How Well Do You Know Your Kidney
Your kidneys are two of the most important organs in your body and contain about a million nephrons. The Kidneys filter your blood and help get rid of waste products from your body. Test out how much you know about your kidneys by taking up the quiz below. All the best!
Questions and Answers
- 1.
When performing a splenctomy, it is important to remember that the lienorenal ligament contains which of the following structures?
- A.
Left gastro-omental artery
- B.
Tail of the pancreas
- C.
Left renal artery
- D.
Left ureter
- E.
Short gastric arteries
Correct Answer
B. Tail of the pancreasExplanation
During a splenectomy, the lienorenal ligament is an important structure to consider. This ligament connects the spleen to the kidney and contains various structures. One of these structures is the tail of the pancreas. Therefore, it is important to be aware of the presence of the tail of the pancreas when performing a splenectomy to avoid any damage or complications.Rate this question:
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- 2.
A 13-year-old boy is brought into the emergency department with blunt trauma to the left side of the chest during an attack by members of a rival street gang. X-ray shows fracture of rib 9th and 10th. Which of the following organs would most likely get damage?
- A.
Ascending colon
- B.
Spleen
- C.
Duodenum
- D.
Left kidney
- E.
Left lobe of the liver
Correct Answer
B. SpleenExplanation
Blunt trauma to the left side of the chest can cause damage to the spleen. The spleen is located in the upper left quadrant of the abdomen, close to the ribcage. Fractures of the 9th and 10th ribs indicate that the trauma occurred in this area. The spleen is a fragile organ and can be easily injured by blunt force, leading to internal bleeding. Therefore, in this case, the spleen is the organ most likely to be damaged.Rate this question:
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- 3.
A surgeon performing an appendectomy makes an incision through the ventrolateral wall. Which of the following corresponds to the order of penetration of the layers of the abdominal wall? 1. Internal oblique 2. External oblique 3. Peritoneum 4. Transversus abdominis
- A.
1-3-4-2
- B.
2-1-3-4
- C.
2-1-4-3
- D.
4-1-2-3
- E.
4-2-1-3
Correct Answer
C. 2-1-4-3Explanation
The correct order of penetration of the layers of the abdominal wall during an appendectomy is as follows: first, the surgeon penetrates the external oblique muscle, then the internal oblique muscle, followed by the transversus abdominis muscle, and finally, the peritoneum. Therefore, the correct answer is 2-1-4-3.Rate this question:
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- 4.
Dr Adebiyi Which of the following immune cells or structures of the gastrointestinal tract form the first line of defense against inhaled or ingested microorganisms?
- A.
G cells
- B.
M cells
- C.
Neutrophils
- D.
Intraepithelial lymphocytes
- E.
Tonsils
Correct Answer
E. TonsilsExplanation
Tonsils are lymphoid tissues located in the back of the throat and are part of the immune system. They act as the first line of defense against inhaled or ingested microorganisms by trapping and filtering out pathogens that enter the body through the mouth or nose. Tonsils contain immune cells, such as lymphocytes, that help in recognizing and destroying these microorganisms, preventing them from further invading the body. Therefore, tonsils play a crucial role in the immune response of the gastrointestinal tract against microorganisms.Rate this question:
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- 5.
Dr Dhiman One named muscle of the anterior abdominal wall arises along the inguinal ligament and iliac crest, whose fibers travel upward and medially to insert in the midline at the linea alba to the lower ribs. What is the name of the muscle described?
- A.
External oblique
- B.
Pyramidalis
- C.
Rectus abdominis
- D.
Transversus abdominis
- E.
Internal oblique
Correct Answer
E. Internal obliqueExplanation
The correct answer is Internal oblique. The description provided matches the characteristics of the internal oblique muscle. It arises along the inguinal ligament and iliac crest, and its fibers travel upward and medially to insert in the midline at the linea alba to the lower ribs.Rate this question:
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- 6.
The presence of nutrients in the distal stomach triggers the release of a hormone that directly stimulates acid secretion from parietal cells. The resulting acidification in turn triggers the release of a paracrine factor that inhibits the release of the hormone. What are the likely identities of the hormone and paracrine factor?
- A.
Acetylcholine and gastrin
- B.
Gastrin and somatostatin
- C.
GRP and acetylcholine
- D.
Histamine and somatostatin
- E.
Gastrin and histamine
Correct Answer
B. Gastrin and somatostatinExplanation
The hormone that is released in response to the presence of nutrients in the distal stomach is likely gastrin. Gastrin directly stimulates acid secretion from parietal cells. The resulting acidification triggers the release of a paracrine factor, which is likely somatostatin. Somatostatin inhibits the release of gastrin, creating a negative feedback loop to regulate acid secretion.Rate this question:
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- 7.
A biopsy specimen of the GI tract shows the transition from simple columnar epithelium with simple tubular glands (predominant cell type is goblet) to non-keratinized stratified squamous epithelium. Which part of GI tract is being examined?
- A.
Junction of upper half and lower half of anal canal
- B.
Junction of esophagus and stomach
- C.
Junction of pylorus and duodenum
- D.
Junction of excretory duct of salivary glands and oral cavity
- E.
Junction of sigmoid colon and rectum
Correct Answer
A. Junction of upper half and lower half of anal canalExplanation
The transition from simple columnar epithelium with simple tubular glands to non-keratinized stratified squamous epithelium suggests a change in the lining of the gastrointestinal (GI) tract. This transition occurs at the junction of the upper half and lower half of the anal canal. The anal canal is the final part of the GI tract where feces are eliminated from the body. The change in epithelial type is necessary to protect the anal canal from the abrasive nature of feces, hence the transition to stratified squamous epithelium.Rate this question:
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- 8.
Periodontal ligaments are derived from which of the following embryonic structures?
- A.
Ectoderm
- B.
Endoderm
- C.
Odontoblasts
- D.
Neural crest
- E.
Mesoderm
Correct Answer
D. Neural crestExplanation
Periodontal ligaments are derived from the neural crest. The neural crest is a group of cells that forms during embryonic development and gives rise to various structures in the body, including the periodontal ligaments. These ligaments play a crucial role in supporting and anchoring the teeth to the surrounding bone. Therefore, the correct answer is neural crest.Rate this question:
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- 9.
Dr White As part of a research study, a volunteer undergoes esophageal manometery. At one recording site, the pressure before a swallow was a few mmHg above atmospheric pressure. Following the initiation of a swallow, the pressure at this site increased transiently, before returning to an above atmospheric pressure. Where was the manometer sensor most likely to be situated?
- A.
The abdominal esophagus
- B.
The cardia of the stomach
- C.
The upper esophageal sphincter
- D.
The thoracic esophagus
- E.
The lower esophageal sphincter
Correct Answer
A. The abdominal esophagusExplanation
The manometer sensor is most likely to be situated in the abdominal esophagus because the pressure before a swallow was a few mmHg above atmospheric pressure, indicating that the sensor is measuring the pressure in the esophagus before the food enters the stomach.Rate this question:
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- 10.
A mouse is genetically engineered so that it lacks a small intestinal hormone that is released in response to the presence of luminal fatty acids. In these mice, what are the likely effects of the absence of this hormone on gall bladder motility?
- A.
Increased rhythmic contractions in the cephalic phase of the response to a meal
- B.
Decreased rhythmic contractions in the cephalic phase of the response to a meal
- C.
Decreased contractions on entry of chyme into the duodenum
- D.
Increased contraction on entry of chyme into the duodenum
- E.
Increases in both rhythmic contractions in the cephalic phase and in contraction on entry of chyme to the duodenum
Correct Answer
C. Decreased contractions on entry of chyme into the duodenumExplanation
The small intestinal hormone that is released in response to the presence of luminal fatty acids plays a role in stimulating contractions in the gall bladder. Therefore, the absence of this hormone would likely lead to a decrease in contractions on entry of chyme into the duodenum.Rate this question:
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- 11.
A patient who has been prescribed the antibiotic amoxicillin for an inner-ear infection complains of increased frequency of defecation and loose, watery stools that began about five days after the start of his treatment. In this patient, which of the following mechanisms is the most likely cause of his symptoms?
- A.
Increased production of short chain fatty acids (SCFAs) and decreased Na+ - dependent absorption
- B.
Decreased production of SCFAs and decreased Na+ - dependent absorption
- C.
Decreased production of SCFAs and decreased Na+ - dependent secretion
- D.
Increased production of SCFAs and decreased Na+ - dependent secretion
- E.
Increased production of SCFAs and increased Na+ - dependent absorption
Correct Answer
B. Decreased production of SCFAs and decreased Na+ - dependent absorptionExplanation
Colonic absorption:
In the proximal colon the absorption of short-chain fatty acids, including acetate, propionate, and butyrate occurs by Na+-dependent symporters related to the Na+-glucose symporter in the small intestine known as sodium-monocarboxylate transporters (SMCTs). Uptake of short-chain fatty acids by SMCTs located in the apical plasma membrane is driven by the low intracellular [Na+] maintained by the basolateral Na+/K+-ATPase. These short-chain fatty acids are used for energy by colonocytes.
Salt absorption also occurs by the parallel arrangement of two exchangers in the luminal membrane.
The other absorptive process of major significance in the distal colon is the absorption of Na+ via the epithelial Na+ channel ENaC, (which is also involved in reabsorption of Na+ in the kidney). Na+ enters the colonocyte and is then transported across the basolateral membrane by Na+,K+-ATPase. These cells are also responsible for secretion of K+ through potassium-selective channels which is the cause of the high level of K+ in the feces. Secretion of potassium is electrically coupled to sodium entry – in a similar way to the collecting duct system of the kidney.
Water and Cl- ions follow passively via the tight junctions to maintain electrical neutrality. This mode of Na+ absorption is the last line of defence against excessive loss of water in feces.
• Patients suffering from bowel inflammation often show markedly diminished expression of ENaC, perhaps contributing to their diarrheal symptoms.
It is also know that expression of ENaC can be acutely regulated in response to whole-body Na+ balance. Thus, in situations of reduced Na+ intake, the hormone aldosterone increases ENaC expression in both the colon and kidney, thereby promoting retention of Na+ and of water, as well as potassium loss.
(slide 49 smll-intst&colon-White)Rate this question:
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- 12.
A scientist genetically engineers a mouse so that it lacks a gastrointestinal hormone that is released upon acidification of the duodenal lumen. Compared with a normal animal, which of the following effects would be expected to be seen in the mutant mouse?
- A.
PH of pancreatic ductular secretion will be high
- B.
Pancreatic enzyme activity will be decreased
- C.
Volume of pancreatic secretion will be increased
- D.
Gastric acid secretion will be decreased
Correct Answer
B. Pancreatic enzyme activity will be decreasedExplanation
The gastrointestinal hormone released upon acidification of the duodenal lumen plays a role in stimulating pancreatic enzyme activity. Therefore, if the mutant mouse lacks this hormone, it would be expected that the pancreatic enzyme activity will be decreased compared to a normal animal.Rate this question:
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- 13.
As part of a diagnostic test, a gastroenterologist uses an endoscopic tube to deliver a bolus of a dilute acid solution into the distal esophagus. The bolus is sufficient to cause distension of the esophageal wall. In a normal individual, the bolus will be cleared from the esophagus predominantly by which of the following?
- A.
Primary peristalsis
- B.
Swallowing
- C.
Relaxation of the upper esophageal sphincter
- D.
Secondary peristalsis
- E.
Receptive relaxation
Correct Answer
D. Secondary peristalsisExplanation
Secondary peristalsis is the correct answer because it is responsible for clearing the bolus from the esophagus in a normal individual. Primary peristalsis is the initial wave of contraction that occurs after swallowing, but it is not sufficient to clear the bolus. Swallowing is the act of moving the bolus from the mouth to the esophagus, but it does not play a major role in clearing the bolus from the esophagus. Relaxation of the upper esophageal sphincter allows the bolus to enter the esophagus, but it does not contribute to its clearance. Receptive relaxation is the relaxation of the lower esophageal sphincter, which allows the bolus to enter the stomach, but it is not involved in clearing the bolus from the esophagus.Rate this question:
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- 14.
The parietal cell is responsible for gastric acid secretion. Which of the following features characterizes acid secretion by these cells?
- A.
Hydrogen ions are secreted passively down an electrochemical gradient into the lumen
- B.
The absorption of a bicarbonate ion into the blood is dependent on cytoplasmic carbonic anhydrase
- C.
Acid secretion is stimulated by activation of CCKB and by muscarinic receptors which give rise to increased production of cAMP within the parietal cell
- D.
Acid secretion is stimulated by increases in intracellular Ca2+ in response to H2 receptor activation
- E.
The stimulatory effect of histamine on acid secretion will be potentiated by somatostatin
Correct Answer
B. The absorption of a bicarbonate ion into the blood is dependent on cytoplasmic carbonic anhydraseExplanation
The key features of the parietal cell model to understand are:
1. The electrochemical gradient opposing secretion of H+ ions from the cell into the lumen is HUGE – about 1 million fold! The pH of the parietal cell cytoplasm is about 7, but in the gastric lumen the pH can be as low as 1.0. Coupled with the negative membrane potential of parietal cells, this is one of the steepest unfavorable electrochemical gradients known in nature.
2. H+ ions are actively secreted across the luminal membrane in exchange for K+ ions from the gastric lumen. This process consumes ATP directly (primary active transport). The transporter responsible is a member of the H+/K+ -ATPase family of proteins. This transporter is also known as “the proton pump”. The K+ ions that are exchanged for H+ come both from nutrients in the lumen, but also by secretion from the cell through channels. Potassium is above equilibrium in these cells and there is a favorable electrochemical gradient for exit into the lumen.
2. Cl- ions are secreted through channels in the luminal membrane. Chloride enters the cell primarily through a basolateral exchange process with intracellular bicarbonate. Chloride is above electrochemical equilibrium in the parietal cell which allows movement of Cl- ions into the gastric lumen through channels. Entry of Cl- ions into the lumen generates a lumen-negative potential difference, which helps reduce the electrochemical gradient opposing the secretion of H+ ions.
3. The Bicarbonate ions (HCO3-) that exit the cell across the basolateral membrane in exchange for Cl- enter the blood – they are absorbed.
4. Both the secreted H+ and absorbed HCO3- ions are derived from intracellular formation and hydrolysis of carbonic acid, a process which is dependent on the enzyme CARBONIC ANHYDRASE .
Thus secretion of H+ and absorption of HCO3- are linked. Because for every H+ secreted into the lumen, one HCO3- ion is absorbed. During stimulated acid secretion, the pH of the portal blood becomes more alkaline (higher pH) due to the increase in the concentration of HCO3-. The so called, “alkaline tide”. Some of this bicarbonate is secreted by surface cells as part of mucosal protection (later slide), it also helps buffer absorbed food acids and so is self-limiting. It lasts for around an hour and a half.
Binding of histamine to H2 receptors activates adenylyl cyclase and elevates the cytosolic concentration of cAMP. These events stimulate H+ secretion by activating by causing more proton pump molecules and Cl- channels to be inserted into the apical plasma membrane (see previous slide). Acetylcholine binds to M3 muscarinic receptors and opens Ca++ channels in the apical plasma membrane. Acetylcholine and gastrin also elevate intracellular calcium by promoting the release of Ca++ from intracellular stores, and entry from the extracellular fluid through channels.
(slide 9,11 gastric phase&Exoc-White)Rate this question:
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- 15.
Bile is a complex solution that results from hepatic and biliary tree secretions. Which of the following properties or processes is critical to bile production?
- A.
Hepatic bile production is dependent on cholecystokinin
- B.
Bile acids are absorbed primarily in the terminal duodenum for return to the liver
- C.
Conjugation of bile acids to bile salts decreases their pK making them more hydrophilic at the ambient pH of the duodenum
- D.
Hepatic bile is concentrated by gall bladder secretion of bile salts
- E.
Uptake of conjugated bile salts across the mucosal membrane in the intestine is via passive diffusion
Correct Answer
C. Conjugation of bile acids to bile salts decreases their pK making them more hydrophilic at the ambient pH of the duodenumExplanation
The process of conjugation of bile acids to bile salts decreases their pK, which means it increases their hydrophilicity. This is important for bile production because bile needs to be hydrophilic in order to emulsify fats and facilitate their digestion and absorption in the duodenum. By increasing the hydrophilicity of bile acids, conjugation allows them to dissolve in the aqueous environment of the duodenum and effectively carry out their role in fat digestion.Rate this question:
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- 16.
Mrs. Lambert Jane is frustrated that in spite of her efforts at weight reduction, she is still gaining weight. Analysis of her dietary intake reveals a daily intake of 3,200 calories. She weighs 67 kgs and has a moderately active lifestyle. Which of these figures truly represent her present total daily energy requirement?
- A.
3,540 cals
- B.
2,346 cals
- C.
1,800 cals
- D.
2,600 cals
- E.
1,500 cals
Correct Answer
B. 2,346 calsExplanation
Your total energy output. (in kcal) per day is the sum of your body's three uses of energy:
1. Resting metabolic rate
2. Thermic effect of food
3_ Physical activity
1. Basal metabolic rate (BMR):
Use general formula: Women: 10-9 kcal/kg/hr
Men: 1.0 kcal/kg/hr
Convert Weight (lb) to kg: 1 kg = 2.2 lb Multiply by formula:
BMR (kcal) = 1 (or 0.9) x kg weight x 24 (hours in clay)
2. Thermic effect of food intake (TEF):
The Thermic effect-of food is the energy the body uses in the processes of digestion and absorption. It averages 10% of the energy in the food.
Record your food intake for one day (24 hours) and calculate approximate energy value (kcal), using either Table of Food Values in Appendix A or a sim¬ple computer program.
Find .energy cost of therrnic effect of food (TEF):
TEF (kcal) = 10% Of total kcal in food con¬sumed
3. Physical activity:
Estimate your general average level of activity.
The energy used by physical activity can he approximated as a percentage of your BMR and varies with the degree of physical activity_ Use this List to select your activity level:
average Activity Level Energy Cost: % of RMR
Sedentary 20%
Very light 30%
Moderate 40%
Heavy 50%
Find the energy cost of your activity level:
Physical activity, energy cost (kcal) BMR, X your activity %
For example. if you are sedentary (mostly sitting): x 20%
4- Calculate your total energy output:
Total energy output (kcal)= BMR+TEF+ physical activity
Example 1:
A woman who weighs 130 lbs (59 kg). who eats an average of 1800 kcal per day, and who has started and maintains a regular physical exercise program_
BMR = 0.9 x 59 x 24 = 1274 kcal TEF = 1800 x 10% = 180 kcal Activity = BMP x 40% = 510 kcal Total energy output = 1964 kcal
Result: This woman will lose weight. Her energy
output is around 150 kcal per clay greater than her food intake. Because 1 lb of body weight equals approximately 3500 kcal, she will lose about 1 lb every 20 to 30 days with the above eating and exer¬cise routine.
Example 2:
A man who weighs 180 lbs (82 kg), who eats an average of 2700 kcal per day, and who has a seden¬tary lifestyle
BMR = 1 x 82 x 24 = 1968 kcal TEF 2700 x 10% =- 270 kcal Activity = BM.R. x 20% =-- 394 kcal
Total energy output 2632 kcal
Result: This man will tend to gain weight slowly over time. What would your clinical advice be to him?
(slide 44 nutrition 1- Lambert)Rate this question:
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- 17.
- A.
A
- B.
B
- C.
C
- D.
D
- E.
E
Correct Answer
E. E -
- 18.
- A.
A
- B.
B
- C.
C
- D.
D
- E.
E
- F.
F
- G.
G
- H.
H
- I.
I
- J.
J
Correct Answer
B. B -
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Mar 21, 2023Quiz Edited by
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Mar 06, 2012Quiz Created by
Chachelly
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