Block 2 Mock Mini II
.JPG "Block 2 Mock Mini II - Quiz")
- 1.
Your patient is unable to dorsiflex and evert his right foot. The nerve most likely damaged is the:
- A.
Common peroneal
- B.
Deep peroneal
- C.
Obturator
- D.
Superficial peroneal
- E.
Tibial
Correct Answer
A. Common peronealExplanation
The common peroneal nerve is responsible for innervating the muscles that dorsiflex and evert the foot. In this case, the patient's inability to perform these movements suggests that the common peroneal nerve is damaged. The deep peroneal nerve innervates the muscles that dorsiflex the foot, but it does not innervate the muscles that evert the foot. The obturator nerve innervates the muscles of the medial thigh and does not affect dorsiflexion or eversion of the foot. The superficial peroneal nerve innervates the muscles that evert the foot, but it does not affect dorsiflexion. The tibial nerve innervates the muscles that plantarflex the foot and does not affect dorsiflexion or eversion.Rate this question:
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- 2.
A 24-year-old female patient was slashed over the popliteal fossa, severing the nerve running vertically through the fossa. Such an injury would most likely result in
- A.
Inability to extend the leg at the knee
- B.
Foot drop
- C.
A dorsiflexed and everted foot
- D.
A plantar flexed and inverted foot
- E.
Total inability to flex the leg at the knee joint
Correct Answer
C. A dorsiflexed and everted footExplanation
The popliteal fossa is located at the back of the knee, and it contains the popliteal artery, vein, and the tibial and common fibular nerves. The tibial nerve is responsible for innervating the muscles that allow dorsiflexion (lifting the foot upwards) and inversion (turning the foot inward). If the nerve is severed, it would result in the inability to perform these actions, leading to a dorsiflexed (foot pointing downwards) and everted (foot turned outward) foot.Rate this question:
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- 3.
Which is NOT CORRECT concerning hematocrit (HCT)?
- A.
Newborns have a higher HCT than adults
- B.
In pregnancy HCT is decreased
- C.
Males generally have a higher HCT than females
- D.
HCT is a measure of the plasma and RBC (Red Blood cells)
- E.
HCT in a male is usually around 40%
Correct Answer
D. HCT is a measure of the plasma and RBC (Red Blood cells)Explanation
Hematocrit (HCT) is not a measure of the plasma and RBC (Red Blood cells). Hematocrit refers to the percentage of the total blood volume that is composed of red blood cells. It is a measure of the proportion of red blood cells in the blood. The correct answer is that HCT is not a measure of the plasma and RBC.Rate this question:
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- 4.
Which of the following substances is NOT transferred across the placenta from the maternal circulation?
- A.
IgG
- B.
Amino acids and glucose
- C.
Electrolytes
- D.
Phospholipids and cholesterol
- E.
Steroidal hormones
Correct Answer
D. pHospHolipids and cholesterolExplanation
Phospholipids and cholesterol are not transferred across the placenta from the maternal circulation. The placenta acts as a barrier between the maternal and fetal circulation, allowing for the transfer of nutrients, oxygen, and waste products, but some substances are not able to pass through. IgG antibodies, amino acids, glucose, electrolytes, and steroidal hormones can all be transferred across the placenta to support fetal development and immune protection, but phospholipids and cholesterol are not able to cross this barrier.Rate this question:
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- 5.
Where in the neuron is the action potential generated?
- A.
Postsynaptic membrane
- B.
Axon Hillock
- C.
Initial segment of the axon
- D.
First internodal segment of axon
- E.
Dendritic membrane
Correct Answer
B. Axon HillockExplanation
The action potential is generated at the axon hillock, which is the initial segment of the axon. This is because the axon hillock contains a high concentration of voltage-gated sodium channels, which are responsible for the initiation and propagation of the action potential. These channels open in response to a depolarization of the membrane potential, allowing sodium ions to enter the cell and triggering the action potential. Therefore, the axon hillock plays a crucial role in the generation of the action potential in a neuron.Rate this question:
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- 6.
Basophils can accurately be described as:
- A.
Mitotic and forming alpha, delta, and lambda granules
- B.
Increasing local vascular permeability via histamine
- C.
The most numerous of the granulocytes
- D.
Phagocytic and containing Major Basic Protein granules
- E.
Antagonistic to the action of mast cells
Correct Answer
B. Increasing local vascular permeability via histamineExplanation
Basophils are a type of white blood cell that play a role in the immune response. They are known for releasing histamine, a chemical that causes blood vessels to become more permeable, allowing immune cells and substances to move more easily to the site of infection or injury. This increased vascular permeability helps to facilitate the immune response by allowing immune cells to reach the affected area more efficiently. Therefore, the statement "increasing local vascular permeability via histamine" accurately describes the role of basophils in the immune system.Rate this question:
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- 7.
A patient with painful swelling in the distal calf cannot plantar flex at the ankle with any power. Which tendon was LIKELY ruptured?
- A.
Calcaneal
- B.
Extensor digitorum longus
- C.
Extensor hallucis longus
- D.
Plantaris
- E.
Tibialis anterior
Correct Answer
A. CalcanealExplanation
The likely ruptured tendon in this case is the Calcaneal tendon. This tendon is responsible for connecting the calf muscles to the heel bone (calcaneus) and allows for plantar flexion at the ankle. The inability to plantar flex with any power suggests a rupture of this tendon, causing the painful swelling in the distal calf.Rate this question:
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- 8.
If the femoral artery is occluded at the beginning of the adductor canal, which artery could help provide viability to the leg through collateral circulation?
- A.
Descending branch of the lateral circumflex femoral
- B.
Descending genicular
- C.
First perforating branch of the deep femoral
- D.
Medial circumflex femoral
- E.
Obturator
Correct Answer
A. Descending branch of the lateral circumflex femoralExplanation
The descending branch of the lateral circumflex femoral artery could help provide viability to the leg through collateral circulation if the femoral artery is occluded at the beginning of the adductor canal. Collateral circulation refers to the alternate pathways that blood can take when a main blood vessel is blocked. The descending branch of the lateral circumflex femoral artery is a branch of the femoral artery that runs along the lateral side of the thigh. It can provide an alternative route for blood flow to the leg, bypassing the occluded femoral artery and ensuring the viability of the leg tissues.Rate this question:
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- 9.
A construction worker lacerates the dorsum of his foot just below his ankle. Profuse bleeding that occurs would result from cutting which vessel?
- A.
Small saphenous vein
- B.
Medial plantar artery
- C.
Dorsalis pedis artery
- D.
Anterior tibial artery
- E.
Fibular artery
Correct Answer
C. Dorsalis pedis arteryExplanation
The dorsalis pedis artery is responsible for supplying blood to the dorsum (top) of the foot. If this artery is cut, it would result in profuse bleeding due to the high pressure and volume of blood flowing through it. The other options, such as the small saphenous vein and fibular artery, may also cause bleeding, but not to the same extent as the dorsalis pedis artery. The medial plantar artery and anterior tibial artery are not located in the dorsum of the foot, so they would not be responsible for the profuse bleeding in this case.Rate this question:
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- 10.
An elderly patient complains of difficulty in walking up stairs. Tests by her doctor reveal weakness in extension at her hip, but no change in hip flexion, or flexion or extension of the knee. Based upon these results, what muscle is most likely not functioning properly
- A.
Adductor magnus
- B.
Gluteus maximus
- C.
Gluteus medius
- D.
Iliopsoas
- E.
Semitendinosus
Correct Answer
B. Gluteus maximusExplanation
Based on the given information, the elderly patient has weakness in hip extension but no change in hip flexion or knee flexion/extension. The gluteus maximus is responsible for hip extension, so if there is weakness in this movement, it suggests that the gluteus maximus muscle is not functioning properly.Rate this question:
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- 11.
Patients with vitamin K deficiency are most likely to have prolonged coagulation time because of problems associated with:
- A.
Factor II
- B.
Factor VIII
- C.
Factor V
- D.
High Molecular Weight Kininogen
- E.
Bradykinin
Correct Answer
A. Factor IIExplanation
Patients with vitamin K deficiency are most likely to have prolonged coagulation time because factor II, also known as prothrombin, is dependent on vitamin K for its activation. Vitamin K is necessary for the synthesis of several clotting factors, including factor II. Without sufficient vitamin K, factor II cannot be properly activated, leading to impaired blood clotting and prolonged coagulation time. Factors VIII and V are not directly affected by vitamin K deficiency, while high molecular weight kininogen and bradykinin are not involved in the coagulation process.Rate this question:
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- 12.
Which of the following muscles is the strongest flexor of the hip?
- A.
Adductor Longus
- B.
Gluteus Medius
- C.
Gracilis
- D.
Iliopsoas
- E.
Pectineus
Correct Answer
D. IliopsoasExplanation
The iliopsoas muscle is the strongest flexor of the hip. It is a combination of the iliacus and the psoas major muscles, which work together to flex the hip joint. The iliopsoas muscle is responsible for bringing the thigh towards the abdomen, allowing for movements such as walking, running, and climbing stairs. It is located deep within the abdomen and attaches to the femur, making it a powerful muscle for hip flexion.Rate this question:
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- 13.
Which of the following muscles is innervated by the obturator nerve?
- A.
Obturator internus
- B.
Rectus femoris
- C.
Obturator externus
- D.
Long head of biceps femoris
- E.
Gemellus inferior
Correct Answer
C. Obturator externusExplanation
The obturator nerve innervates the obturator externus muscle.Rate this question:
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- 14.
Fetal hemoglobin binds oxygen with greater affinity than adult hemoglobin due primarily to which statement?
- A.
The heme iron of fetal hemoglobin is Fe3+
- B.
Fetal hemoglobin binds 2 hemes per subunit
- C.
Fetal hemoglobin is monomeric
- D.
Fetal hemoglobin forms more salt bridges
- E.
Fetal hemoglobin has a lower affinity for 2,3-BPG
Correct Answer
E. Fetal hemoglobin has a lower affinity for 2,3-BPGExplanation
Fetal hemoglobin has a lower affinity for 2,3-BPG. 2,3-BPG is a molecule that binds to adult hemoglobin and reduces its affinity for oxygen, allowing for the release of oxygen in tissues. Fetal hemoglobin, on the other hand, has a lower affinity for 2,3-BPG, meaning it is less likely to bind to it. This results in fetal hemoglobin having a higher affinity for oxygen, allowing for efficient oxygen transfer from the mother to the fetus in the placenta.Rate this question:
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- 15.
Which of the following is produced by endothelial cells in order to keep platelets from activating?
- A.
Seretonin
- B.
Thromboxane
- C.
Prostacylcin
- D.
Leukotrienes
- E.
Phosphatidyl Inositol
Correct Answer
C. ProstacylcinExplanation
Prostacyclin is produced by endothelial cells to prevent platelet activation. It acts as a vasodilator and inhibits platelet aggregation, thereby preventing the formation of blood clots. Serotonin, thromboxane, and leukotrienes are all involved in platelet activation and aggregation, while phosphatidyl inositol is not directly related to platelet activation. Therefore, prostacyclin is the correct answer as it is specifically produced by endothelial cells to inhibit platelet activation.Rate this question:
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- 16.
(the rest of the quiz is from mini 1 part 2) The correct property of glucokinase is:
- A.
It is inhibited by its product glucose-6-phosphate
- B.
It is present in muscle cells
- C.
It has a lower affinity for glucose than hexokinase
- D.
It has a higher affinity for glucose than hexokinase
- E.
It is fully saturated at normal blood glucose levels
Correct Answer
C. It has a lower affinity for glucose than hexokinaseExplanation
Glucokinase has a lower affinity for glucose than hexokinase. This means that glucokinase requires a higher concentration of glucose to be activated compared to hexokinase. Glucokinase is primarily found in the liver and pancreatic beta cells, where it plays a key role in glucose metabolism. Its lower affinity for glucose allows the liver to efficiently remove excess glucose from the blood and store it as glycogen. In contrast, hexokinase, which is found in most tissues, has a higher affinity for glucose and is involved in the initial steps of glucose metabolism.Rate this question:
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- 17.
Which of the following does not inhibit PFK-1?
- A.
Glucagon
- B.
ATP
- C.
Citrate
- D.
Low pH
- E.
Insulin
Correct Answer
E. InsulinExplanation
Insulin does not inhibit PFK-1. PFK-1 is an enzyme involved in glycolysis, the process by which glucose is converted into energy. Insulin, a hormone released by the pancreas, promotes glucose uptake by cells and stimulates glycolysis. Therefore, it does not inhibit PFK-1. On the other hand, glucagon, ATP, citrate, and low pH can all inhibit PFK-1 activity.Rate this question:
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- 18.
A baby had suffered damage to ERB’s point of the brachial plexus, by excessive traction on its head at birth. Which of the following is not likely to be evident?
- A.
Inability to abduct the arm at the shoulder
- B.
Inability to supinate the forearm
- C.
Weakened flexion of the elbow
- D.
Paralysis of the abductors of the fingers
- E.
Loss of sensation on the lateral side of the arm and forearm
Correct Answer
D. Paralysis of the abductors of the fingers -
- 19.
Inability to supinate the forearm could result from an injury of which of the following pairs of nerves?
- A.
Suprascapular and axillary
- B.
Musculocutaneous and median
- C.
Axillary and radial
- D.
Radial and musculocutaneous
- E.
Median and ulnar
Correct Answer
D. Radial and musculocutaneousExplanation
Inability to supinate the forearm refers to the inability to turn the palm of the hand upward. The muscles responsible for supination are primarily innervated by the radial nerve, which controls the movement of the forearm and wrist. The musculocutaneous nerve also plays a role in supination by innervating some of the muscles involved. Therefore, injury to both the radial and musculocutaneous nerves would result in the inability to supinate the forearm.Rate this question:
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- 20.
Which of the processes establishes the three definitive germ layers?
- A.
Angiogenesis
- B.
Cranio caudal folding
- C.
Gastrulation
- D.
Neurulation
- E.
Notochordal formation
Correct Answer
C. GastrulationExplanation
Gastrulation is the process that establishes the three definitive germ layers, which are the ectoderm, mesoderm, and endoderm. During gastrulation, the blastula undergoes a series of cell movements and rearrangements to form the three germ layers. The ectoderm gives rise to the nervous system, epidermis, and other structures. The mesoderm develops into muscles, bones, and connective tissues. The endoderm forms the lining of the digestive tract, respiratory system, and other internal organs. Therefore, gastrulation is the correct process that leads to the formation of these three germ layers.Rate this question:
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- 21.
The formation of the neural tube:
- A.
Begins in the extreme caudal end of the embryonic disk and proceeds to the cranial end.
- B.
Is not completed until the posterior neural tube closes on about the 34th day
- C.
Begins early in the Fifth week.
- D.
Is affected by the fusion of the neural folds
- E.
Is unique in that it is not affected by inductive effects of adjacent tissues such as the notochord
Correct Answer
D. Is affected by the fusion of the neural foldsExplanation
The formation of the neural tube is affected by the fusion of the neural folds. This process begins in the extreme caudal end of the embryonic disk and proceeds to the cranial end. It is not completed until the posterior neural tube closes on about the 34th day. The fusion of the neural folds is a crucial step in the development of the neural tube, as it allows for the proper formation and closure of the tube. This fusion is unique in that it is not influenced by the inductive effects of adjacent tissues such as the notochord.Rate this question:
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- 22.
In which of the following congenital malformations is there usually NO associated neurological problem.
- A.
Spina bifida myelocele
- B.
Spina bifida occulta
- C.
Anencephaly
- D.
Spina bifida meningomyelocele
- E.
Spina bifida meningocele
Correct Answer
B. Spina bifida occultaExplanation
Spina bifida occulta is a form of spina bifida where the spinal cord and the surrounding structures are not exposed. It is considered a mild form of spina bifida and usually does not cause any neurological problems. Unlike other forms of spina bifida, such as myelocele and meningomyelocele, which involve the protrusion of the spinal cord and its covering membranes, spina bifida occulta is often asymptomatic and does not typically require surgical intervention. Therefore, there is usually no associated neurological problem with spina bifida occulta.Rate this question:
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- 23.
Which of the following is correct concerning the intervillous spaces of the placenta?
- A.
Cytotrophoblast cells line the inside of the intervillous spaces and thus form the placental barrier.
- B.
Syncytiotrophoblastic cells line the inside of the intervillous space
- C.
Blood flows from endometrial veins into the intervillous space
- D.
Fetal capillaries are in direct contact with the intervillous space
- E.
The intervillous spaces are found in the basal layer of the endometrium
Correct Answer
B. SyncytiotropHoblastic cells line the inside of the intervillous spaceExplanation
Syncytiotrophoblastic cells line the inside of the intervillous space. The intervillous spaces are located in the placenta and are filled with maternal blood. These spaces are lined by syncytiotrophoblastic cells, which are multinucleated cells that form the outer layer of the placenta. These cells play a crucial role in nutrient and gas exchange between the maternal and fetal blood. They also help in the removal of waste products from the fetal circulation. Therefore, the correct answer is that syncytiotrophoblastic cells line the inside of the intervillous space.Rate this question:
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- 24.
Which of the following statements about the heart is false?
- A.
Contraction is initiated by a nerve impulse
- B.
Impulses are conducted from one muscle cell to the next
- C.
The heart contains a number of cells with an unstable membrane potential
- D.
The heart contains a number of cells with a stable membrane potential
- E.
The ventricles are inexcitable for most of the contraction period
Correct Answer
A. Contraction is initiated by a nerve impulseExplanation
The statement "Contraction is initiated by a nerve impulse" is false because the contraction of the heart is actually initiated by the sinoatrial (SA) node, which is a group of specialized cells located in the right atrium. The SA node acts as the natural pacemaker of the heart, generating electrical impulses that stimulate the contraction of the heart muscles. These impulses are then conducted through the atria and ventricles, causing them to contract in a coordinated manner. Therefore, the initiation of contraction in the heart is not dependent on a nerve impulse.Rate this question:
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- 25.
Positive inotropism refers to:
- A.
Decreased stroke volume
- B.
Movement of blood elements related to a change in chemical gradient
- C.
Repolarization of the SA node
- D.
Increased contractility of the heart
- E.
Decreased end-diastolic volume
Correct Answer
D. Increased contractility of the heartExplanation
Positive inotropism refers to an increased contractility of the heart. This means that the force of contraction of the heart muscle is increased, leading to a more efficient pumping of blood. This can be caused by various factors such as sympathetic nervous system stimulation or the use of certain medications. Increased contractility helps to improve cardiac output and can be beneficial in conditions where the heart is not pumping effectively.Rate this question:
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- 26.
What degree of AV heart block (if any) is present in the following EKG?
- A.
1st degree AV block
- B.
2nd degree AV block
- C.
3rd degree AV block
- D.
No AV block
- E.
This is a complete AV disassociation
Correct Answer
B. 2nd degree AV blockExplanation
The given EKG shows a 2nd degree AV block. This can be identified by the presence of P waves that are not followed by QRS complexes. In this type of AV block, some of the electrical signals from the atria fail to reach the ventricles, resulting in occasional dropped beats. This is different from a 1st degree AV block where there is a delay in the conduction of the electrical signals, and from a 3rd degree AV block where there is a complete blockage of the signals between the atria and ventricles.Rate this question:
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- 27.
Which of the values is closest to the mean electrical axis of this patient?
- A.
+180 Degrees
- B.
+90 Degrees
- C.
+30 Degrees
- D.
-30 Degrees
- E.
-150 Degrees
Correct Answer
C. +30 DegreesExplanation
The mean electrical axis represents the average direction of electrical activity in the heart. A positive value indicates that the electrical axis is shifted towards the left side of the heart, while a negative value indicates a shift towards the right side. In this case, +30 degrees is closest to the mean electrical axis because it represents a slight shift towards the left side of the heart.Rate this question:
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- 28.
Which of the following mean electrical axes is most consistent with right ventricular hypertrophy?
- A.
+60 Degrees
- B.
-60 Degrees
- C.
+120 Degrees
- D.
0 Degrees
Correct Answer
C. +120 DegreesExplanation
A mean electrical axis of +120 degrees is most consistent with right ventricular hypertrophy. In right ventricular hypertrophy, the right ventricle becomes enlarged and the electrical activity is shifted towards the right side of the heart. This results in a deviation of the mean electrical axis towards the right side, which is represented by a positive value. Therefore, a mean electrical axis of +120 degrees is the most consistent with right ventricular hypertrophy.Rate this question:
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.JPG "Which of the values is closest to the mean electrical axis of this patient?")
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Mar 21, 2023Quiz Edited by
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Oct 20, 2011Quiz Created by
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