq4 n=5 p5 + 5p4q + 10p3q2
+ 10p2q3 + 5pq4 + q5 etc.
What is the probability of rolling one six-sided die and obtaining the following numbers?
(a) 2
Since 2 is only found on one side of a six-sided die, then there is a 1/6 chance of
rolling a two.
(b) 1 or 2 (notice it’s OR not AND therefore you must use the sum rule)
The probability of rolling a 1 on a six-sided die is 1/6. Similarly, the probability of
rolling a 2 on a six-sided die is 1/6. Because the question asks what is the probability
of rolling a 1 or a 2, and these are mutually exclusive events, we should use the
additive rule of probability to determine the probability of rolling a 1 or a 2:(p of
rolling a 1) + (p of rolling a 2) = p of rolling either a 1 or a 2.
1/6 + 1/6 = 2/6 = 1/3 probability of rolling either a 1 or a 2.
(c) An even number
The probability of rolling an even number depends on the number of even numbers
found on the die. A single die contains three even numbers (2, 4, 6). The probability of
rolling any one of these three numbers on a six-sided die is 1/6. To determine the
pg. 3
139 SOLVED TUTORIAL EXAMPLES Tutorial 1 –
Example problems
In cucumbers, dull fruit (D) is dominant over glossy fruit (d), orange fruit (R) is dominant over
cream fruit (r), and bitter cotyledons (B) are dominant over nonbitter cotyledons (b). The
three characters are encoded by genes located on different pairs of chromosomes. A plant
homozygous for dull, orange fruit and bitter cotyledons is crossed with a plant that has
glossy, cream fruit and nonbitter cotyledons.
All of the F1 plants have dull, orange fruit and bitter cotyledons (DdRrBb). By
intercrossing the F1, the F2 are produced. Give the phenotypes and their expected
proportions in the F2.
To answer this question, the expected phenotypic ratios in the F2 can be calculated more easily
by examining the phenotypic ratios produced by the individual crosses of each gene locus.
F1 are intercrossed: DdRrBb x DdRrBb
Locus (gene) 1: Dd x Dd = 3/4 dull (DD and Dd); 1/4 glossy (dd)
Locus (gene) 2: Rr x Rr = 3/4 orange (RR and Rr); 1/4 cream (rr)
Locus (gene) 3:Bb x Bb = 3/4 bitter (BB and Bb); 1/4 nonbitter (bb)
dull, orange, bitter: 3/4 dull x 3/4 orange x 3/4 bitter = 27/64 dull,
orange, nonbitter:3/4 dull x 3/4 orange x 1/4 nonbitter = 9/64 dull,
cream, bitter: 3/4 dull x 1/4 cream x 3/4 bitter = 9/64 dull, cream,
nonbitter: 3/4 dull x 1/4 cream x 1/4 nonbitter = 3/64 glossy,
orange, bitter: 1/4 glossy x 3/4 orange x 3/4 bitter = 9/64 glossy,
orange, nonbitter: 1/4 glossy x 3/4 orange x 1/4 nonbitter = 3/64
glossy, cream, bitter: 1/4 glossy x 1/4 cream x 3/4 bitter = 3/64
glossy, cream, nonbitter: 1/4 glossy x 1/4 cream x 1/4 nonbitter = 1/64
In chickens, comb shape is determined by alleles at two loci (R, r and P, p). A walnut comb is
produced when at least one dominant allele R is present at one locus and at least one
dominant allele P is present at a second locus (genotype R_P_). A rose comb is produced
when at least one dominant allele is present at the first locus and two recessive alleles are
present at the second locus (genotype R_pp). A pea comb is produced when two recessive
alleles are present at the first locus and at least one dominant allele is present at the second
(genotype rrP_). If two recessive alleles are present at the first and at the second locus
pg. 1
, 139 SOLVEDTUTORIALEXAMPLES
(rrpp), a single comb is produced. Progeny with what types of combs and in what
proportions will result from the following crosses?
(a) RRPP x rrpp
All walnut (RrPp)
(b) RrPp x rrpp
1/4 walnut (RrPp), 1/4 rose (Rrpp), 1/4 pea (rrPp), 1/4 single (rrpp) (c)
RrPp x RrPp
9/16 walnut (R_P_), 3/16 rose (R_pp), 3/16 pea (rrP_), 1/16 single (rrpp)
(d) Rrpp x Rrpp
3/4 rose (R_pp), 1/4 single (rrpp)
(e) Rrpp x rrPp
1/4 walnut (RrPp), 1/4 rose (Rrpp), 1/4 pea (rrPp), 1/4 single (rrpp)
(f) Rrpp x rrpp
1/2 rose (Rrpp), 1/2 single (rrpp)
pg. 2
, 139 SOLVEDTUTORIALEXAMPLES
Tutorial 2 – Example problems
There is often more than one way to approach probability questions. Several of the following
examples will give alternative approaches to solving the same problem. The binomial
probability using Pascal’s triangle is often a fast way to calculate probabilities involving the
product rule. Below is the derivation of Pascals’ triangle. You may find it useful when
answering the tutorial questions.
p+q
p+q
2
pq+q
p2 + pq
n=2 p2 + 2pq +q2
p+q
p2q + 2pq2 + q3
p3 + 2p2q + pq2
n=3 p3 + 3p2q + 3pq2 + q3
p+q
3 2 2 3 4
p q + 3p q + 3pq +q
p4 + 3p3q + 3p2q2 + pq3
n=4 p + 4p q + 6p2q2 + 4pq3 +q4
4 3
p+q p4q +
3 2 2 3
4p q + 6p q + 4pq + q4 5 p + 4p4q +
5
6p3q2 + 4p2q3 + pq4 n=5 p5 + 5p4q + 10p3q2
+ 10p2q3 + 5pq4 + q5 etc.
What is the probability of rolling one six-sided die and obtaining the following numbers?
(a) 2
Since 2 is only found on one side of a six-sided die, then there is a 1/6 chance of
rolling a two.
(b) 1 or 2 (notice it’s OR not AND therefore you must use the sum rule)
The probability of rolling a 1 on a six-sided die is 1/6. Similarly, the probability of
rolling a 2 on a six-sided die is 1/6. Because the question asks what is the probability
of rolling a 1 or a 2, and these are mutually exclusive events, we should use the
additive rule of probability to determine the probability of rolling a 1 or a 2:(p of
rolling a 1) + (p of rolling a 2) = p of rolling either a 1 or a 2.
1/6 + 1/6 = 2/6 = 1/3 probability of rolling either a 1 or a 2.
(c) An even number
The probability of rolling an even number depends on the number of even numbers
found on the die. A single die contains three even numbers (2, 4, 6). The probability of
rolling any one of these three numbers on a six-sided die is 1/6. To determine the
pg. 3
