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Quiz: Non-Mendelian Inheritance and Chromosomal Genetics

Test your understanding of incomplete dominance, codominance, sex-linked inheritance, polygenic traits, epistasis, linkage, and chromosomal abnormalities with these review questions.

1. What term describes a heterozygote phenotype that appears as an intermediate blend between the two homozygous phenotypes?

  1. Codominance
  2. Multiple alleles
  3. Incomplete dominance
  4. Epistasis
Show Answer

The correct answer is C. In incomplete dominance, neither allele fully masks the other, so the heterozygote displays a phenotype intermediate between the two homozygous phenotypes, such as pink flowers produced by crossing red and white snapdragons. This differs from codominance, where both alleles are fully expressed rather than blended.

Concept Tested: Incomplete dominance

2. What is the term for the failure of chromosomes to separate properly during meiosis, producing gametes with an abnormal chromosome number?

  1. Nondisjunction
  2. Genetic linkage
  3. Pleiotropy
  4. Recombination
Show Answer

The correct answer is A. Nondisjunction occurs when homologous chromosomes (in meiosis I) or sister chromatids (in meiosis II) fail to separate properly, resulting in gametes with one extra or one missing chromosome. When such a gamete is fertilized, the resulting zygote is aneuploid, as seen in conditions like Down syndrome (trisomy 21).

Concept Tested: Nondisjunction

3. How does codominance, as seen in AB blood type, differ from incomplete dominance?

  1. Codominance produces an intermediate blended phenotype, just like incomplete dominance
  2. Codominance only occurs in traits controlled by a single allele
  3. Codominance and incomplete dominance are identical mechanisms with different names
  4. In codominance, both alleles are fully and simultaneously expressed in the heterozygote, producing a phenotype showing both traits distinctly rather than a blend
Show Answer

The correct answer is D. In codominance, both alleles in a heterozygote are fully expressed at the same time, as in the AB blood type where both A and B antigens appear on the red blood cell surface. This is distinct from incomplete dominance, where the two alleles blend to produce a single intermediate phenotype, such as pink flowers rather than distinctly red and white regions.

Concept Tested: Codominance vs. incomplete dominance

4. Why do X-linked recessive traits appear far more frequently in males than in females?

  1. Because the Y chromosome carries a dominant allele that always masks the X-linked recessive trait
  2. Because males are hemizygous for X-linked genes, so a single recessive allele on their one X chromosome is sufficient to produce the recessive phenotype, while females need two copies
  3. Because X-linked traits can only be inherited from the father
  4. Because the SRY gene increases the mutation rate on the X chromosome in males
Show Answer

The correct answer is B. Males carry only one X chromosome, making them hemizygous for any gene located there. A single copy of a recessive X-linked allele is therefore sufficient to produce the recessive phenotype in males, whereas females carry two X chromosomes and must inherit the recessive allele from both parents to express the trait.

Concept Tested: X-linked inheritance

5. Why does polygenic inheritance produce continuous, bell-curve phenotypic variation rather than discrete categories?

  1. Multiple genes each contribute a small additive effect to the phenotype, and the many possible combinations of alleles blend into a continuous, bell-shaped distribution
  2. A single gene with many alleles produces discrete phenotypic classes
  3. Polygenic traits are controlled entirely by environmental factors, not genes
  4. Epistatic interactions between genes eliminate all phenotypic classes except one
Show Answer

The correct answer is A. Polygenic traits, such as human height or skin color, are influenced by many genes, each contributing a small additive effect to the overall phenotype. Because so many allele combinations are possible across all the contributing genes, the population-level distribution of the phenotype smooths into a continuous bell curve rather than a small number of discrete categories.

Concept Tested: Polygenic inheritance

6. A cross of BbEe × BbEe in Labrador retrievers produces a 9:3:4 phenotypic ratio instead of the standard 9:3:3:1. What does the "4" category represent, and why?

  1. Chocolate dogs, because bb is epistatic to E
  2. Black dogs, because E is epistatic to B
  3. Yellow dogs, because the ee genotype is epistatic and masks pigment color regardless of the B/b genotype, combining the 3 B_ee and 1 bbee classes into one phenotype
  4. A genotyping error that should be excluded from analysis
Show Answer

The correct answer is C. The ee genotype at the pigment-deposition locus is epistatic to the B/b pigment-color locus, meaning that regardless of whether an ee dog carries B or b alleles, no pigment is deposited and the dog appears yellow. This combines the 3 B_ee and 1 bbee genotypic classes (which would otherwise be separate categories under standard dihybrid inheritance) into a single yellow phenotypic class of 4 out of 16.

Concept Tested: Epistasis

7. A female who is a carrier for an X-linked recessive disorder (X^H X^h) has children with an unaffected male (X^H Y). What proportion of their sons and daughters is expected to be affected?

  1. 1/4 of sons affected; 1/4 of daughters affected
  2. All sons affected; no daughters affected
  3. No sons affected; half of daughters affected
  4. Half of sons will be affected; no daughters will be affected, though half of daughters will be carriers
Show Answer

The correct answer is D. The carrier mother passes the X^h allele to half of her offspring. Sons who inherit X^h have no second X chromosome to mask it and are affected, so half of sons are affected. Daughters always receive a normal X^H from their unaffected father, so none are affected, though half of the daughters inherit X^h from their mother and become carriers.

Concept Tested: X-linked recessive inheritance patterns

8. A test cross produces 36 recombinant offspring out of 400 total offspring for two linked genes. What is the recombination frequency, and approximately how far apart are the genes on the chromosome?

  1. RF = 9%; the genes are approximately 9 cM apart
  2. RF = 36%; the genes are approximately 36 cM apart
  3. RF = 0.09%; the genes are essentially unlinked
  4. RF = 91%; the genes are on different chromosomes
Show Answer

The correct answer is A. Recombination frequency is calculated as (recombinant offspring / total offspring) × 100%, which gives (36/400) × 100% = 9%. Because 1% recombination frequency corresponds to approximately 1 centimorgan (map unit) of genetic distance, these two genes are estimated to be about 9 cM apart on the chromosome.

Concept Tested: Recombination frequency and genetic mapping

9. A pedigree shows a trait skipping generation II, with unaffected parents in generation I producing an affected child, and both males and females equally affected across the pedigree. What is the most likely mode of inheritance?

  1. X-linked recessive
  2. Autosomal recessive
  3. Autosomal dominant
  4. X-linked dominant
Show Answer

The correct answer is B. A trait that skips generations, arising in children of unaffected parents, indicates recessive inheritance, since both parents must be unaffected carriers. Because the trait affects males and females with roughly equal frequency rather than showing a strong sex bias, an autosomal (rather than X-linked) location is the most consistent interpretation.

Concept Tested: Pedigree analysis

10. How does the chromosomal composition of gametes differ when nondisjunction occurs during meiosis I versus meiosis II?

  1. Both meiosis I and meiosis II nondisjunction produce identical proportions of normal and abnormal gametes
  2. Nondisjunction in meiosis II always produces four abnormal gametes, while meiosis I nondisjunction produces four normal gametes
  3. Nondisjunction in meiosis I causes homologs to fail to separate, so all four resulting gametes are abnormal (two n+1 and two n-1), whereas nondisjunction in meiosis II affects only one of the two secondary cells, producing two normal gametes plus one n+1 and one n-1 gamete
  4. Nondisjunction can only occur during mitosis, not during either meiotic division
Show Answer

The correct answer is C. When homologous chromosomes fail to separate in meiosis I, both homologs travel to the same secondary cell, so both resulting cells are already abnormal before meiosis II even begins, ultimately producing four abnormal gametes (two with an extra chromosome, two missing one). When nondisjunction instead occurs in meiosis II, only one of the two secondary cells from a normal meiosis I is affected, so the final result is two normal gametes plus one n+1 and one n-1 gamete from the affected cell.

Concept Tested: Nondisjunction in meiosis I vs. meiosis II