Genetics Quiz: Multiplication And Addition Rules

15 questions · exam conditions

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Multiplication And Addition RulesQuestion 1 of 15

Phenylketonuria (PKU) and Tay-Sachs disease are unlinked autosomal recessive disorders. A couple are both carriers for both diseases (genotype PpTt). What is the probability that their first child will be a carrier for PKU or a carrier for Tay-Sachs, but not a carrier for both?

1/4

1/2

3/4

1

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Genetics Quiz

Genetics Quiz: Multiplication And Addition Rules

Practice Multiplication And Addition Rules in Genetics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Multiplication And Addition Rules, giving you a quick way to practice the rules, question types, and explanations that matter most for Genetics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$1/4$
$1/2$ (correct answer)
$3/4$
$1$

Explanation: For the PKU gene (Pp x Pp), P(carrier Pp) =

1/2

and P(not carrier PP or pp) =

1/2

. Similarly, for the Tay-Sachs gene (Tt x Tt), P(carrier Tt) =

1/2

and P(not carrier TT or tt) =

1/2

. The question asks for an exclusive OR (XOR) condition. There are two ways this can happen:

Carrier for PKU (Pp) AND not a carrier for Tay-Sachs: $P(Pp) \times P(\text{not } Tt) = (1/2) \times (1/2) = 1/4$ .
Not a carrier for PKU AND a carrier for Tay-Sachs (Tt): $P(\text{not } Pp) \times P(Tt) = (1/2) \times (1/2) = 1/4$ . Since these are mutually exclusive events, we add their probabilities: $1/4 + 1/4 = 1/2$ .

Question 2

In Labrador retrievers, coat color is determined by two genes. The first gene (B/b) determines pigment color (B for black, b for brown). The second gene (E/e) controls pigment deposition (E allows deposition, e prevents it, resulting in a yellow coat regardless of the B/b genotype). Two dogs, both with genotype BbEe, are mated. What is the probability that a puppy in the litter will be yellow OR brown?

$3/16$
$4/16$
$6/16$
$7/16$ (correct answer)

Explanation: From a BbEe x BbEe cross, the phenotypic ratio is 9 B_E_ (black) : 3 bbE_ (brown) : 4 _ee (yellow). The probability of a brown puppy (bbE) is

P(bb) \times P(E_) = (1/4) \times (3/4) = 3/16

. The probability of a yellow puppy (__ee) is

P(ee) = 1/4

4/16

. The question asks for the probability of a puppy being yellow OR brown. Since these are mutually exclusive phenotypes, we use the addition rule:

P(\text{brown}) + P(\text{yellow}) = 3/16 + 4/16 = 7/16

Question 3

Red-green color blindness is an X-linked recessive trait. A woman with normal vision, whose father was colorblind, marries a man with normal vision. They plan to have two children, a son and a daughter. What is the probability that their son will be colorblind AND their daughter will be a carrier?

$1/16$
$1/8$
$1/4$ (correct answer)
$1/2$

Explanation: Let X^b be the allele for color blindness and X^B be the normal allele. The woman's father was X^bY, so she must have inherited his X^b. Since she has normal vision, her genotype is X^B X^b. The man has normal vision, so his genotype is X^B Y. The cross is X^B X^b x X^B Y. For a son, the genotypes are

1/2

X^B Y (normal) and

1/2

X^b Y (colorblind). So, P(son is colorblind) =

1/2

. For a daughter, the genotypes are

1/2

X^B X^B (normal) and

1/2

X^B X^b (carrier). So, P(daughter is a carrier) =

1/2

. Since the outcomes for the two children are independent events, we use the multiplication rule:

P(\text{son colorblind}) \times P(\text{daughter carrier}) = (1/2) \times (1/2) = 1/4

Question 4

In pea plants, tall (T) is dominant to short (t), and purple flowers (P) are dominant to white (p). The genes are unlinked. From a cross between two TtPp plants, what is the probability that an offspring will be tall or have purple flowers (or both)?

$9/16$
$7/16$
$3/8$
$15/16$ (correct answer)

Explanation: The probability of being tall (T_) is

3/4

. The probability of having purple flowers (P_) is

3/4

. The question asks for the probability of being tall OR having purple flowers. These events are not mutually exclusive, so we use the formula

P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)

. The probability of being tall AND purple is

P(T_) \times P(P_) = (3/4) \times (3/4) = 9/16

. Therefore,

P(\text{tall or purple}) = 3/4 + 3/4 - 9/16 = 6/4 - 9/16 = 24/16 - 9/16 = 15/16

. Alternatively, one can calculate the probability of the only outcome that does not fit the criteria (short and white, ttpp), which is

(1/4) \times (1/4) = 1/16

, and subtract this from 1:

1 - 1/16 = 15/16

Question 5

A phenotypically normal couple has a child with a rare autosomal recessive disorder. They plan to have a second child. If this second child is phenotypically normal, what is the probability that this child is a carrier of the recessive allele?

$1/4$
$1/3$
$1/2$
$2/3$ (correct answer)

Explanation: Since the couple has a child with an autosomal recessive disorder (genotype aa), both parents must be heterozygous carriers (Aa). The possible genotypes for their offspring are

1/4

AA,

1/2

Aa, and

1/4

aa. The condition is that the second child is phenotypically normal, which excludes the aa genotype. The new sample space consists of AA and Aa genotypes. The probability of being in this sample space is

P(AA) + P(Aa) = 1/4 + 1/2 = 3/4

. The probability of being a carrier (Aa) within this new sample space is

P(Aa) / P(phenotypically normal) = (1/2) / (3/4) = 2/3

Question 6

In dragons, the ability to breathe fire (F) is a dominant trait. However, the allele has incomplete penetrance, meaning only 80% of individuals with at least one F allele actually breathe fire. Two heterozygous dragons (Ff) are crossed. What is the probability that their offspring will breathe fire?

$3/5$ (correct answer)
$3/4$
$4/5$
$13/20$

Explanation: First, determine the offspring genotypes from an Ff x Ff cross:

1/4

FF,

1/2

Ff, and

1/4

ff. The genotypes capable of producing the fire-breathing phenotype are FF and Ff. The total probability of inheriting one of these genotypes is

1/4 + 1/2 = 3/4

. The penetrance of the F allele is 80% (or

4/5

). To find the probability of an offspring both having the genotype and expressing the phenotype, we use the multiplication rule:

P(\text{fire-breathing genotype}) \times P(\text{penetrance}) = (3/4) \times (4/5) = 12/20 = 3/5

Question 7

A couple are both heterozygous carriers for cystic fibrosis, an autosomal recessive condition. They have three children. What is the probability that exactly one of their three children is affected by the disease?

$9/64$
$27/64$ (correct answer)
$37/64$
$1/4$

Explanation: For a cross between two carriers (Cc x Cc), the probability of having an affected child (cc) is

1/4

, and the probability of having an unaffected child (C_) is

3/4

. There are three possible birth orders for having exactly one affected child in three: (Affected, Unaffected, Unaffected), (Unaffected, Affected, Unaffected), and (Unaffected, Unaffected, Affected). The probability for each specific order is

(1/4) \times (3/4) \times (3/4) = 9/64

. Since these are mutually exclusive events, we use the addition rule:

9/64 + 9/64 + 9/64 = 27/64

. Alternatively, using the binomial probability formula:

P(k=1) = \binom{n}{k}p^k q^{n-k} = \binom{3}{1}(1/4)^1(3/4)^2 = 3 \times (1/4) \times (9/16) = 27/64

Question 8

A man's maternal grandfather has an autosomal recessive disease. The man's maternal grandmother and the man's father are both from families with no history of the disorder and are considered homozygous dominant. The man marries a woman who is also homozygous dominant. What is the probability that their first child will be a carrier?

$1/2$
$1/4$ (correct answer)
$1/8$
$0$

Explanation: Let 'a' be the recessive allele. The maternal grandfather's genotype is 'aa'. The maternal grandmother is 'AA'. Therefore, the man's mother must be a carrier ('Aa') as she inherited 'a' from her father and 'A' from her mother. The man's father is 'AA'. The man's parents' cross is Aa x AA. The probability that the man is a carrier (Aa) is

1/2

. The man marries a homozygous dominant woman (AA). For their child to be a carrier (Aa), the man must be a carrier (prob

1/2

) and he must pass on the 'a' allele (prob

1/2

). The woman will pass on 'A'. So, the total probability is

P(\text{man is Aa}) \times P(\text{child is Aa | man is Aa}) = (1/2) \times (1/2) = 1/4

Question 9

Two individuals are heterozygous for three unlinked genes with simple dominant/recessive alleles (genotype AaBbCc). If they have an offspring, what is the probability that the offspring will exhibit a dominant phenotype for exactly two of the three traits?

$9/64$
$3/8$
$27/64$ (correct answer)
$27/32$

Explanation: For any single gene (e.g., Aa x Aa), the probability of a dominant phenotype (A_) is

3/4

and the probability of a recessive phenotype (aa) is

1/4

. There are three mutually exclusive ways to have exactly two dominant phenotypes:

Dominant for A and B, recessive for C: $(3/4) \times (3/4) \times (1/4) = 9/64$
Dominant for A and C, recessive for B: $(3/4) \times (1/4) \times (3/4) = 9/64$
Dominant for B and C, recessive for A: $(1/4) \times (3/4) \times (3/4) = 9/64$ Using the addition rule, the total probability is the sum of these possibilities: $9/64 + 9/64 + 9/64 = 27/64$ .

Question 10

Two genes in corn, C/c and Sh/sh, are linked with a recombination frequency of 3.0%. A plant with genotype CSh/csh is crossed with a plant with genotype csh/csh. What is the probability that an offspring will have the phenotype C_shsh?

$0.015$ (correct answer)
$0.030$
$0.485$
$0.970$

Explanation: The heterozygous parent (CSh/csh) produces four types of gametes. The parental gametes are CSh and csh. The recombinant gametes are Csh and cSh. The recombination frequency is 3%, so the total probability of recombinant gametes is 0.03. This is split equally between the two types: P(Csh) = 0.015 and P(cSh) = 0.015. The remaining 97% of gametes are parental, so P(CSh) = 0.485 and P(csh) = 0.485. The other parent (csh/csh) only produces csh gametes (P=1). To get a C_shsh phenotype, the offspring must inherit a Csh gamete from the heterozygous parent and a csh gamete from the homozygous parent. The probability of this is

P(\text{Csh gamete}) \times P(\text{csh gamete}) = 0.015 \times 1 = 0.015

Question 11

In a certain flower, a two-step biochemical pathway produces purple pigment: Precursor --(Gene A)--> Intermediate --(Gene B)--> Purple. Dominant alleles A and B produce functional enzymes, while recessive alleles a and b do not. To get a purple flower, an individual must have at least one dominant allele for both genes. What is the probability of a purple-flowered offspring from a cross of AaBb x aabb?

$1/4$ (correct answer)
$3/4$
$1/2$
$9/16$

Explanation: The cross is AaBb x aabb. For an offspring to be purple, its genotype must be A_B_. We analyze each gene independently. For gene A, the cross is Aa x aa, so the probability of an A_ (specifically, Aa) genotype is

1/2

. For gene B, the cross is Bb x bb, so the probability of a B_ (specifically, Bb) genotype is

1/2

. Since both conditions must be met simultaneously and the genes are unlinked, we use the multiplication rule:

P(A_) \times P(B_) = (1/2) \times (1/2) = 1/4

Question 12

A couple are both carriers of an autosomal recessive allele for a genetic disorder. They have two children. What is the probability that at least one child is affected AND at least one child is phenotypically normal?

$6/16$ (correct answer)
$7/16$
$9/16$
$1/2$

Explanation: From a carrier x carrier cross (Aa x Aa), P(affected, aa) =

1/4

and P(normal, A_) =

3/4

. For a family of two, there are four outcomes: (Normal, Normal), (Normal, Affected), (Affected, Normal), and (Affected, Affected). We want the probability of having at least one of each. The outcomes that satisfy this are (Normal, Affected) and (Affected, Normal).

P(Child 1 Normal, Child 2 Affected) = $(3/4) \times (1/4) = 3/16$ .
P(Child 1 Affected, Child 2 Normal) = $(1/4) \times (3/4) = 3/16$ . Using the addition rule for these mutually exclusive events, the total probability is $3/16 + 3/16 = 6/16$ . The other outcomes, (Normal, Normal) with P=9/16 and (Affected, Affected) with P=1/16, do not satisfy the condition.

Question 13

In humans, the ABO blood group is determined by three alleles ( $I^A$ , $I^B$ , i), and the Rh factor is determined by a separate, unlinked gene (R/r). A man with blood type A-positive (genotype $I^A$ i Rr) and a woman with blood type B-positive (genotype $I^B$ i Rr) have a child. What is the probability the child will have blood type O-negative?

$1/4$
$1/8$
$1/16$ (correct answer)
$3/16$

Explanation: This problem requires calculating the probability for each blood group system and then multiplying them. For the ABO system, the cross is I^A i x I^B i. The probability of type O (genotype ii) is

P(i \text{ from father}) \times P(i \text{ from mother}) = (1/2) \times (1/2) = 1/4

. For the Rh system, the cross is Rr x Rr. The probability of Rh-negative (genotype rr) is

1/4

. To find the probability of having both type O and Rh-negative blood, we use the multiplication rule for these independent events:

P(\text{type O}) \times P(\text{Rh-negative}) = (1/4) \times (1/4) = 1/16

Question 14

In a cross between two organisms of genotype AaBb, what is the probability of producing an offspring with genotype AABB, given that the offspring exhibits at least one dominant phenotype?

$1/16$
$1/15$ (correct answer)
$1/9$
$15/16$

Explanation: The unconditional probability of an AABB offspring from an AaBb x AaBb cross is

P(AA) \times P(BB) = (1/4) \times (1/4) = 1/16

. The condition is that the offspring has at least one dominant phenotype. The only phenotype that does not satisfy this is the double recessive (aabb). The probability of an aabb offspring is

P(aa) \times P(bb) = (1/4) \times (1/4) = 1/16

. Therefore, the probability of the condition being met is

1 - P(aabb) = 1 - 1/16 = 15/16

. Using the conditional probability formula,

P(A|B) = P(A \text{ and } B) / P(B)

, where A is genotype AABB and B is at least one dominant phenotype. Since AABB results in a dominant phenotype for both traits, it is a subset of B, so

P(A \text{ and } B) = P(A) = 1/16

. The final probability is

(1/16) / (15/16) = 1/15

Question 15

A plant of genotype AABbCcDd is self-fertilized. All four genes assort independently. What is the probability that an offspring will be heterozygous for at least one of these four genes?

$1/8$
$1/2$
$7/8$ (correct answer)
$15/16$

Explanation: This problem is best solved by calculating the probability of the complementary event: the offspring being homozygous for all four genes. Then, subtract this from 1. For each gene:

Gene A (AA x AA): Offspring are all AA. P(homozygous) = 1.
Gene B (Bb x Bb): P(homozygous, BB or bb) = $1/4 + 1/4 = 1/2$ .
Gene C (Cc x Cc): P(homozygous, CC or cc) = $1/2$ .
Gene D (Dd x Dd): P(homozygous, DD or dd) = $1/2$ . The probability of being homozygous for all genes is the product of these probabilities: $1 \times 1/2 \times 1/2 \times 1/2 = 1/8$ . The probability of being heterozygous for at least one gene is $1 - P(\text{homozygous for all}) = 1 - 1/8 = 7/8$ .