Block 1 Part 2
1st semester mini practice exam
- 1.
Dr. Thompson A 14-year-old female presents to you with a case of severe scoliosis present since birth. Radiology shows that the curvature of her spine has caused a marked elevation of her right 1st rib. What type of (i) motor and (ii) sensory deficits do you expect to see in her presentation? Answer: Motor Loss in:, Sensory Loss In
- A.
Abduction of the wrist ; Medial forearm
- B.
Flexion of the forearm ; Lateral forearm
- C.
Abduction of the arm ; Lateral arm
- D.
Extension of the forearm ; Dorsum of hand
- E.
Adduction of the wrist ; Medial arm
Correct Answer
E. Adduction of the wrist ; Medial armExplanation
The severe scoliosis and marked elevation of the right 1st rib are likely causing compression or impingement on the nerves in the affected area. This can result in motor and sensory deficits in the corresponding regions supplied by these nerves. In this case, the motor deficit would involve the loss of abduction of the wrist and abduction of the arm. The sensory deficit would involve the loss of sensation in the medial forearm, lateral forearm, lateral arm, dorsum of the hand, and medial arm.Rate this question:
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- 2.
Rafael Nadal is vying for a sixth French Open title this week. Coming into the championship match against Novak Djokavic, Rafa is feeling some pain during his “back hand” movement (extension of the forearm). What muscles are least likely to be irritated in his forearm?
- A.
Extensor carpi radialis brevis
- B.
Extensor carpi radialis longus
- C.
Extensor digitorum
- D.
Extensor digiti minimi
- E.
Extensor carpi ulnaris
Correct Answer
B. Extensor carpi radialis longusExplanation
The extensor carpi radialis longus muscle is least likely to be irritated in Rafael Nadal's forearm. This is because the pain is specifically mentioned to be during his "back hand" movement, which involves the extension of the forearm. The extensor carpi radialis longus muscle primarily assists in wrist extension and abduction, and is not directly involved in the backhand movement. Therefore, it is less likely to be irritated compared to the other muscles listed.Rate this question:
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- 3.
Dr. Kirera A five-year-old girl is brought to the doctor’s office after her left upper limb was violently pulled by another kid while playing at the park. The girl is screaming and appears to be in excruciating pain. She holds her injured limb in a flexed position close to her belly. Radiographic examination reveals radial head dislocation from the anular ligament of radius. The radius is also retracted proximally on the lateral side. Which of the following explains why the girl is in great pain?
- A.
Trapped radial nerve
- B.
Pinched anular ligament of radius
- C.
Strained anconeus muscle tendons
- D.
Trapped radial recurrent collateral artery
- E.
Lacerated synovial membrane
Correct Answer
B. Pinched anular ligament of radiusExplanation
The correct answer is "Pinched anular ligament of radius". In this scenario, the girl is experiencing excruciating pain because her radial head is dislocated from the anular ligament of the radius, and the ligament is pinched. This pinching causes compression and irritation of the surrounding nerves and tissues, leading to severe pain.Rate this question:
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- 4.
Dr. Kalliecharan Which of the following cell types has an abundance of rough endoplasmic reticulum and displays a basophilic cytoplasm?
- A.
Mast cell
- B.
Adipocyte
- C.
Plasma cell
- D.
Macrophage
- E.
Mesenchymal cell
Correct Answer
C. Plasma cellExplanation
Plasma cells have an abundance of rough endoplasmic reticulum and display a basophilic cytoplasm. The rough endoplasmic reticulum is responsible for the production of proteins, and plasma cells produce a large amount of antibodies, which are proteins. This high protein synthesis activity leads to the abundance of rough endoplasmic reticulum in plasma cells. The basophilic cytoplasm refers to the staining properties of the cytoplasm, which appear blue-purple when stained with basic dyes. This is due to the presence of RNA and ribosomes in the rough endoplasmic reticulum, which gives the cytoplasm a basophilic appearance.Rate this question:
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- 5.
During the 3rd week of development
- A.
Epiblast cells migrating into the cranial end of the primitive streak forms paraxial mesoderm
- B.
The embryonic disk is composed of two layers
- C.
The endodermal layer migrates into the primitive streak to form the somites
- D.
The endodermal layer gives rise to the skin
- E.
The cardiogenic area lies at the base of the primitive streak
Correct Answer
A. Epiblast cells migrating into the cranial end of the primitive streak forms paraxial mesodermExplanation
During the 3rd week of development, epiblast cells migrate into the cranial end of the primitive streak and form the paraxial mesoderm. This process is crucial in the formation of various structures in the embryo, including the skeletal muscles, vertebrae, and dermis of the skin. The paraxial mesoderm gives rise to somites, which are segmented blocks of tissue that further differentiate into different cell types. Therefore, the correct answer is that epiblast cells migrating into the cranial end of the primitive streak forms paraxial mesoderm.Rate this question:
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- 6.
Which of the following basic components of connective tissue will stain acidophilic ?
- A.
Type I collagen fibers
- B.
Mast cells
- C.
Elastic fibers
- D.
Ground substance
- E.
Proteoglycans and glycosaminoglycans
Correct Answer
A. Type I collagen fibersExplanation
Type I collagen fibers will stain acidophilic because they have a high affinity for acid dyes. Acidophilic staining is a method used to visualize connective tissue components, where acid dyes are used to stain acidic components such as collagen fibers. This staining technique helps to differentiate and identify different components of connective tissue under a microscope.Rate this question:
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- 7.
During the 1st and 2nd week of development
- A.
The cytotrophoblasts lack plasma membranes
- B.
The blastocyst is composed only of epiblast and syncytiotrophoblast
- C.
The morula implants into the endometrium
- D.
The syncytiotrophoblast degrades the endometrium of the uterus
- E.
The prechordal plate is composed of epiblast cells
Correct Answer
D. The syncytiotropHoblast degrades the endometrium of the uterusExplanation
During the 1st and 2nd week of development, the syncytiotrophoblast degrades the endometrium of the uterus. This is an important process in implantation, where the blastocyst attaches itself to the endometrial lining of the uterus. The syncytiotrophoblast, which is a layer of cells in the outer part of the blastocyst, secretes enzymes that break down the endometrial tissue, allowing the blastocyst to invade and establish a connection with the maternal blood supply. This degradation of the endometrium is crucial for successful implantation and subsequent development of the embryo.Rate this question:
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- 8.
Which structure has desmosine that is involved in the cross-linking of molecules?
- A.
Collagen type I
- B.
Collagen type II
- C.
Elastic fibers
- D.
Reticular fibers
- E.
Microfibrils
Correct Answer
C. Elastic fibersExplanation
Elastic fibers have desmosine, which is involved in the cross-linking of molecules. Desmosine is a unique amino acid found in elastin, the main component of elastic fibers. It forms covalent bonds between elastin molecules, allowing them to stretch and recoil, giving tissues elasticity. Collagen type I and type II are not involved in the cross-linking of molecules with desmosine. Reticular fibers provide structural support, while microfibrils are involved in the formation of elastic fibers but do not contain desmosine.Rate this question:
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- 9.
Which of the following is derived from ectoderm?
- A.
Bones of the arm
- B.
Cartilage of the trachea
- C.
Simple columnar cells of the stomach
- D.
Medullary region of the Adrenal gland
- E.
Epidermis of skin
Correct Answer
E. Epidermis of skinExplanation
The epidermis of the skin is derived from the ectoderm. During embryonic development, the ectoderm gives rise to the outermost layer of the skin, which is known as the epidermis. The epidermis serves as a protective barrier for the body and is responsible for functions such as regulating temperature, preventing water loss, and protecting against pathogens.Rate this question:
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- 10.
The basal lamina is composed of
- A.
Laminins and reticular fibers
- B.
Fibronectin and lamina lucida.
- C.
Lamina lucida and densa
- D.
Lamina lucida and lamina reticularis.
- E.
Cadherins and intermediate filaments.
Correct Answer
C. Lamina lucida and densaExplanation
The basal lamina is a thin layer of extracellular matrix that separates epithelial cells from connective tissue. It is composed of two layers: the lamina lucida, which is the more superficial layer closer to the epithelial cells, and the lamina densa, which is the deeper layer closer to the connective tissue. These two layers work together to provide structural support and regulate the movement of molecules between the epithelial cells and the underlying connective tissue.Rate this question:
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- 11.
Dr. Larsen In the pediatric hospital, you are seeing at a one-year-old girl. In her file, there is a karyotype showing five small acrocentric chromosomes. What would be the most likely karyotype of this girl?
- A.
47XYY
- B.
46XX, r(16)
- C.
47XX, +21
- D.
45XY, rob(14;21)
- E.
47XX, +22
Correct Answer
C. 47XX, +21Explanation
The most likely karyotype of this girl is 47XX, +21. This is because the presence of an extra chromosome 21, also known as trisomy 21 or Down syndrome, is indicated by the "+21" in the karyotype. The other options do not involve an extra chromosome 21.Rate this question:
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- 12.
In the pedigree below is shown a family where two people are affected by benign neonatal epilepsy, a disease that shows a carrier risk in the population of 1/100. Person II-3 marries a healthy woman and has a healthy daughter. The daughter marries a healthy, unrelated man; based on your best estimate of the inheritance pattern of this disease, what is the best estimate of the risk that they will have an affected child?
- A.
1 in 800
- B.
1 in 50
- C.
1 in 100
- D.
1 in 200
- E.
1 in 400
Correct Answer
E. 1 in 400Explanation
Based on the given information, Person II-3 is affected by benign neonatal epilepsy, which is a disease that shows a carrier risk in the population of 1/100. This means that there is a 1% chance that Person II-3 is a carrier of the disease. Person II-3 marries a healthy woman and they have a healthy daughter, which suggests that Person II-3 is likely a carrier and not affected by the disease. The daughter then marries a healthy, unrelated man, which means that there is a 1/100 chance that she is a carrier as well. Based on this information, the best estimate of the risk that they will have an affected child is 1 in 400, as it is the product of the carrier risk for both parents (1/100 * 1/100 = 1/10,000) and the chance that a carrier couple will have an affected child (1/10,000 * 1/4 = 1/40,000 ≈ 1/400).Rate this question:
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- 13.
The q-arm of an acrocentric chromosome has exchanged places with the p-arm of a submetacentric chromosome; both breakpoints are in band 12 of the respective chromosome arm. There are 46 chromosomes in the karyotype. The individual has a normal phenotype. This rearrangement represents which of the following?
- A.
Balanced Robertsonian translocation
- B.
Paracentric inversion
- C.
Balanced reciprocal translocation
- D.
Unbalanced reciprocal translocation
- E.
Unbalanced Robertsonian translocation
Correct Answer
C. Balanced reciprocal translocationExplanation
The given scenario describes a balanced reciprocal translocation. In this type of chromosomal rearrangement, two nonhomologous chromosomes exchange segments, resulting in a balanced rearrangement of genetic material. In this case, the q-arm of an acrocentric chromosome and the p-arm of a submetacentric chromosome have swapped places. Both breakpoints are in band 12 of the respective chromosome arm. Despite this rearrangement, the individual has a normal phenotype because there is no gain or loss of genetic material.Rate this question:
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- 14.
A study reported three families with several members showing loss or eyesight, type I microtia, and microdontia with widely spaced teeth. Most patients in addition have heart problems. This description best fits with which genetic concept?
- A.
Locus heterogeneity
- B.
Contigous gene disorder
- C.
Allelic heterogeneity
- D.
Pleiotropy
- E.
Fitness
Correct Answer
D. PleiotropyExplanation
The given description suggests that multiple traits and conditions are associated with a single genetic concept. This is known as pleiotropy, where a single gene or genetic variant influences multiple phenotypic traits. In this case, the gene or genetic variant is responsible for causing loss of eyesight, type I microtia, microdontia with widely spaced teeth, and heart problems in the affected individuals. Therefore, the best fitting genetic concept for this scenario is pleiotropy.Rate this question:
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- 15.
Dr. Meisenberg Coffee is a stimulant that can enhance effects of the autonomic nervous system. It is especially likely to enhance effects that are mediated by
- A.
β-adrenergic receptors
- B.
Nicotinic acetylcholine receptors
- C.
Muscarinic acetylcholine receptors
- D.
α1-adrenergic receptors
- E.
α2-adrenergic receptors
Correct Answer
A. β-adrenergic receptorsExplanation
Coffee contains caffeine, which is a stimulant that can activate β-adrenergic receptors. β-adrenergic receptors are found in various tissues, including the heart, lungs, and blood vessels. Activation of these receptors can lead to increased heart rate, bronchodilation, and vasodilation. Therefore, the consumption of coffee can enhance the effects mediated by β-adrenergic receptors, such as increased heart rate and improved alertness.Rate this question:
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- 16.
A pharmaceutical company is developing a new anti-obesity drug. In-vitro screening tests show that one of the side effects of this drug is the inhibition of the constitutive nitric oxide synthase. One likely side effect of this drug is
- A.
Hypoglycemia
- B.
Insulin resistance
- C.
Diarrhea
- D.
Hypertension
- E.
Long-lasting erections
Correct Answer
D. HypertensionExplanation
The inhibition of constitutive nitric oxide synthase can lead to hypertension. Nitric oxide is a vasodilator, meaning it helps relax and widen blood vessels, which in turn helps lower blood pressure. By inhibiting the production of nitric oxide, the drug can cause blood vessels to constrict, leading to increased blood pressure and hypertension.Rate this question:
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- 17.
Dr. Sands Fred is married to Helen, who was previously married to George, now deceased. George and Helen conceived one child together and adopted one child. Fred and Helen have also conceived one child. All members of Helen’s current family have had DNA fingerprinting done at a single VNTR locus. Unfortunately, the sheet that identified each child has been misplaced. Identify which fingerprint in each lane (in lanes 5, 6, and 7) correspond to each child. (see question 37 for image) Which child was adopted by George and Helen?
- A.
Child 2
- B.
Child 3
- C.
Child 1
- D.
Not enough information to tell
Correct Answer
C. Child 1Explanation
Based on the information given, it is stated that George and Helen adopted one child together. Therefore, Child 1 must be the child adopted by George and Helen.Rate this question:
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- 18.
Which of the following mutations is likely to prevent cell-cycle progression?
- A.
A mutation in a cell-surface mitogen receptor that made it active even in the absence of its mitogen ligand
- B.
A mutation that destroyed the kinase activity of the S-cyclin-Cdk complex
- C.
A mutation that allowed the kinase subunit of the G1-Cdk complex to be active independent of its phosphorylation status
- D.
A mutation that prevented binding of Rb to gene regulatory proteins
Correct Answer
B. A mutation that destroyed the kinase activity of the S-cyclin-Cdk complexExplanation
A mutation that destroyed the kinase activity of the S-cyclin-Cdk complex is likely to prevent cell-cycle progression because the S-cyclin-Cdk complex plays a crucial role in regulating the progression of the cell cycle. The kinase activity of this complex is responsible for phosphorylating target proteins that are necessary for cell-cycle progression. If the kinase activity is destroyed due to a mutation, the complex will not be able to phosphorylate its target proteins, leading to a disruption in the cell cycle and preventing further progression.Rate this question:
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- 19.
Dr. Smolanoff A deficiency of each of the following enzymes of the Embden-Myerhof glycolytic pathway leads to a nonspherocytic hemolytic anemia. A deficiency of which of the following enzymes leads to an increase in 2,3-bisphosphoglycerate levels in erythrocytes?
- A.
Aldolase
- B.
Pyruvate kinase
- C.
Phosphoglucose isomerase
- D.
Hexokinase
- E.
Triose phosphate isomerase
Correct Answer
B. Pyruvate kinaseExplanation
A deficiency of pyruvate kinase leads to an increase in 2,3-bisphosphoglycerate levels in erythrocytes. Pyruvate kinase is an enzyme involved in the final step of glycolysis, where it converts phosphoenolpyruvate to pyruvate. This step also generates ATP, which is important for the energy needs of the erythrocytes. Without sufficient pyruvate kinase activity, there is a buildup of upstream metabolites, including 2,3-bisphosphoglycerate. Increased levels of 2,3-bisphosphoglycerate can lead to a shift in the oxygen-hemoglobin dissociation curve, causing a decrease in the affinity of hemoglobin for oxygen and impairing oxygen delivery to tissues.Rate this question:
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- 20.
Streptococcus mutans, found in dental plaque, produces lactic acid from the metabolism of carbohydrates. Topical fluoride treatment in the dental office can slow the production of lactic acid, resulting in the accumulation of which metabolite?
- A.
Glucose-6-phosphate
- B.
Fructose-1,6-bisphosphate
- C.
Glyceraldehyde-3-phosphate
- D.
2-phosphoglycerate
- E.
Phosphoenolpyruvate
Correct Answer
D. 2-pHospHoglycerateExplanation
Topical fluoride treatment in the dental office can slow the production of lactic acid by Streptococcus mutans. This is because fluoride inhibits the enzyme enolase, which is involved in the conversion of 2-phosphoglycerate to phosphoenolpyruvate. Therefore, when lactic acid production is slowed, there is an accumulation of 2-phosphoglycerate, which is the metabolite mentioned in the question.Rate this question:
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- 21.
What is meant by the steady-state assumption that underlies the Michaelis-Menton relationship between substrate concentration and reaction velocity?
- A.
) The reaction velocity is linearly related to substrate concentration
- B.
The rate of breakdown of the enzyme-substrate complex equals the rate of formation of the complex
- C.
The reaction velocity is independent of substrate concentration
- D.
The rate of formation of product equals the rate of disappearance of substrate
- E.
The amount of enzyme remains constant
Correct Answer
B. The rate of breakdown of the enzyme-substrate complex equals the rate of formation of the complexExplanation
The steady-state assumption in the Michaelis-Menton relationship means that the rate of breakdown of the enzyme-substrate complex is equal to the rate of formation of the complex. This assumption implies that the concentration of the enzyme-substrate complex remains constant over time. It is this steady-state condition that allows for the formation of the Michaelis-Menton equation, which describes the relationship between substrate concentration and reaction velocity.Rate this question:
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- 22.
The statin drugs, e.g, Mevacor(Lovastatin), bind noncovalently to the enzyme HMG-CoA reductase because they mimic the transition state of the enzyme. This means they:
- A.
Are irreversible in their action.
- B.
Are considered suicide inhibitors.
- C.
Bind to the allosteric site of the enzyme.
- D.
Make the binding of the substrate appear tighter, i.e., appear to make Km of the enzyme smaller.
- E.
Compete with the substrate HMG-CoA for binding to the active site
Correct Answer
E. Compete with the substrate HMG-CoA for binding to the active siteExplanation
Statin drugs, such as Mevacor, bind to the enzyme HMG-CoA reductase in a way that allows them to compete with the substrate HMG-CoA for binding to the active site of the enzyme. This means that statin drugs can effectively block the active site and prevent the enzyme from carrying out its normal function of converting HMG-CoA to mevalonate. By competing with the substrate, statin drugs reduce the enzyme's ability to catalyze the reaction, ultimately leading to a decrease in cholesterol synthesis.Rate this question:
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Mar 21, 2023Quiz Edited by
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Oct 01, 2011Quiz Created by
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