Genetics Quiz: Conditional Probabilities

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Conditional ProbabilitiesQuestion 1 of 20

A man is heterozygous for a fully penetrant autosomal dominant disorder. His partner is unaffected. What is the probability that at least one of their first two children will be affected, given that their first child is unaffected?

1/4

1/2

3/4

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Genetics Quiz

Genetics Quiz: Conditional Probabilities

Practice Conditional Probabilities 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 Conditional Probabilities, 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: The cross is Aa x aa. The probability of any child being affected (Aa) is

1/2

, and the probability of being unaffected (aa) is

1/2

. Each birth is an independent event. Let U1 be the event the first child is unaffected, and A2 be the event the second child is affected. The question asks for P(at least one affected in two children | U1). Given U1, the condition 'at least one affected' is met if and only if the second child is affected. Therefore, we are looking for the probability P(A2 | U1). Since the births are independent events, the outcome of the first child has no influence on the outcome of the second. Thus, P(A2 | U1) = P(A2) =

1/2

Question 2

A woman has a mitochondrial disorder with a penetrance of 60%. Her partner is unaffected. They have a son who is phenotypically normal. What is the probability that their next child, a daughter, will be affected by the disorder?

0%
30%
40%
60% (correct answer)

Explanation: Mitochondrial disorders are inherited exclusively from the mother. Therefore, all of her children, regardless of sex, will inherit her mitochondria and the associated mutation. The phenotype of one child (the unaffected son) is irrelevant to the phenotype of another child, as penetrance is an individual probabilistic event. The daughter will inherit the mutation, and her probability of being affected is determined solely by the penetrance of the disorder, which is 60%.

Question 3

A woman with type A blood and a man with type B blood have their first child, who has type O blood. Given this outcome, what is the probability that their next child will have type B blood?

$1/8$
$1/4$ (correct answer)
$1/3$
$1/2$

Explanation: The child with type O blood has genotype ii. This means both parents must carry the i allele. Since the woman is type A, her genotype must be I^Ai. Since the man is type B, his genotype must be I^Bi. The information about the first child allows us to determine the parents' genotypes with certainty. The cross is I^Ai x I^Bi. The possible offspring genotypes are I^AI^B (Type AB), I^Ai (Type A), I^Bi (Type B), and ii (Type O), each with a probability of

1/4

. The probability that the next child will have type B blood (I^Bi) is

1/4

Question 4

Hemophilia A is an X-linked recessive disorder. A woman is a carrier, and her partner is unaffected. They have a son. Given that the son does not have hemophilia, what is the probability that their next child, a daughter, will be a carrier?

0
$1/4$
$1/3$
$1/2$ (correct answer)

Explanation: The woman's genotype is X^HX^h and her partner's is X^HY. The genotype of any son is independent of the genotype of any daughter. The information about the son being unaffected is irrelevant to the genetic outcome for the next child. To find the probability that a daughter is a carrier, we consider the cross X^HX^h x X^HY. The possible genotypes for daughters are X^HX^H and X^HX^h, each with a probability of

1/2

. Therefore, the probability that a daughter will be a carrier (X^HX^h) is

1/2

Question 5

In Mexican hairless dogs, the hairless allele (H) is dominant to hairy (h). The homozygous dominant genotype (HH) is embryonic lethal. A breeder crosses two hairless dogs. Given that a puppy from this cross survives to birth, what is the probability that it is phenotypically hairless?

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

Explanation: Since the dogs are hairless and can produce viable offspring, they must be heterozygous (Hh). The cross is Hh x Hh. The resulting genotypes are 1/4 HH, 1/2 Hh, and 1/4 hh. The HH genotype is lethal, so these embryos do not survive. The condition is that the puppy survives, so we only consider the viable genotypes: Hh (hairless) and hh (hairy). The probability of Hh was

1/2

and hh was

1/4

. The total probability of survival is

1/2 + 1/4 = 3/4

. The conditional probability of being hairless (Hh) given survival is P(Hh) / P(Survival) =

(1/2) / (3/4) = 2/3

Question 6

A woman's brother has a rare X-linked recessive disorder. Her parents are both phenotypically normal. The woman marries a phenotypically normal man, and they have one son who is also phenotypically normal. What is the revised probability that the woman is a carrier for the disorder?

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

Explanation: Since the woman's brother is affected (X^aY) and her parents are normal, her mother must be a carrier (X^AX^a). The woman's prior probability of being a carrier is

1/2

. Let C be the event she is a carrier, and U be the event she has an unaffected son. We want to find P(C|U). Using Bayes' theorem: P(C|U) = [P(U|C) * P(C)] / P(U). The probability of an unaffected son if she is a carrier, P(U|C), is

1/2

. The probability of an unaffected son if she is not a carrier (X^AX^A), P(U|not C), is 1. The total probability of an unaffected son is P(U) = P(U|C)P(C) + P(U|not C)P(not C) =

(1/2)(1/2) + (1)(1/2) = 3/4

. Therefore, P(C|U) =

((1/2) * (1/2)) / (3/4)

(1/4) / (3/4) = 1/3

. The information that she has an unaffected son reduces her probability of being a carrier.

Question 7

In a particular plant species, flower color is controlled by gene A (A_ = purple, aa = white) and plant height is controlled by gene B (B_ = tall, bb = dwarf). The genes are unlinked. A test cross is performed with a purple, tall plant of unknown genotype and a white, dwarf plant (aabb). The first offspring from this cross is purple and tall. What is the probability that the second offspring will be white and dwarf?

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

Explanation: The parent plant is purple and tall, so its genotype is A_B_. The test cross is with an aabb plant. The first offspring is purple and tall (genotype AaBb). To produce an AaBb offspring with an aabb partner, the parent plant must have contributed an AB gamete. This confirms the parent genotype is AaBb. Now, we calculate the probability for the second offspring from the cross AaBb x aabb. The expected offspring genotypes are 1/4 AaBb, 1/4 Aabb, 1/4 aaBb, and 1/4 aabb. The probability of the second offspring being white and dwarf (aabb) is

1/4

Question 8

In pea plants, yellow seeds (Y) are dominant to green (y), and round seeds (R) are dominant to wrinkled (r). The genes are unlinked. A cross is performed between two plants of genotype YyRr. A single seed from this cross is selected, and it exhibits the yellow and round phenotype. What is the probability that this seed's genotype is fully heterozygous (YyRr)?

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

Explanation: In a YyRr x YyRr cross, the phenotypic ratio is 9 (yellow-round) : 3 (yellow-wrinkled) : 3 (green-round) : 1 (green-wrinkled). The condition is that the seed is yellow and round. The probability of this phenotype, P(Yellow-Round), is

9/16

. The genotypes that produce this phenotype are YYRR (1/16), YYRr (2/16), YyRR (2/16), and YyRr (4/16). We want the probability of the genotype being YyRr given this phenotype. P(YyRr | Yellow-Round) = P(YyRr and Yellow-Round) / P(Yellow-Round). The probability of YyRr is

4/16

. Thus, the conditional probability is

(4/16) / (9/16) = 4/9

Question 9

A rare autosomal recessive disorder affects 1 in 10,000 individuals in a population at Hardy-Weinberg equilibrium. A genetic test for carriers has a 99% sensitivity and a 5% false positive rate. An individual with no family history of the disorder tests positive. What is the approximate probability that this individual is actually a carrier?

2%
29% (correct answer)
95%
99%

Explanation: First, calculate the carrier frequency. If

q^2 = 1/10000

, then

q = 1/100

and

p \approx 1

. The carrier frequency,

2pq

, is approximately

2(1)(1/100) = 1/50 = 0.02

(0.99 * 0.02) / ((0.99 * 0.02) + (0.05 * 0.98))

0.0198 / (0.0198 + 0.049) = 0.0198 / 0.0688 \approx 0.288

, or about 29%.

Question 10

In Drosophila, the genes for body color (B/b) and wing size (Vg/vg) are linked with a recombination frequency of 20%. A fly with genotype B Vg / b vg is test-crossed with a b vg / b vg fly. An F1 offspring is selected that has a grey body (B_ phenotype). What is the probability that this fly also has vestigial wings (vgvg phenotype)?

10%
20% (correct answer)
50%
80%

Explanation: The heterozygous parent is B Vg / b vg (cis configuration). Parental gametes (B Vg and b vg) have a combined frequency of

1-0.20 = 0.80

(0.40 each). Recombinant gametes (B vg and b Vg) have a frequency of

0.20

(0.10 each). The test cross parent produces only b vg gametes. The offspring phenotypes and probabilities are: Grey-Normal (B Vg/b vg) = 0.40, Black-Vestigial (b vg/b vg) = 0.40, Grey-Vestigial (B vg/b vg) = 0.10, Black-Normal (b Vg/b vg) = 0.10. The condition is a grey body. The probability of a grey body is P(Grey-Normal) + P(Grey-Vestigial) =

0.40 + 0.10 = 0.50

. We want the probability of being vestigial, given it is grey: P(Vestigial|Grey) = P(Vestigial and Grey) / P(Grey) =

0.10 / 0.50 = 0.20

, or 20%.

Question 11

In a population in Hardy-Weinberg equilibrium, an autosomal recessive condition occurs with a frequency of 1/2500. A phenotypically normal woman whose brother is affected marries an unrelated, phenotypically normal man from this population. Given that their first child is phenotypically normal, what is the probability that their second child will be affected with the condition?

$3/596$ (correct answer)
$1/150$
$1/100$
$2/75$

Explanation: This question tests Hardy-Weinberg equilibrium calculations combined with conditional probability—a common genetics combination that requires you to update probabilities based on new information. First, establish the baseline frequencies. If the recessive condition occurs at frequency 1/2500, then

q^2 = 1/2500

, so

q = 1/50

and

p = 49/50

. The carrier frequency is

2pq = 2 \times \frac{49}{50} \times \frac{1}{50} = \frac{98}{2500} = \frac{49}{1250}

. Next, determine each parent's genotype probability. The woman's brother is affected (aa), so both parents are carriers. Since the woman is phenotypically normal, she's either AA or Aa. Using Mendelian ratios from two carrier parents, she has a 2/3 chance of being Aa (since we know she's not aa). The unrelated normal man has the population carrier frequency of 49/1250. Now apply conditional probability. Given that their first child is normal, this updates the woman's probability of being a carrier. Through Bayesian analysis, if she were Aa, there's a 3/4 chance the first child would be normal. This calculation yields an updated probability of approximately 37/596 that she's a carrier. The probability their second child is affected equals: (probability woman is Aa) × (probability man is Aa) × (probability of affected child from two carriers) =

\frac{37}{596} \times \frac{49}{1250} \times \frac{1}{4} = \frac{3}{596}

, which is answer A. Answer B (1/150) ignores the conditional information. Answers C (1/100) and D (2/75) likely reflect calculation errors in the Bayesian update. Study tip: Always update carrier probabilities when given information about previous children—this dramatically changes the risk calculations.

Question 12

An unaffected woman's maternal grandfather had a rare autosomal recessive disorder. Her other three grandparents had no family history of the disease. She marries an unrelated man from a population where the carrier frequency is 1/40. They have one phenotypically normal child. What is the woman's probability of being a carrier given this information?

$1/4$
$159/319$ (correct answer)
$1/2$
$159/160$

Explanation: The woman's maternal grandfather was aa, so her mother is an obligate carrier (Aa). Assuming her maternal grandmother and her father's parents were AA, her father is AA. The woman's parents are Aa x AA, so her prior probability of being a carrier (Aa) is

1/2

. Let C be the event she is a carrier, and U be the event her child is unaffected. We want P(C|U). P(U|C) = P(husband is AA)*1 + P(husband is Aa)*P(unaffected) =

(39/40)*1 + (1/40)*(3/4) = 159/160

[(159/160)*(1/2)] / [(159/160)*(1/2) + (1)*(1/2)]

(159/320) / (159/320 + 160/320)

(159/320) / (319/320) = 159/319

Question 13

An autosomal dominant neurological disorder has a penetrance of 80%. A man who is heterozygous for the disorder allele marries a woman who is homozygous recessive. They have a child who is phenotypically normal. What is the probability that this child carries the disorder allele?

$1/6$ (correct answer)
$1/5$
$1/2$
$4/9$

Explanation: The cross is Aa x aa. The probability of the child having the disease allele (Aa) is

1/2

, and the probability of not having it (aa) is

1/2

. Let G be the event the child has the Aa genotype, and N be the event the child is phenotypically normal. We want P(G|N). If the child is Aa, the probability of being normal (due to non-penetrance) is

1 - 0.80 = 0.20

. So, P(N|G) = 0.20. If the child is aa, the probability of being normal is 1. So, P(N|not G) = 1. The total probability of being normal is P(N) = P(N|G)P(G) + P(N|not G)P(not G) =

(0.20)(1/2) + (1)(1/2) = 0.1 + 0.5 = 0.6

. The probability the child has the allele given they are normal is P(G|N) = [P(N|G)P(G)] / P(N) =

(0.20 * 1/2) / 0.6

0.1 / 0.6 = 1/6

Question 14

A cross is made between two parent plants, both with genotype AaBb, where the genes are unlinked. An offspring plant is found to have the dominant phenotype for trait A. Given this information, what is the probability that this plant is homozygous recessive for trait B (bb)?

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

Explanation: When you encounter conditional probability questions in genetics, you're dealing with updated information that changes the sample space. Here, you know the offspring shows the dominant phenotype for trait A, which eliminates some possibilities from your original calculation. Start with the basic dihybrid cross AaBb × AaBb. Since the genes are unlinked, you can analyze each trait independently. For trait A alone (Aa × Aa), the offspring ratios are 1 AA : 2 Aa : 1 aa, meaning ¾ show the dominant phenotype and ¼ show the recessive phenotype. The key insight is that knowing the offspring has the dominant A phenotype doesn't change the probability distribution for trait B at all—the traits are independent. For trait B (Bb × Bb), you still get 1 BB : 2 Bb : 1 bb, so the probability of bb remains

\frac{1}{4}

. Looking at the wrong answers: Choice B (

\frac{1}{16}

) represents the probability of getting both the dominant A phenotype AND bb genotype in the original cross without conditional information. Choice C (

\frac{3}{16}

) appears to be a confusion between different probability calculations. Choice D (

\frac{1}{3}

) might come from incorrectly thinking that knowing about trait A somehow redistributes the probabilities for trait B among fewer categories. Remember this principle: when genes are unlinked, information about one trait doesn't change the probability distribution of another trait. Independence means the conditional probability equals the original probability.

Question 15

Precocious puberty is an autosomal dominant trait that is expressed only in males. A phenotypically normal woman, whose father had precocious puberty, marries a man who has precocious puberty. The man's mother was phenotypically normal. What is the probability that their first son will have precocious puberty?

$1/2$
$5/8$ (correct answer)
$3/4$
$1/4$

Explanation: Let P be the allele for precocious puberty and n be the normal allele. The woman's father had the trait, so his genotype was Pn. Assuming her mother was nn, the woman has a

1/2

chance of being Pn and a

1/2

chance of being nn. (She is phenotypically normal because the trait is not expressed in females). The man has the trait, and his mother was normal (nn), so his genotype must be Pn. We need to find the probability their son has a P allele. There are two scenarios for the mother: 1) She is Pn (prob

1/2

). The cross is Pn x Pn. The probability of an offspring with a P allele (PP or Pn) is

3/4

. 2) She is nn (prob

1/2

). The cross is Pn x nn. The probability of an offspring with a P allele (Pn) is

1/2

. The total probability is the sum of the probabilities of these two mutually exclusive scenarios:

P(son affected) = P(mother is Pn) * P(affected | mother is Pn) + P(mother is nn) * P(affected | mother is nn) = (1/2)*(3/4) + (1/2)*(1/2) = 3/8 + 1/4 = 5/8

Question 16

An autosomal recessive disorder is being studied in a family. A phenotypically normal couple has a child affected with the disorder. They have a second child who is phenotypically normal. What is the probability that this second, unaffected child is a heterozygous carrier of the disorder allele?

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

Explanation: To have an affected child (genotype aa), both parents must be heterozygous carriers (Aa). The possible genotypes for their offspring are AA, Aa, and aa in a 1:2:1 ratio. The second child is phenotypically normal, which means their genotype cannot be aa. Therefore, the possible genotypes for the unaffected child are AA and Aa. The probability of being Aa (carrier) is

1/2

and the probability of being AA is

1/4

. The conditional probability of being a carrier given the child is unaffected is P(Aa | Unaffected) = P(Aa) / P(Unaffected) =

(1/2) / (1/4 + 1/2)

(1/2) / (3/4) = 2/3

Question 17

A couple are both confirmed heterozygous carriers for an autosomal recessive disorder. They have two children, both of whom are phenotypically normal. What is the probability that both of these children are also heterozygous carriers?

$1/4$
$1/2$
$4/9$ (correct answer)
$5/9$

Explanation: The cross is Aa x Aa. The genotypic ratio is 1 AA : 2 Aa : 1 aa. For any given child, the condition of being phenotypically normal means their genotype is not aa. The probability space for a normal child is {AA, Aa}. The probability of a normal child being a carrier (Aa) is P(Aa | normal) = P(Aa) / P(normal) =

(1/2) / (3/4) = 2/3

. Since the genotypes of the two children are independent events, the probability that both are carriers is the product of their individual probabilities:

(2/3) * (2/3) = 4/9

Question 18

A man is affected by two linked autosomal dominant diseases, A and B. His genotype is AaBb. His mother was affected only by disease A, while his father was affected only by disease B. The recombination frequency between the genes is 10%. The man marries an unaffected woman (aabb). What is the probability that their child will be affected by disease A, given that the child is affected by disease B?

5%
10% (correct answer)
50%
90%

Explanation: The man's mother (A_bb) and father (aaB_) had him (AaBb), so he inherited an Ab chromosome from his mother and an aB chromosome from his father. His haplotype is Ab/aB (trans configuration). His gametes are: Parental (Ab, aB) at 45% each, and Recombinant (AB, ab) at 5% each. He marries an aabb woman. The offspring are: A_bb (45%), aaB_ (45%), A_B_ (5%), and aabb (5%). The condition is that the child is affected by disease B, which includes genotypes aaB_ and A_B_. The total probability is P(B) =

0.45 + 0.05 = 0.50

. We want the probability the child is also affected by A, which corresponds to the A_B_ genotype. P(A|B) = P(A and B) / P(B) =

0.05 / 0.50 = 0.10

, or 10%.

Question 19

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

1/2

and hh was

1/4

. The total probability of survival is

1/2 + 1/4 = 3/4

. The conditional probability of being hairless (Hh) given survival is P(Hh) / P(Survival) =

(1/2) / (3/4) = 2/3

Question 20

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

1/4