Quiz 4 - GI System Practice Problems Flashcards

Question

Answer 1

The pressures would go up; the thoracic pressure goes up especially when the glottis is closed (the vocal cords would close of the larynx and not allow air out the airway). Increase pressure when deficate, urinate, or giving birth!

Answer 2

Discussion: The abnormality shown is likely to be due to partial rotation of the midgut. Midgut herniation is accompanied by a 90° rotation, which positions the cecum close to the duodenum (see panels C-D in the diagram provided). If this is the only rotation that occurs, the colon remains predominantly on the left side of the body. Normally, the ascending colon becomes secondarily retroperitoneal, forming an attachment to the posterior/dorsal abdominal wall. It is not entirely clear how Ladd’s bands form, but the most likely explanation is that they reflect an abnormal fixation of the ascending colon, followed by leftward repositioning of this region of the GI tract due to the extensive growth of the small intestines. Often this condition is associated with

Answer 3

The hindgut originally terminates in a structure known as the cloaca, that is separated from the surface of the body by the cloacal membrane (see diagram). The uterus forms by insertion of the paramesonephric ducts into the urogenital sinus. Normally the cloaca is partitioned by descent of the urorectal septum, that divides the space into the urogenital sinus (the future urinary bladder) and the rectum. If the urorectal septum does not completely descend, the future bladder, rectum and (in females) the vagina have a common opening in the position of the cloacal membrane.

Answer 4

Muscularis Mucosa The large vessels in the GI tract closest to the mucosa are in the submucosa. In order to penetrate into the submucosa, an ulcer must completely penetrate the muscularis mucosa. Complete penetration of the submucosa is not required to erode large vessels and create a significant bleed.

Answer 5

Enteric Nerve plex

Answer 6

Epithelium The enteroendocrine cells of the GI tract are found in the epithelium sublayer of the mucosa and produce a number of hormones that produce both local and systemic effects.

Answer 7

Muscularis mucosa Lymphatic aggregates in the GI tract are often found in the lamina propria or in the submucosa, so they are on one side or the other of the muscularis mucosa.

Answer 8

Surface mucous cells Surface mucous cells line the gastric pits, neck mucous cells are found in the gastric glands.

Answer 9

Parietal cells produce intrinsic factor, which is required for vitamin B12 absorption. These cells also secrete HCl into the stomach’s lumen, using the H+,K+-ATPase to energize that export. Thus, if these cells are killed by an autoimmune condition, intrinsic factor would be lost and a macrocytic anemia due to vitamin B12 deficiency would result. This is known as pernicious anemia, or Addison’s anemia, or Addison-Biermer anemia. Pernicious anemia is a megaloblastic anemia caused by low levels of circulating B12 due to the loss of parietal cells

Answer 10

Parietal cells contain H+,K+-ATPase, which is used to actively export H+; Cl- follows through chloride channels, resulting in HCl secretion. These cells are strongly stimulated by histamine. Gastrin can directly stimulate parietal cells, but it also stimulates the cells that release histamine and that creates a stronger signal. Parietal cells are also stimulated by parasympathetic fibers from the vagus nerve and submucosal enteric neurons.

Answer 11

Surface mucous cells face the lumen of the stomach, which will contain 0.1 N HCl and activated pepsin protease after a meal – this is a rough environment. Their half-life is very short, on the order of 3 days. Cells in the glands of the stomach have a much longer life-span; parietal cells, for example, live about two months in mice.

Answer 12

Chief Cell ## Footnote Chief cells secrete pepsinogen into the lumen of the glands; therefore, they contain apically located secretory granules.

Answer 13

surface mucous cell Surface mucous cells face the lumen of the stomach, which will contain 0.1 N HCl and activated pepsin protease after a meal – this is a rough environment. Their half-life is very short, on the order of 3 days. Cells in the glands of the stomach have a much longer life-span; parietal cells, for example, live about two months in mice.

Answer 14

Fungiform Papillae

Answer 15

A A identifies myoepithelial cells. These are contractile epithelial cell, which contain a smooth muscle-like myosin and are able to contract to force the secretions from the secretory acini down the ductal system.

Answer 16

C. It was absorbed by tissues with sufficient oxygen and either oxidized to CO2 and water or, in the liver, converted to glucose.

Answer 17

B. Increase

Answer 18

A. Apple Juice

Answer 19

B. Breast Milk

Answer 20

B. It occurs in most cells of the body and must be attached to the nucleotide UDP in order to be ultimately converted to glucose 6-phosphate.

Answer 21

C. Trap the sugar inside the cell.

Answer 22

## Footnote C. produces NADPH, a cofactor required for effectively reducing oxygen free radicals

Answer 23

C. Muscularis Mucosa the vessels are located in the submucosa just below the muscularis mucosa. This is the definition of a peptic ulcer

Answer 24

A. Enteric nerve plexi things don't pass and the colon swells and is infected

Answer 25

1/4... I think!

Answer 26

25% -- the offspring of the affected grandparent is an obligate carrier, and therefore their offspring all have a one half chance of being a carrier (heterozygote) and if you want two of them to both be heterozygote just use the 'and' rule of prob. and do the multiplication of 1/2\*1/2!

Answer 27

this is the glycolysis pathway. and during exercise you are cranking up the glucose pathway to lead to pyruvate, and during exercise the pyruvate would lead to lactate therefore you would have lower levels of blood lactate!

Answer 28

about 4 grams since we have abut 70mg/dL in normal blood, and about 6-7 liters of blood

Answer 29

Glucose 6-phosphatase -- remove the phoshpate from the 6 postioin on Glucose (while it is phosphorilated it can't escape the cell)

Answer 30

glucose 6 phosphatase because it is involved in glycogen breakdown and gluconeogenis -- couldn't get the glucose out of the hepatocyte!

Answer 31

Phosphoenolpyruvate carboxykinase

Answer 32

Coelom is a sac; Mesothelium is the material of the sac; paratanium is the name of the sac in the abdomonal; Parietal is the sac around the body wall serious is the sac around the organs mesentery is the sac betwee the wall and the organs.

Answer 33

C. can't transport the fats into the mitochondria for use in the liver!

Answer 34

B! MCAD are medium chain

Answer 35

B . Ketone bodies supply the brain when fasting.

Answer 36

1. Anne has insufficient insulin to control her blood glucose. This has most likely been brought on by metabolic stress from the flu and/or insufficient insulin injection. The insufficient insulin results in: a. elevated hepatic: i. glycogen breakdown ii. gluconeogenesis b. reduced glucose uptake into muscle and adipose tissue.

Answer 37

2. Metabolic acidosis. Her blood pH of 7.23 indicates acidosis. Her low PaCO2 indicates that it is metabolic and not respiratory acidosis. An elevated anion gap of 20 indicates increased levels of a metabolic acid.

Answer 38

3. Acetoacetate and β-hydroxybutyrate (ketone bodies)

Answer 39

4. Anne is in metabolic acidosis caused by the excessive ketone body production. The body tries to compensate by exhaling more CO2. (Recall that CO2 + H2OH2CO3 or carbonic acid. Therefore reducing CO2 reduces the acid load and helps to restore the pH back to normal). Bicarbonate is one of the main buffers in the blood. As it combines with protons to form carbonic acid (which dissociates to form CO2 + H2O), its levels in the blood are reduced.

Answer 40

5. Insulin is required to inhibit lipolysis (the hydrolysis of triacylglycerol to fatty acids and glycerol) in adipose tissue. Anne’s lack of insulin results in very rapid release of fatty acids and glycerol into the blood. The liver takes up much of the fatty acids and oxidizes them to acetyl CoA. The entry of acetyl CoA into the TCA cycle is limited by the energy demands of the hepatocytes so the extra acetyl CoA produced is converted to ketone bodies. The ketone bodies are released into the blood. The kidney does not reabsorb all of the ketone bodies, so some end up in the urine.

Answer 41

6. When the concentration of glucose in the blood exceeds about 220 mg/dL, the reabsorption of glucose by the kidney becomes saturated. Glucose in the urine pulls water into the urine by osmosis resulting in dehydration (osmotic diuresis).

Answer 42

7. She should be given an insulin infusion. a. Insulin will cause her muscle to take up more glucose, her liver to stop releasing glucose (by inhibiting gluconeogenesis and stimulating both glycogen synthesis and glycolysis). b. The insulin will inhibit adipose lipolysis, rapidly reducing the levels of fatty acids in the blood. The insulin will also inhibit hepatic fatty acid oxidation and stimulate hepatic fatty acid synthesis. This combination will drastically reduce the amount of acetyl CoA being used for ketone body production. c. Since the ketone bodies are the major source of Anne’s metabolic acidosis, the reduction in their production will return her blood pH to normal. d. Returning her blood glucose levels to normal will decrease the osmotic diuresis she is experiencing and will therefore relieve the dehydration. See answer 6.

Answer 43

1. Type 2 diabetes mellitus based on repeated fasting blood glucose values above 125 mg/dL. (He also has the metabolic syndrome, as defined by a combination of high blood pressure, abdominal obesity, dyslipidemia (elevated triglycerides and total/LDL- cholesterol) and elevated fasting blood glucose.)

Answer 44

i. He is insulin resistant. His skeletal muscle, adipose, liver and other tissues no longer respond to insulin as they should. The normal insulin signaling pathways in these tissues are disrupted. ii. His pancreatic beta cells are not producing enough insulin to overcome this resistance. iii. As a result of i. and ii. above, gluconeogenesis in his liver is not appropriately inhibited and glucose uptake by skeletal/cardiac muscle and adipose is not appropriately stimulated by insulin.

Answer 45

3. His beta cells continue to produce enough insulin to prevent abnormally high levels of adipose lipolysis.

Answer 46

4. i. vascular disease (atherosclerosis, cardiovascular disease, capillary problems) ii. neuropathy (nerve problems, often first showing up in the feet) iii. nephropathy (kidney problems) iv. retinopathy/cataracts

Answer 47

5. i. He should be referred to a dietitian for individualized Medical Nutrition Therapy ii. Modest weight reduction (5-10% weight loss) by a combination of calorie reduction and exercise is often effective in reducing insulin resistance. iii. A dietary pattern similar to that recommended for everyone (carbohydrates primarily coming from fresh/frozen fruits and vegetables, whole grains, legumes and low fat milk, consumption of 2 serving of fish per week, moderate sodium intake). iv. Reducing saturated fat to \<7% of total calories. v. Monitoring carbohydrate intake to achieve glycemic control. Note: there are no specific recommendations for vitamin or mineral supplements. Supplementation with antioxidants (vitamins C, E, beta-carotene) is NOT advised. NO specific recommendations for protein intake UNLESS the patient has impaired renal function (when protein would need to be limited)

Answer 48

6. i. Weight reduction (see 5 above) ii. Exercise stimulates glucose uptake into working muscles by mechanisms independent of insulin. Therefore, even people with insulin-resistant muscles can increase muscle uptake of blood glucose through exercise.

Answer 49

An anabolic peptide hormone

Answer 50

D. Skeletal muscle protein

Answer 51

C. maintian blood glucose levels while fasting or stress

Answer 52

D. Phosphoenolpyruvate carboxykinase (PEPCK)

Answer 53

B. not the liver

Answer 54

D. fatty acids

Answer 55

‘D’ represents zone 3 of the hepatic acinus model of Rappaport and cells in this zone receive blood that has passed through zones 1 and 2. As a consequence, this blood has the lowest oxygen concentration, putting the cells of zone 3 at greatest risk of damage by hypoxia. It is in this region surrounding the terminal hepatic venules that hypoxic cell damage occurs first, which can lead to the replacement of dead hepatocytes by connective tissue, or fibrosis. If the damage occurs due to insufficient blood flow, usually the result of right-side heart failure, the fibrosis is termed cardiac cirrhosis.

Answer 56

‘B’ represents zone 1 cells of the liver, which are the most active in endocytosis and, therefore, the cells that will be expected to take up the greatest amount of the radioactive metal oxides.

Answer 57

Bilirubin, a heme-containing breakdown product of hemoglobin, is typically conjugated with glucuronic acid in the liver and secreted into the bile for elimination from the body. If gallstones have blocked efflux from the gallbladder, bile components will begin to leak into the blood and bilirubin levels will then be elevated. Some cholesterol (choice c) is also likely to be present in bile, so this is a possible answer, but I’d always select the most likely and medically relevant choice when answering questions. Cholesterol is a normal component of blood, bilirubin is not. The other choices, except for amylase, are made by hepatocytes, but these are secreted into the blood not the bile, while amylase is an exocrine product of the pancreas.

Answer 58

The other organ upstream of the hepatopancreatic ampulla is the pancreas; therefore amylase is the best answer to this question. Amylase is an exocrine product of the pancreas and the only pancreas-specific answer listed. In fact, clinical labs test for elevated serum amylase in cases such as this. Tests for serum lipase, another exocrine product of the pancreas, are considered more sensitive and specific for identifying acute pancreatitis than those for amylase, but amylase analysis is more commonly available.

Answer 59

Both the parenchyma of the exocrine pancreas and parotid gland consist of serous cells arranged in acini, so that’s the closest match. The other choices all include some mucous cells, which you should be able to notice. Of course, the parotid gland will have striated ducts, which aren’t present in the pancreas, so once you recognized one of those; you’d know you were looking at a salivary gland.

Answer 60

CCK stimulates the release of pancreatic enzymes from the acinar cells.

Answer 61

Pancreatitis Secretin acts on the centroacinar cells and intercalated ducts to secrete a bicarbonate-rich fluid that helps dilute and keeps the proenzymes released from the acinar cells inactive. A failure to stimulate this secretion would be expected to increase the chance that the pancreatic enzymes could become active in the organ, leading to damage and inflammation. Please note, however, that this is reasonable speculation and not an established relationship, as I can’t find any reports of secretin deficiency linked to pancreatitis.

Answer 62

1-3-2-5-4-6-7 The canals of Hering, or bile ductules, are the small tubes into which the bile canaliculi empty. After that, the bile goes into bile ducts in portal triads, on to the gall bladder for storage and concentration, then it is released into the duodenum, absorbed in the ileum, and returned to hepatocytes via the portal vein.

Answer 63

Hepatic stellate cells, or fat cells, or Vitamin A cells, reside in the Space of Disse. They are activated by damage and transform into myofibroblasts, which synthesize extracellular matrix and secrete cytokines to support regeneration. In cases of chronic or extensive damage, these myofibroblasts may produce too much matrix and the resulting fibrosis can interfere with fluid flow through the organ and degrade its function

Answer 64

‘D’ represents a bile canaliculus, the site where the components that make up bile are secreted.

Answer 65

E ‘E’ represents an area of plasma membrane just outside a bile canaliculus, which is where tight junctions are formed between adjacent cells. These junctions serve to seal the canaliculus. There would also be tight junctions below the canaliculus in the drawing that are not shown.

Answer 66

‘B’ represents the space of Disse. This is where plasma can leave the circulatory system, so it is where you’d expect that much of the lymph would originate. The liver produces a large amount of lymph.

Answer 67

A. Alanine and Glutamine Alanine to feed into pyruvate, and glutamine to transport the amine group to urea cycle in the liver.

Answer 68

C. catalyzing the transfer of amino groups

Answer 69

F. the urea Cycle since there is extra urea cycle pre-cursers and high ammonium!

Answer 70

Insulin -- the ultimate anabolic hormone!

Answer 71

A. remember that acetoacete is a ketone!

Answer 72

D. The first step! get rid of the amino

Answer 73

A. Urea is bad for the Brain!

Answer 74

Albumin produced in the liver as the main osmotic element of the blood.

Answer 75

Lacking in essential amin acids

Answer 76

Negative nitrogen balance due to cachexia

Answer 77

Triglycerides Ingested lipids will be processed by enterocytes into chylomicrons, which are released to the lamina propria, picked up by lymphatic capillaries, and sent via the thoracic duct to the left subclavian vein. Thus, they enter the heart before they get a chance to travel to the liver. Note: fat soluble vitamins may concentrate in chylomicrons and be carried with the lymph, so I was careful to make the last choice water soluble vitamins.

Answer 78

The absorptive columnar cells (enterocytes) are the principal source of bicarbonate that protects the mucosa of the duodenum from stomach acid. Centroacinar cells also secrete bicarbonate, but that’s delivered below the pyloroduodenal opening, where the acid is delivered, and serves largely to keep the pancreatic zymogens inactive. So, A is the best answer.

Answer 79

Plicae circulares are formed by permanent folds of the submucosa.

Answer 80

IgA molecules are picked up at the basal surface of enterocytes, transported across the cells, and released from apical membranes. They are typically stuck in the mucus layer and can bind bacteria and their products in the lumen. The other substances listed move in the opposite direction.

Answer 81

Taeniae coli are only present in the wall of the colon itself, not in the appendix.

Answer 82

The brush border of enterocytes contains membrane bound hydrolases, such as amino peptidase and disaccharidase, which chop peptides and carbohydrates into smaller units that can serve as substrates for sodium-dependent uptake systems on these cells.

Answer 83

Enterocytes take in fatty acids released from the digestion of lipid by pancreatic lipase. They then assemble large structures, chylomicrons, which contain protein and fatty acids and other lipids, and release these from the basolateral membranes to be picked up by lymphatic capillaries in the lamina propria.

Answer 84

DHEA is made in the zona reticularis, the innermost layer of the adrenal cortex.

Answer 85

The adrenal cortex produces cortisol and aldosterone, so it’s destruction would explain the symptoms. Loss of the pars distalis of the pituitary would eliminate ACTH, but many other hormones as well, producing a far more complicated set of symptoms.

Answer 86

The zona glomerulosa produces aldosterone, which acts on the kidney to increase sodium recovery. The pars nervosa of the pituitary is the site of release of ADH, which acts on the kidney to increase water recovery.

Answer 87

An adenoma of the pituitary cells that produce TSH could over-stimulate the thyroid gland, producing the feelings of anxiety. The increased mass of the pituitary gland can result in compression of the overlying neural tissue, which includes the optic nerve, producing headaches and visual problems.

Answer 88

D Some cells in the pars distalis (adenohypophysis) produce and secrete prolactin, which is named for its ability to stimulate milk production, though it has many other activities, as well. ‘D’ is pointing at a cell in the pars distalis that is secreting into the blood; ‘E’ indicates the pars distalis as a region of the pituitary gland.

Answer 89

Lipoprotein lipase is required to break down chylomicrons in the blood so the fatty acids can be absorbed by adipocytes. The failure to digest chylomicrons will leave the blood appearing milky long after feeding.

Answer 90

Adipocytes synthesize lipoprotein lipase and secrete it. It is taken up by endothelial cells and displayed in the lumens of capillaries. Cardiac and skeletal muscles also produce and secrete this enzyme to supply capillary endothelial cells in those tissues.

Answer 91

Leptin levels increase with increasing fat storage in adipose tissue while adiponectin levels decrease. The fall of leptin production with increasing fat storage seems contradictory. A possible explanation is that adiponectin serves to help the body handle periods of starvation, which is likely a feature of the early evolution of humans and our ancestors, so it’s production increases with falling fat stores. A possibly related explanation is that mitochondria activities become disrupted as adipocytes grow large and adiponectin synthesis is correlated to mitochondrial activity. This is clearly a complicated business.

Answer 92

The inability to buffer lipid in adipocytes (principally absorbing lipid that enters from the diet) leads to increased deposition of lipid in liver and skeletal muscle that likely contributes to the perturbation in signaling processes that results in insulin resistance.

Answer 93

? Posed mostly as a thought question. The reduction of lipid buffering and dysregulated endocrine signaling associated with excess adipose tissue will be different than that produced by no adipose tissue at all, so while the statement is partially true, “identical” is too strong a term and the best answer is likely False.

Answer 94

Maintain blood glucose levels during fasting

Answer 95

glycogen phosphorylase