Meiosis and genetic recombination US Medical PG Practice Questions and MCQs
Practice US Medical PG questions for Meiosis and genetic recombination. These multiple choice questions (MCQs) cover important concepts and help you prepare for your exams.
Meiosis and genetic recombination US Medical PG Question 1: A 12-year-old girl with a recently diagnosed seizure disorder is brought to the physician by her mother for genetic counseling. She has difficulties in school due to a learning disability. Medications include carbamazepine. She is at the 95th percentile for height. Genetic analysis shows a 47, XXX karyotype. An error in which of the following stages of cell division is most likely responsible for this genetic abnormality?
- A. Paternal meiosis, anaphase I
- B. Maternal meiosis, telophase II
- C. Maternal meiosis, metaphase II
- D. Paternal meiosis, metaphase II
- E. Maternal meiosis, anaphase II (Correct Answer)
Meiosis and genetic recombination Explanation: ***Maternal meiosis, anaphase II***
- A 47,XXX karyotype indicates an extra X chromosome, which most commonly results from **nondisjunction during maternal meiosis**.
- **Nondisjunction in anaphase II** occurs when sister chromatids fail to separate properly, leading to an ovum containing two X chromosomes, which upon fertilization by an X-bearing sperm results in an XXX zygote.
- While nondisjunction can occur in either meiosis I or II, both produce the same 47,XXX outcome; anaphase is the critical stage where chromosome separation occurs.
*Paternal meiosis, anaphase I*
- Nondisjunction during paternal meiosis I would lead to sperm with either no sex chromosomes or both X and Y chromosomes, resulting in **45,X0 (Turner syndrome) or 47,XXY (Klinefelter syndrome)** upon fertilization, not 47,XXX.
- This involves improper segregation of **homologous chromosomes**.
*Maternal meiosis, telophase II*
- While telophase II is the final stage of meiosis II, the critical event of nondisjunction (failure of sister chromatids to separate) occurs during **anaphase II**, not telophase.
- Errors in telophase II would affect nuclear reformation but do not cause the chromosomal separation failure responsible for **aneuploidy**.
*Maternal meiosis, metaphase II*
- Metaphase II is when sister chromatids align at the metaphase plate; while improper alignment could theoretically contribute to aneuploidy, the actual **physical separation failure** defining nondisjunction occurs during **anaphase II**.
- Nondisjunction is specifically an anaphase phenomenon.
*Paternal meiosis, metaphase II*
- Paternal meiosis II nondisjunction would result in sperm carrying either two sex chromosomes (XX or YY) or none, leading to karyotypes like **47,XYY or 47,XXY**, not 47,XXX.
- Paternal contribution of two X chromosomes would require the rare scenario of an X-bearing sperm with nondisjunction fertilizing an X-bearing ovum that also had nondisjunction, which is extremely unlikely.
Meiosis and genetic recombination US Medical PG Question 2: A laboratory physician investigates the chromosomes of a fetus with a suspected chromosomal anomaly. She processes a cell culture obtained by amniocentesis. Prior to staining and microscopic examination of the fetal chromosomes, a drug that blocks cell division is added to the cell culture. In order to arrest chromosomes in metaphase, the physician most likely added a drug that is also used for the treatment of which of the following conditions?
- A. Trichomonas vaginitis
- B. Testicular cancer
- C. Herpes zoster
- D. Polycythemia vera
- E. Acute gouty arthritis (Correct Answer)
Meiosis and genetic recombination Explanation: ***Acute gouty arthritis***
- The drug used to arrest chromosomes in metaphase is likely **colchicine**, which inhibits **microtubule polymerization** and spindle formation, thus arresting cells in metaphase.
- **Colchicine** is a well-established treatment for **acute gouty arthritis** due to its anti-inflammatory properties, primarily through disrupting neutrophil functions.
*Trichomonas vaginitis*
- This condition is typically treated with **metronidazole** or **tinidazole**, which are antibiotics targeting protozoa and anaerobic bacteria.
- These drugs do not inhibit microtubule assembly or arrest cells in metaphase.
*Testicular cancer*
- Testicular cancer is primarily treated with **BEP regimen** (bleomycin, etoposide, cisplatin), which does not include microtubule-inhibiting agents.
- While vinca alkaloids (vincristine, vinblastine) do arrest cells in metaphase via microtubule inhibition similar to colchicine, they are not standard first-line agents for testicular cancer.
- The question specifically asks about the primary clinical use of colchicine, which is gout, not cancer chemotherapy.
*Herpes zoster*
- Herpes zoster (shingles) is a viral infection treated with **antiviral medications** like acyclovir, valacyclovir, or famciclovir.
- These antivirals work by interfering with viral DNA replication and do not target microtubule formation or cell division.
*Polycythemia vera*
- Polycythemia vera is a myeloproliferative neoplasm often managed with **phlebotomy**, **hydroxyurea**, or ruxolitinib.
- These treatments aim to reduce blood cell counts or inhibit specific signaling pathways, none of which primarily involve arresting cells in metaphase by disrupting microtubules.
Meiosis and genetic recombination US Medical PG Question 3: A 19-year-old male from rural West Virginia presents to his family medicine doctor to discuss why he is having trouble getting his wife pregnant. On exam, he is 6 feet 2 inches with a frail frame and broad hips for a male his size. He is noted to have mild gynecomastia, no facial hair, and small, underdeveloped testes. He claims that although he has a lower libido than most of his friends, he does have unprotected sex with his wife. His past medical history is notable for developmental delay and difficulties in school. What is the most likely chromosomal abnormality in this patient?
- A. Trisomy 13
- B. 45: XO
- C. Trisomy 21
- D. 47: XYY
- E. 47: XXY (Correct Answer)
Meiosis and genetic recombination Explanation: ***47: XXY***
- The patient's presentation with **infertility**, small testes, **gynecomastia**, eunuchoid body habitus (tall, frail frame, broad hips), lack of facial hair, and **developmental delay** are classic features of **Klinefelter syndrome (47, XXY)**.
- This chromosomal abnormality leads to primary **hypogonadism** due to the presence of an extra X chromosome in males.
*Trisomy 13*
- Trisomy 13, or **Patau syndrome**, is characterized by severe developmental anomalies, including **cleft lip and palate**, polydactyly, and severe neurological defects.
- Infants with Trisomy 13 rarely survive beyond the first year and do not present with the described signs of hypogonadism or gynecomastia in adolescence.
*45: XO*
- **45, XO** or **Turner syndrome** affects females and is characterized by **short stature**, primary amenorrhea, webbed neck, and **gonadal dysgenesis (streak gonads)**.
- This karyotype is incompatible with a male phenotype and the symptoms described.
*Trisomy 21*
- Trisomy 21, or **Down syndrome**, is associated with distinct facial features, intellectual disability, and congenital heart defects.
- While individuals with Down syndrome may have fertility issues, they do not typically present with the specific combination of **gynecomastia**, eunuchoid habitus, and **small testes** seen in this patient.
*47: XYY*
- **47, XYY syndrome** is associated with increased height and potentially some learning difficulties, but typically does not cause the significant **hypogonadism**, **gynecomastia**, or **small testes** seen in this patient.
- Men with 47, XYY usually have normal sexual development and fertility, though some may experience learning disabilities or behavioral problems.
Meiosis and genetic recombination US Medical PG Question 4: A 5-year-old boy is brought to a pediatrician by his parents for evaluation of learning difficulties in school. He has short stature, a flat face, low-set ears, a large tongue, and a single line on the palm. He was born to his parents after 20 years of marriage. You ordered karyotyping which will likely reveal which of the following?
- A. 47, XXX
- B. 47, XY, +18
- C. 47, XXY
- D. 45, XO
- E. 47, XY, +21 (Correct Answer)
Meiosis and genetic recombination Explanation: ***47, XY, +21***
- The patient's presentation with **short stature**, a **flat face**, **low-set ears**, a **large tongue**, a **single palmar crease (Simian crease)**, and **learning difficulties** are classic diagnostic features of **Down syndrome**.
- **Down syndrome** is caused by the presence of an extra copy of chromosome 21, leading to a karyotype of **47, XY, +21** (if male) or 47, XX, +21 (if female). The mention of the parents' age (born after 20 years of marriage implies an older maternal age) is a significant risk factor for Down syndrome.
*47, XXX*
- This karyotype describes **Triple X syndrome**, which affects females. Individuals usually present with normal appearance, learning difficulties, and often do not have distinct physical features like those described in the case.
- The patient is a 5-year-old boy, immediately ruling out Triple X syndrome.
*47, XY, +18*
- This karyotype indicates **Edwards syndrome (Trisomy 18)**. While it presents with developmental delay and distinctive physical features, these typically include a **rocker-bottom feet**, **clenched hands**, and other severe abnormalities often leading to early demise.
- The specific features described in the patient, such as a **flat face** and **single palmar crease**, are not characteristic of Edwards syndrome.
*47, XXY*
- This karyotype describes **Klinefelter syndrome**, which affects males. This condition is characterized by **tall stature**, **hypogonadism**, and often **learning difficulties**, but the patient's features like **short stature**, **flat face**, and **single palmar crease** are not consistent with Klinefelter syndrome.
- The phenotype of Klinefelter syndrome becomes more evident in adolescence and adulthood.
*45, XO*
- This karyotype describes **Turner syndrome**, which affects females. Features include **short stature**, **webbed neck**, and **gonadal dysgenesis**.
- The patient is a 5-year-old boy, which rules out Turner syndrome.
Meiosis and genetic recombination US Medical PG Question 5: As part of a clinical research study, the characteristics of neoplastic and normal cells are being analyzed in culture. It is observed that neoplastic cell division is aided by an enzyme which repairs progressive chromosomal shortening, which is not the case in normal cells. Due to the lack of chromosomal shortening, these neoplastic cells divide more rapidly than the normal cells. Which of the following enzymes is most likely involved?
- A. Topoisomerase
- B. DNA polymerase
- C. Reverse transcriptase
- D. Protein kinase
- E. Telomerase (Correct Answer)
Meiosis and genetic recombination Explanation: ***Telomerase***
- **Telomerase** is an enzyme that adds repetitive nucleotide sequences (telomeres) to the ends of chromosomes, counteracting their progressive shortening during DNA replication. This activity is crucial for the continuous division of neoplastic cells.
- In normal somatic cells, **telomerase activity is typically low or absent**, leading to telomere shortening with each division, eventually triggering cellular senescence or apoptosis. The presence of telomerase in neoplastic cells allows them to bypass these natural limits on proliferation.
*Topoisomerase*
- **Topoisomerases** are enzymes that regulate the supercoiling of DNA by breaking and rejoining DNA strands, which is essential during replication and transcription to relieve torsional stress.
- They do not directly repair chromosomal shortening but rather manage the topological state of DNA.
*DNA polymerase*
- **DNA polymerase** is primarily responsible for synthesizing new DNA strands by adding nucleotides, thereby elongating the DNA molecule during replication and DNA repair processes.
- While essential for DNA replication, it cannot fully replicate the very ends of linear chromosomes, leading to the **end-replication problem** and telomere shortening.
*Reverse transcriptase*
- **Reverse transcriptase** is an enzyme that synthesizes DNA from an RNA template, a process central to retroviruses and some eukaryotic elements like retrotransposons.
- Although telomerase itself is a specialized reverse transcriptase (using an RNA template to synthesize DNA telomeres), the general term "reverse transcriptase" does not specifically refer to the enzyme that repairs chromosomal shortening in the context of cell division.
*Protein kinase*
- **Protein kinases** are enzymes that add phosphate groups to proteins, a process known as phosphorylation. This modification can alter protein activity, localization, or stability, playing a critical role in signal transduction pathways.
- They are involved in regulating various cellular processes, including cell growth and division, but do not directly repair chromosomal shortening.
Meiosis and genetic recombination US Medical PG Question 6: An 11-year-old male with light purple eyes presents with gradual loss of bilateral visual acuity. Over the past several years, vision has worsened from 20/20 to 20/100 in both eyes. He also has mild nystagmus when focusing on objects such as when he is trying to do his homework. He is diagnosed with a disease affecting melanin production in the iris. If both of his parents are unaffected, which of the following represents the most likely probabilities that another male or female child from this family would be affected by this disorder?
- A. Male: 25% Female: 25%
- B. Same as general population
- C. Male: 50% Female: 50%
- D. Male: 50% Female: 0% (Correct Answer)
- E. Male: 100% Female: 0%
Meiosis and genetic recombination Explanation: ***Male: 50% Female: 0%***
- The symptoms (light purple eyes, gradual vision loss, nystagmus, defective melanin production) are characteristic of **ocular albinism**. This condition is typically **X-linked recessive**.
- If the patient's mother is a **carrier** (XAXa) and the father is unaffected (XAY), there is a **50% chance** that a male child will inherit the affected X chromosome (XaY) and thus be affected, and a **0% chance** for a female child to be affected if the father is unaffected (all female children would either be carriers XAXa or unaffected XAXA).
*Male: 25% Female: 25%*
- This probability pattern would typically suggest an **autosomal recessive** inheritance pattern, where both parents are carriers (Aa x Aa), and there is a 25% chance for any child to be affected regardless of sex.
- However, ocular albinism most commonly follows an X-linked recessive pattern, and the described clinical features (e.g., light purple eyes due to melanin defect in the iris) are highly suggestive of ocular albinism.
*Same as general population*
- This would only be true if the disorder was not genetic or if the parents' carrier status did not increase the risk for subsequent children.
- Given the heritable nature of albinism and the specific family history (parents unaffected, one affected child), the risk for subsequent children is significantly higher than the general population.
*Male: 50% Female: 50%*
- This pattern would occur in an **autosomal dominant** disorder with 100% penetrance, where one parent is affected (Aa x aa), or in some specific scenarios of X-linked inheritance if the father was affected and the mother was a carrier.
- Ocular albinism is X-linked recessive, not autosomal dominant, and the father is stated to be unaffected.
*Male: 100% Female: 0%*
- This genetic pattern is highly unlikely unless the mother was fully mosaic for the condition or an extremely rare and specialized inheritance pattern was at play.
- In a typical X-linked recessive inheritance with an unaffected father and a carrier mother, there is always a 50% chance for a male child to be unaffected.
Meiosis and genetic recombination US Medical PG Question 7: A Caucasian 32-year-old woman has an uncomplicated vaginal delivery, giving birth to male and female fraternal twins at term. At 2 days of life, the twin sister develops abdominal distension without emesis, and the mother states that she has not noticed the passage of stool for this infant. Genetic testing identifies deletion of an amino acid in a membrane channel for the girl. Both parents are healthy. Assuming that twin brother's disease status/symptomatology is unclear, which of the following best approximates the probability that the twin brother is a carrier of the disease allele?
- A. 100%
- B. 67% (Correct Answer)
- C. 0%
- D. 50%
- E. 25%
Meiosis and genetic recombination Explanation: ***67%***
- The sister's symptoms of **abdominal distension** without emesis and lack of stool passage, along with genetic testing identifying a **deletion of an amino acid in a membrane channel**, strongly suggest **Cystic Fibrosis (CF)**. CF is an **autosomal recessive disorder**.
- Since the affected twin sister has CF (genotype **aa**), and both parents are healthy, both parents must be **heterozygous carriers (Aa)**. When two carriers (Aa x Aa) have offspring, the probability of any child being a carrier (Aa) is **2/3** among the unaffected offspring. The twin brother is currently unaffected (phenotypically healthy), so the probability of him being a carrier is 2/3, or approximately 67%.
*100%*
- This would only be true if one or both parents were **homozygous affected (aa)**, or if the disease inheritance was **dominant** and the parents were carriers, which is not the case for this autosomal recessive disorder where the parents are healthy carriers and the brother is phenotypically unaffected.
- While both parents *are* carriers, the brother, being unaffected, has a chance of being **homozygous dominant (AA)**, meaning he is not a carrier.
*0%*
- This is incorrect because we know both parents are **obligate carriers** (heterozygous, Aa) for the recessive allele, given their affected child (aa). Therefore, their children have a 75% chance of inheriting at least one disease allele (50% carrier, 25% affected).
- The twin brother being unaffected means he has a 2/3 chance of being a carrier, not 0%.
*50%*
- This probability (1/2) is the chance of a child inheriting a specific allele from one parent, or the chance of being a carrier if one parent is affected and the other is homozygous dominant.
- In an **autosomal recessive** inheritance pattern where both parents are carriers (Aa x Aa) and the offspring is unaffected, the probability of being a carrier is **2/3**, not 1/2.
*25%*
- This is the probability of a child being **homozygous dominant (AA)** from two carrier parents (Aa x Aa), meaning they would neither have the disease nor be carriers.
- It is also the probability of a child being affected (aa) if both parents are carriers. Neither of these scenarios matches the question asking for the probability of the *unaffected* twin brother being a carrier.
Meiosis and genetic recombination US Medical PG Question 8: A healthy 30-year-old woman comes to the physician with her husband for preconception counseling. Her husband is healthy but she is concerned because her brother was recently diagnosed with a genetic liver condition for which he takes penicillamine. Her father-in-law has liver cirrhosis and a tremor. The results of genetic testing show that both the patient and her husband are carriers of a mutation in the ATP7B gene. Which of the following is the chance that this patient’s offspring will eventually develop the hereditary condition?
- A. 0%
- B. 25% (Correct Answer)
- C. 100%
- D. 50%
- E. 75%
Meiosis and genetic recombination Explanation: ***25%***
- The familial history (brother with a genetic liver condition, father-in-law with cirrhosis and tremor) and the **ATP7B gene mutation** indicate **Wilson's disease**, which is typically inherited in an **autosomal recessive** pattern.
- If both parents are carriers (heterozygous for the mutation), the probability that their offspring will inherit two copies of the mutated gene (one from each parent) and, therefore, develop the condition is **25%** as per Mendelian inheritance.
*0%*
- This is incorrect because both parents are identified as carriers, meaning there is a definite risk of passing on the mutated genes to their offspring.
- For the risk to be 0%, at least one parent would need to be completely free of the mutation or the inheritance pattern would need to be dominant with no penetrance.
*100%*
- This would only be the case if both parents had the disease (were homozygous for the mutation) or if the condition were dominant and at least one parent had the disease and passed on the dominant allele.
- Since both are carriers, the chance of inheriting two mutated alleles is not 100%.
*50%*
- A 50% chance would apply if one parent had the disease (homozygous recessive) and the other was a carrier, or if it were an autosomal dominant condition with one affected heterozygous parent.
- This does not reflect the inheritance pattern for two carrier parents in an autosomal recessive condition.
*75%*
- A 75% chance is not typical for a single genetic outcome in standard Mendelian inheritance patterns from carrier parents.
- In the context of two carriers for an autosomal recessive trait, 75% represents the chance of the offspring either being a carrier (50%) or being completely unaffected (25%), but not the chance of developing the condition.
Meiosis and genetic recombination US Medical PG Question 9: Although nucleotide addition during DNA replication in prokaryotes proceeds approximately 20-times faster than in eukaryotes, why can much larger amounts of DNA be replicated in eukaryotes in a time-effective manner?
- A. Eukaryotes have multiple origins of replication (Correct Answer)
- B. Eukaryotes have helicase which can more easily unwind DNA strands
- C. Eukaryotes have fewer polymerase types
- D. Eukaryotes have less genetic material to replicate
- E. Eukaryotes have a single, circular chromosome
Meiosis and genetic recombination Explanation: ***Eukaryotes have multiple origins of replication***
- Eukaryotic chromosomes are much larger than prokaryotic chromosomes and require multiple origins of replication to complete DNA synthesis within a reasonable timeframe.
- Each origin of replication initiates simultaneously, allowing DNA synthesis to occur at many sites along the chromosome, effectively increasing the overall speed of replication.
- This compensates for the slower rate of nucleotide addition by DNA polymerase in eukaryotes compared to prokaryotes.
*Eukaryotes have helicase which can more easily unwind DNA strands*
- While helicase activity is crucial for unwinding DNA, there is no evidence to suggest that eukaryotic helicases are significantly more efficient or faster at unwinding DNA compared to prokaryotic helicases in a way that would account for the large difference in overall replication time.
- The rate of DNA unwinding by helicase is a factor in replication speed, but it does not overcome the fundamental limitation of a single origin of replication in prokaryotes.
*Eukaryotes have fewer polymerase types*
- Eukaryotic cells actually have **more** types of DNA polymerases than prokaryotic cells, each specialized for different functions like replication, repair, and mitochondrial DNA synthesis.
- The number of polymerase types does not directly relate to the speed or efficiency of overall DNA replication in terms of replicating large amounts of DNA.
*Eukaryotes have less genetic material to replicate*
- Eukaryotic organisms typically have significantly **more** genetic material (a larger genome size) than prokaryotic organisms, not less.
- If eukaryotes had less genetic material, the question itself about effective replication of "much larger amounts of DNA" would be contradictory.
*Eukaryotes have a single, circular chromosome*
- Eukaryotic cells have **multiple, linear chromosomes** within a membrane-bound nucleus, not a single circular chromosome.
- Prokaryotic cells typically have a single, circular chromosome located in the nucleoid region.
- The linear structure of eukaryotic chromosomes with multiple origins is actually what enables efficient replication of large genomes, making this statement both factually incorrect and contradictory to the mechanism in question.
Meiosis and genetic recombination US Medical PG Question 10: A 2-month-old boy is presented to the clinic for a well-child visit by his parents. They are concerned with his weak cry and difficulty with feeding. Birth history reveals that the boy was born at the 37th week of gestation by cesarean section due to poor fetal movement and fetal distress. His Apgar scores were 3 and 5 at 1st and 5th minute respectively and his birth weight was 2.5 kg (6 lb). His vital signs include heart rate 120/min, respiratory rate 40/min, blood pressure 90/50 mm Hg, and temperature 37.0°C (98.6°F). Physical examination reveals a malnourished boy with a small narrow forehead and a small jaw. His mouth is small and he has comparatively small genitals. He has a poor muscle tone. After repeated follow-up, he gains weight rapidly but his height fails to increase. Developmental milestones are delayed at the age of 3 years. Genetic testing reveals Prader-Willi syndrome. Which of the following is the most common mechanism for the development of this patient’s condition?
- A. Anticipation
- B. Heteroplasmy
- C. Incomplete penetrance
- D. Maternal uniparental disomy
- E. Paternal deletion of 15q11-q13 (Correct Answer)
Meiosis and genetic recombination Explanation: ***Paternal deletion of 15q11-q13***
- This is the **most common genetic mechanism** (occurring in about 70-75% of cases) for Prader-Willi syndrome, involving the loss of genetic material from the paternally inherited chromosome 15 in the specified region
- The deletion affects genes that are **normally expressed only from the paternal chromosome** due to genomic imprinting, leading to the characteristic features of hypotonia, feeding difficulties in infancy, subsequent hyperphagia with obesity, hypogonadism, and developmental delays
*Incorrect: Anticipation*
- Anticipation describes a genetic phenomenon where a disorder appears earlier or symptoms become more severe with each successive generation
- This is typically seen in disorders caused by expanding **trinucleotide repeats** (e.g., Huntington's disease, myotonic dystrophy), not applicable to Prader-Willi syndrome
*Incorrect: Heteroplasmy*
- Heteroplasmy refers to the presence of more than one type of mitochondrial DNA within a cell or individual
- This concept is relevant to **mitochondrial genetic disorders** which are maternally inherited, not to Prader-Willi syndrome which is a nuclear chromosomal imprinting disorder
*Incorrect: Incomplete penetrance*
- Incomplete penetrance occurs when individuals carrying a pathogenic mutation do not express the associated clinical phenotype
- Prader-Willi syndrome typically presents with a **consistent set of features** when the genetic defect is present; incomplete penetrance is not the mechanism of disease development
*Incorrect: Maternal uniparental disomy*
- Maternal uniparental disomy (UPD) of chromosome 15 is the **second most common mechanism** for Prader-Willi syndrome (occurring in about 20-25% of cases)
- This involves inheriting **both copies of chromosome 15 from the mother** and none from the father, leading to absence of paternal gene expression in the critical 15q11-q13 region
- While less common than paternal deletion, this is still a significant cause of the syndrome
More Meiosis and genetic recombination US Medical PG questions available in the OnCourse app. Practice MCQs, flashcards, and get detailed explanations.
