NUMS — Medical & Dental Entry Test

NUMS Biology — Reproduction, Inheritance, Evolution and Biotechnology (Set 7)

Take the NUMS Biology — Reproduction, Inheritance, Evolution and Biotechnology (Set 7): 30 multiple-choice questions (40 minutes) with instant scoring and…

What this test covers

Sample questions with answers

The first 5 questions of this test, with the correct answer and explanation.

  1. Q1. Two phenotypically normal parents have a child with an autosomal recessive disease. Their other child is phenotypically normal. What is the probability that this normal child is a carrier?

    • A. 2/3 ✓
    • B. 3/4
    • C. 1/4
    • D. 1/2

    Answer: A. 2/3

    Explanation: Both parents are Aa. Among normal offspring the genotypes are AA : Aa : Aa = 1 : 2, so the chance of being a carrier is 2/3 (not 1/2).

  2. Q2. Two parents are both heterozygous for two unlinked recessive disease genes (AaBb × AaBb). What is the probability that a child is affected by at least one of the two diseases?

    • A. 9/16
    • B. 1/16
    • C. 7/16 ✓
    • D. 3/16

    Answer: C. 7/16

    Explanation: P(not affected by A) = 3/4 and same for B, so P(unaffected by both) = 9/16; P(at least one disease) = 1 − 9/16 = 7/16.

  3. Q3. A colour-blind man (XᶜY) and a woman who is a carrier of the recessive colour-blindness allele (XᶜX) have children. What is the probability that a child (sex not specified) is a colour-blind girl?

    • A. 1/4 ✓
    • B. 1/2
    • C. 1/8
    • D. 3/4

    Answer: A. 1/4

    Explanation: Cross XᶜY × XᶜX gives XᶜXᶜ : XᶜX : XᶜY : XY in equal proportion, so a colour-blind daughter (XᶜXᶜ) has probability 1/4.

  4. Q4. One parent has blood group A and the other has blood group B. Their first child has group O. What is the probability that their next child has group AB?

    • A. 1/4 ✓
    • B. 1/8
    • C. 0
    • D. 1/2

    Answer: A. 1/4

    Explanation: The O child shows that the parents are IᴬI and IᴮI. Gametes Iᴬ or i crossed with Iᴮ or i give IᴬIᴮ with probability 1/4.

  5. Q5. Red-flowered snapdragons (RR) are crossed with white-flowered (rr), giving pink F₁ plants showing incomplete dominance. If an F₁ plant is crossed with a white plant, what offspring ratio is expected?

    • A. 1 red : 2 pink : 1 white
    • B. 1 red : 1 white
    • C. 1 pink : 1 white ✓
    • D. 3 pink : 1 white

    Answer: C. 1 pink : 1 white

    Explanation: Rr × rr gives 1/2 Rr (pink) and 1/2 rr (white).

The other 25 questions

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  1. Q6. A test cross of AaBb with aabb gives 420 AaBb, 420 aabb, 80 Aabb and 80 aaBb among 1000 offspring. What is the recombination frequency?

    • A. 84%
    • B. 16%
    • C. 8%
    • D. 32%
  2. Q7. Recombination frequencies between three linked genes are: A–B 10%, B–C 6% and A–C 16%. What is the order of the genes?

    • A. B–A–C
    • B. Cannot be determined
    • C. A–C–B
    • D. A–B–C
  3. Q8. Cystic fibrosis (autosomal recessive) affects 1 in 2500 people in a population in Hardy–Weinberg equilibrium. What percentage of the population are carriers?

    • A. About 7.8%
    • B. About 0.04%
    • C. About 2%
    • D. About 3.9%
  4. Q9. In a population of 1000 plants there are 160 AA, 480 Aa and 360 aa. Is the population in Hardy–Weinberg equilibrium for this gene?

    • A. No, because heterozygotes are not the majority in the sample
    • B. No, because there are fewer AA than aa plants in the sample
    • C. Yes, as p = 0.4 and expected numbers match the observed ones
    • D. Yes, because p = 0.5 and the genotype numbers are all equal
  5. Q10. Three consecutive nucleotides that make up exactly one codon are deleted from the middle of a coding region. What is the most likely result?

    • A. No change at all because of the degeneracy of the code
    • B. A premature stop codon in every case without exception
    • C. Loss of one amino acid, with the reading frame kept
    • D. A frameshift changing all the following amino acids
  6. Q11. How many Barr bodies are found in the nuclei of the somatic cells of a person with the karyotype 47, XXY and of a person with 45, X? (Number of Barr bodies = number of X chromosomes − 1.)

    • A. None and one
    • B. One and none
    • C. Two and one
    • D. One and one
  7. Q12. A cell with 2n = 8 is in metaphase II of meiosis. How many chromosomes and how many chromatids does it contain?

    • A. 8 chromosomes and 16 chromatids
    • B. 8 chromosomes and 8 chromatids
    • C. 4 chromosomes and 4 chromatids
    • D. 4 chromosomes and 8 chromatids
  8. Q13. A diploid cell has 6 pg of DNA in G₁. How much DNA is in one cell at metaphase of meiosis I, and in one gamete?

    • A. 6 pg and 3 pg
    • B. 6 pg and 1.5 pg
    • C. 12 pg and 6 pg
    • D. 12 pg and 3 pg
  9. Q14. From 100 primary spermatocytes and 100 primary oocytes, how many mature gametes of each kind can result?

    • A. 400 sperm and 400 ova
    • B. 400 sperm and 100 ova
    • C. 100 sperm and 100 ova
    • D. 200 sperm and 100 ova
  10. Q15. A woman has a regular 32-day cycle. Assuming that the luteal phase (ovulation to the next menstruation) is always 14 days, on which day of the cycle (day 1 = first day of menstruation) does ovulation occur?

    • A. Day 16
    • B. Day 18
    • C. Day 20
    • D. Day 14
  11. Q16. A woman who breastfeeds frequently often has no ovulation for several months after delivery. Which mechanism best explains this?

    • A. Suckling raises prolactin and reduces GnRH release, so FSH and LH stay low
    • B. Suckling raises progesterone and corpus luteum growth, so FSH stays high
    • C. Suckling lowers prolactin and reduces GnRH release, so FSH and LH stay low
    • D. Suckling raises oestrogen and increases GnRH release, so FSH and LH stay high
  12. Q17. In a pregnant woman the corpus luteum is surgically removed at week 5 in one case and at week 16 in another. What is expected?

    • A. Pregnancy continues at both times
    • B. Miscarriage at week 5, but pregnancy continues at week 16
    • C. Pregnancy continues at week 5, but miscarriage at week 16
    • D. Miscarriage at both times
  13. Q18. Sea urchin eggs have their cortical granules removed. After the sperm is added, what is expected?

    • A. No sperm enter, because the acrosome reaction fails
    • B. One sperm enters and the egg divides normally afterwards
    • C. The egg develops by parthenogenesis without any sperm
    • D. Several sperm enter the egg, as the polyspermy block fails
  14. Q19. After a vasectomy, a man's sperm count in semen is zero, but his testosterone level and sexual drive are normal. Why?

    • A. Sertoli cells still release testosterone into the semen, though sperm cannot be made
    • B. Leydig cells stop making testosterone, though sperm are still made
    • C. Leydig cells still make testosterone, though sperm cannot pass the cut vas
    • D. Seminiferous tubules still release testosterone, though FSH is absent
  15. Q20. In a flowering plant with 2n = 14, how many chromosomes are present in the endosperm nucleus and in the embryo cells?

    • A. 14 in the endosperm and 21 in the embryo
    • B. 21 in the endosperm and 14 in the embryo
    • C. 7 in the endosperm and 14 in the embryo
    • D. 21 in the endosperm and 7 in the embryo
  16. Q21. A flower is small and green with no scent, has long hanging stamens producing huge amounts of light, dry pollen, and a large feathery stigma. It is most likely pollinated by:

    • A. moths
    • B. birds
    • C. bees
    • D. wind
  17. Q22. E. coli grown for many generations in ¹⁵N medium is moved to ¹⁴N medium and allowed to replicate three times. What fraction of the DNA molecules are hybrid (¹⁵N/¹⁴N)?

    • A. 1/2
    • B. 1/8
    • C. 3/4
    • D. 1/4
  18. Q23. A mutant cell lacks DNA ligase. Short DNA fragments accumulate during replication. What are these fragments?

    • A. RNA primers that were not made by the enzyme primase
    • B. Okazaki fragments of the lagging strand that have not been joined
    • C. Leading strand pieces that were wrongly cut by nucleases
    • D. Exons that were not spliced from the primary transcript
  19. Q24. In Marfan syndrome a mutation in one gene causes tall stature, long fingers, lens dislocation in the eye and weakness of the aorta wall. Which term describes one gene affecting several traits?

    • A. Polygenic inheritance, many genes together shaping a single trait
    • B. Codominance, both alleles being fully expressed in a heterozygote
    • C. Epistasis, one gene masking the expression of a different gene
    • D. Pleiotropy, one gene influencing several different traits
  20. Q25. A DNA mixture containing fragments of 500, 1500 and 3000 base pairs is separated by agarose gel electrophoresis. Which statement is correct?

    • A. The 3000 bp fragment is nearest the well and all fragments move towards the positive electrode
    • B. The 500 bp fragment is nearest the well and all fragments move towards the positive electrode
    • C. All fragments stay at the well because DNA is neutral
    • D. The 3000 bp fragment is farthest from the well and all fragments move towards the negative electrode
  21. Q26. A circular plasmid of 4000 bp is digested with a restriction enzyme and electrophoresis shows two bands, of 1000 bp and 3000 bp. What does this show?

    • A. The enzyme cuts the plasmid at one site
    • B. The plasmid is linear with one cut site
    • C. The enzyme cuts the plasmid at two sites
    • D. The enzyme does not cut the plasmid
  22. Q27. To make human insulin in bacteria, the gene is usually obtained as cDNA made from mRNA rather than from genomic DNA. Why?

    • A. Bacteria cannot translate exons, which cDNA lacks
    • B. Bacteria cannot remove introns, which cDNA lacks
    • C. Bacteria cannot use promoters, which cDNA contains
    • D. Bacteria cannot transcribe DNA, which cDNA lacks
  23. Q28. In regions where malaria is common, the sickle-cell allele remains at a high frequency even though homozygotes (HbˢHbˢ) are often severely ill. What is the best explanation?

    • A. Heterozygotes cannot be infected, so they have no cost
    • B. Heterozygotes are more resistant to malaria, so selection favours them
    • C. Homozygotes are more resistant to malaria, so selection favours them
    • D. The mutation rate is high, so the allele keeps arising
  24. Q29. The eye of an octopus and the eye of a human are both camera-type eyes, but they evolved independently. These eyes are an example of:

    • A. vestigial structures resulting from divergent evolution
    • B. analogous structures resulting from divergent evolution
    • C. homologous structures resulting from convergent evolution
    • D. analogous structures resulting from convergent evolution
  25. Q30. A new island population was founded by five individuals from a large mainland population. An allele that is rare on the mainland (1%) is found at 20% on the island. This is an example of:

    • A. gene flow bringing the allele from the mainland
    • B. the founder effect, a type of genetic drift
    • C. mutation increasing the allele by chance
    • D. natural selection favouring the allele on the island