Free MCAT-style practice, 1C
MCAT Mendelian Genetics, Pedigrees and Linkage Practice Questions
Nine free MCAT-style genetics practice questions on Punnett squares, probability rules, dihybrid and test crosses, dominance patterns, X-linked pedigrees and recombination mapping, with full explanations.
Quick concept primer
Most genetics questions reduce to probability. Mendel's law of segregation says each parent passes on one of its two alleles at random, and the law of independent assortment says alleles of genes on different chromosomes sort independently. Treat each gene as its own one-gene cross, then multiply: in an AaBb by AaBb cross, the chance of a dominant A phenotype is 3/4 and of a recessive b phenotype is 1/4, so the combined chance is 3/16. Use the product rule for "and" and the sum rule for mutually exclusive "or" outcomes, and the binomial expression when the question asks about exactly k of n children.
Genes on the same chromosome break the rule. Linked genes travel together unless crossing over in prophase I separates them, so recombinant offspring appear less often than the 50 percent expected for unlinked genes. Recombination frequency (recombinants divided by total) gives map distance, with 1 percent equal to 1 centimorgan. In a test cross the homozygous recessive parent contributes only recessive alleles, so the offspring phenotypes read out the gametes of the other parent directly.
Traps: incomplete dominance blends (pink) while codominance shows both (AB blood), and an X-linked recessive trait passes from a carrier mother to half her sons but never from father to son.
9 practice questions
Question 1
In a plant, purple flowers (P) are dominant to white (p) and round seeds (R) are dominant to wrinkled (r). The two genes are on different chromosomes. Two PpRr plants are crossed. What fraction of the offspring is expected to have purple flowers and wrinkled seeds?
- 1/16
- 3/16
- 9/16
- 1/4
Show answer and explanation
Answer: B. 3/16
Treat each gene separately. Pp by Pp gives purple in 3/4 of offspring, and Rr by Rr gives wrinkled (rr) in 1/4. Because the genes assort independently, multiply: 3/4 x 1/4 = 3/16. The value 9/16 is the fraction with both dominant phenotypes, 1/16 is the doubly recessive class, and 1/4 ignores the flower gene.
Question 2
A fruit fly heterozygous for two genes, carrying alleles G and W on one chromosome and g and w on its homolog, is crossed to a gg ww fly. The offspring are:
| Offspring phenotype class | Number |
|---|---|
| G W | 412 |
| g w | 398 |
| G w | 96 |
| g W | 94 |
What is the approximate map distance between the two genes?
- 9.5 centimorgans
- 41 centimorgans
- 81 centimorgans
- 19 centimorgans
Show answer and explanation
Answer: D. 19 centimorgans
The parental classes are the two most common, G W and g w, matching the chromosomes in the heterozygote. The recombinant classes are G w and g W: 96 + 94 = 190 of 1000 offspring, a recombination frequency of 19 percent, or 19 centimorgans. Using only one recombinant class gives 9.5, and 81 percent is the parental fraction, not a distance.
Question 3
Recombination frequencies between three linked genes on one chromosome were measured in separate crosses: genes J and K, 12 percent; genes K and L, 5 percent; genes J and L, 17 percent. Which gene lies between the other two?
- K
- J
- L
- The order cannot be determined from pairwise frequencies
Show answer and explanation
Answer: A. K
The two smaller distances add up to the largest one: 12 + 5 = 17, which is the J to L distance. That fits only the order J, K, L, with K in the middle. If J were in the middle, the K to L distance would be the sum of the other two, which it is not, and the same reasoning excludes L. Pairwise distances are exactly how gene order is inferred.
Question 4
Hemophilia A is an X-linked recessive disorder. A woman who is a carrier and a man with normal clotting plan to have a child. What is the probability that their first child is a boy with hemophilia?
- 1/2
- 1/8
- 1/4
- 0
Show answer and explanation
Answer: C. 1/4
The father gives a Y chromosome to every son, so a son's X comes from his carrier mother, who passes the mutant X half the time. The chance of a boy is 1/2 and the chance that a boy is affected is 1/2, so the combined probability is 1/2 x 1/2 = 1/4. The value 1/2 is the probability for a son already known to be male, and no daughter can be affected because she receives the father's normal X.
Question 5
In a family, two unaffected parents have an affected daughter and two unaffected sons. Assuming complete penetrance and no new mutations, which inheritance pattern is most consistent with this family?
- Autosomal dominant
- Autosomal recessive
- X-linked recessive
- X-linked dominant
Show answer and explanation
Answer: B. Autosomal recessive
An affected child of two unaffected parents means the trait is recessive and both parents are carriers. It cannot be X-linked recessive, because an affected daughter needs a mutant X from her father, and an affected father would show the trait. Both dominant patterns require at least one affected parent when penetrance is complete, so autosomal recessive is the only consistent choice.
Question 6
Each of the following is an example of codominance EXCEPT:
- A person with type AB blood whose red cells carry both A and B antigens
- Roan cattle whose coats contain both red hairs and white hairs
- A person heterozygous for the M and N blood group alleles whose red cells display both M and N antigens
- Pink flowers produced when a red-flowered and a white-flowered plant are crossed
Show answer and explanation
Answer: D. Pink flowers produced when a red-flowered and a white-flowered plant are crossed
In codominance both alleles are fully expressed side by side, as with A and B antigens, M and N antigens, or separate red and white hairs in roan cattle. A pink flower is an intermediate phenotype in which neither allele is fully expressed, which is incomplete dominance. Crossing two pink plants would give red, pink and white in a 1:2:1 ratio.
Question 7
A tall pea plant of unknown genotype is crossed with a short plant, and short height is recessive. The cross produces 51 tall and 49 short offspring. What is the genotype of the tall parent?
- TT
- tt
- Tt
- It cannot be determined because the sample is too small
Show answer and explanation
Answer: C. Tt
This is a test cross. The short parent is tt and gives only t gametes, so any short offspring must have received a t from the tall parent. A roughly 1:1 ratio of tall to short is the expected result for Tt by tt. A TT parent would give only tall offspring, and tt is impossible for a tall plant. One hundred offspring with many short ones is decisive.
Question 8
A man with an autosomal recessive condition (aa) and an unaffected woman who is a carrier (Aa) have four children. What is the probability that exactly two of the four children are affected?
- 1/4
- 1/16
- 1/2
- 3/8
Show answer and explanation
Answer: D. 3/8
Each child of an aa by Aa couple has a 1/2 chance of being affected. Two affected out of four can occur in 6 different birth orders (4 choose 2), and each order has probability (1/2)^4 = 1/16, so the total is 6/16 = 3/8. The value 1/16 counts only one order, and 1/2 is the risk for a single child.
Question 9
A plant is heterozygous at three genes, each on a different chromosome (AaBbCc). Ignoring crossing over, how many genetically different gametes can it produce, and which meiotic event accounts for this variety?
- 8, because each homologous pair orients randomly at the metaphase plate in meiosis I
- 6, because each gene contributes two alleles that add together
- 8, because sister chromatids exchange segments during anaphase II
- 3, because each gamete receives one chromosome of each pair
Show answer and explanation
Answer: A. 8, because each homologous pair orients randomly at the metaphase plate in meiosis I
Each of the three genes contributes one of two alleles to a gamete, and because the pairs line up independently at metaphase I, every combination is possible: 2 x 2 x 2 = 8 gamete types, such as ABC, ABc and abc. This independent orientation is the physical basis of independent assortment. Allele choices multiply rather than add, and anaphase II separates sister chromatids without exchanging segments.