Quiz Practice Flashcards

1
Q

The five classes of steroid hormones (progestagens, mineralocorticoids, glucocorticoids, androgens, and estrogens):

A.are stored in the cells before release
B.bind G-protein Coupled Receptors
C.are each synthesized by unique enzymes
D.use carrier proteins for transport in the serum
E.all has the same number of carbons

A

The five classes of steroid hormones (progestagens, mineralocorticoids, glucocorticoids, androgens, and estrogens):

D.use carrier proteins for transport in the serum

a) is incorrect because steroid hormones are lipophillic and readily cross the plasma membrane. They are not stored.
b) is incorrect. ATCH/LH/FSH use G-protein coupled receptors to initiate steroid hormone biosynthesis in their target tissues (adrenal/gonadal cells). Steroid hormones use intracellular receptors that act as transcription factors.
c) is not correct because many of the enzymes used to synthesize the various classes are redundant.
*d) is correct. Because steroid hormones are lipophillic and hydrophobic, they need carrier proteins be to transported in the blood.
e) is not correct. Progestagens (C21), mineralocorticoids (C21), glucocorticoids (C21), androgens (C19), estrogens (C18).

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
2
Q

A 22-year-old woman has been taking daily doses of prednisone for several years to suppress her sarcoidosis, a potentially life-threatening a multisystem granulomatous inflammatory disease. Which of the following side effects is she MOST LIKELY to experience?

A.acromegaly
B.hypoglycemia
C.hyponatremia
D.lymphocytosis
E.osteoporosis

A

E.osteoporosis

Glucocorticoids have nothing to do with acromegaly so “A” is incorrect.
Glucocorticoids cause HYPERglycemia, so “B” is obviously incorrect.
Glucocorticoids cause sodium retention, i.e. hypernatremia, so “C” is incorrect.
Glucocorticoid are immunosuppressive and caused decreased lymphyocyte counts so “D” is incorrect.
Glucocorticoids stimulate osteoclast and inhibit osteoblast activities, so “E” is the only correct answer.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
3
Q

Which of the following statements is CORRECTLY paired with the associated corticosteroid or corticosteroid antagonist?

A.dexamethasone – potent, long-acting synthetic glucocorticoid with low mineralocorticoid activity
B.beclomethasone – corticosteroid antagonist used for induction of early-term abortions
C.cortisone – endogenous glucocorticoid that is inactivated by 11β-hydroxysteroid dehydrogenase (HSD11B2)
D.eplerenone – inhaled steroid used for treatment of asthma
E.mifepristone – corticosteroid antagonist used for treatment of hypertension

A

A.dexamethasone – potent, long-acting synthetic glucocorticoid with low mineralocorticoid activity

Dexamethasone is a high affinity, long-acting agonist with low affinity for the mineralocorticoid receptor. “A” is correct.
Beclomethasone is used primarily as an inhaled steroid because of its extensive hepatic metabolism. It is mifepristone that is used to induce early-term abortions. “B” and “E” are therefore incorrect.
Cortisol, not cortisone, is the main endogenous glucocorticoid. Cortisone is inactive and must be activated by 11βHSD1. “C” is therefore incorrect as well.
Eplerenone is a specific mineralocorticoid antagonist used for treating hypertension. It has no glucocorticoid activity and no effect in asthma. “D” is incorrect. “E” is incorrect. Mifepristone in an antagonist at both the progesterone receptor and the glucocorticoid receptor that is primarily used to induce early-term abortion; it may be used to treat Cushing’s Syndrome, but it is not used to treat hypertension.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
4
Q

Your patient comes to you complaining that over the last year he has noticed stiffness in his joints and a deeper voice. Interestingly, he says his feet and nose seem to be getting bigger. You do a blood test and it reveals that his

A.Growth hormone levels are below normal.
B.Somatostatin levels are below normal.
C.Prolactin levels are above normal.
D.Growth hormone releasing hormone levels are below normal.
E.Thyroid releasing hormone levels are below normal.

A

D.Growth hormone releasing hormone levels are below normal.

The patient is suffering from an excess of growth hormone, most likely from a pituitary tumor, so A is incorrect. Somatostatin levels will go up to attempt to inhibit the release of GH, so B is incorrect. Prolactin and TRH should not be involved in the tumor secretion so C and E are incorrect. Growth hormone will normally inhibit GHRH so GHRH should be very low. However, the tumor is not dependent on GHRH for stimulation of secretion of GH. GHRH secretion goes down, but without effect on the tumor cells. D is correct.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
5
Q

A 35-year-old woman presents with an elongated mandible and an unusually broad nose. She also complains of joint stiffness and carpal tunnel syndrome. She tells you that she has been lactating even though she has never given birth. Laboratory tests reveal that she has elevated growth hormone and prolactin levels. Which of the follow drugs would most likely cause a lowering of her GH and prolactin levels?

A.glucagon
B.somatropin
C.pegvisomant
D.bromocriptine
E.mecasermin

A

D.bromocriptine

a) Glucagon is incorrect because glucagon acts on the liver to increase gluconeogenesis and glycogenolysis resulting an increased blood glucose levels. This would do nothing to alter GH and prolactin levels.
b) Somatropin is incorrect. Somatropin is GH, and its administration would worsen the problem.
c) Pegvisomant is incorrect. Although pegvisomant is an effective GH antagonist and would improve the symptoms of acromegaly, it would not improve the galactorrhea due to excess prolactin.
d) Bromocriptine is CORRECT. Most pituitary adenomas that secrete both GH and prolactin express D2 dopamine receptors. Bromocriptine is a potent D2 agonist that will inhibit the release of both GH and prolactin.
e) Mecasermin incorrect. Mecasermin is an IGF-1 (somatomedin C) analog that stimulates growth (growth of cartilage in adults). It would make the symptoms of GH excess even worse

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
6
Q

Of the following hormones, which one is most likely to act through a mechanism that involves increased binding of its receptors to DNA?

A.progesterone
B.insulin
C.corticotropin releasing hormone
D.melatonin
E.luteinizing hormone

A

A.progesterone

a. correct –progesterone, a steroid hormone, is lipid soluble and readily diffuse through the cell membrane into target cells where it acts in the nucleus, via nuclear receptors, to regulate gene expression. The rest of the choices (b – e) are water soluble hormones that are excluded by the cell plasma membrane and therefore act on cell surface receptors.
b. incorrect – insulin is a water soluble protein hormone that acts through cell-surface receptors
c. incorrect - corticotropin releasing hormone is a water soluble protein hormone that acts through cell-surface receptors
d. melatonin is a water soluble derivative of an amino acid that acts through cell-surface receptors
e. luteinizing hormone is a water soluble glycoprotein hormone that acts through cell-surface receptors

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
7
Q

During an examination of a pregnant patient in her first trimester, you detect that her heart rate is rapid, her skin is hot and sweaty, and she has a symmetrical goiter. Her blood tests indicate that she has low levels of TSH and high levels of free T4. What would be the best course of treatment of this patient?

A.octreotide to increase TSH release from her anterior pituitary gland
B.sodium levothyroxine to bring about a negative feedback inhibition of hypothalamic TRH release
C.propylthiouracil to inhibit thyroid peroxidase
D.immediate radioactive iodine (131I) treatment, followed by levothyroxine
E.thyroidectomy

A

C.propylthiouracil to inhibit thyroid peroxidase

A. Octreotide is a somatostatin analogue used to suppress growth hormone secretion.
B. Levothyroxine is synthetic thyroid hormone; the patient already has high T4 levels, thus further supplementation would exacerbate the problem.
C. Propylthiouracil is the preferred anti-thyroid peroxidase treatment for pregnant hyperthyroid patients during the first trimester because it does not pass the placenta as efficiently as methimazole. The lowest dose possible is suggested to avoid suppressing the thyroid hormone production in the developing fetus.
D. Radioactive iodine is never recommended for pregnant women.
E. Who do you think you are? A surgeon? Surgery may cause loss of the fetus.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
8
Q

Hyperprolactinemia may be caused by

A.primary hyperthyroidism.
B.bromocriptine therapy.
C.excess hypothalamic dopamine release.
D.haloperidol therapy
E.low serum estrogen.

A

D.haloperidol therapy

a. incorrect. TRH could stimulate an elevation of prolactin levels, but high levels of T3 and T4 that occur in hyperthyroidism would decrease TRH (negative feedback).
b. incorrect. Bromocriptine is a D2 dopamine receptor agonist. Since dopamine acting at a D2 dopamine receptor inhibits prolactin synthesis and release, bromocriptine would lower, not raise, prolactin levels.
c. incorrect. Excess hypothalamic dopamine would decrease prolactin synthesis and release (just like bromocriptine).
d. correct. Haloperidol is an antipsychotic that blocks D2 dopamine receptors. So, haloperidol would block the inhibitory influence of endogenous dopamine resulting hyperprolactinemia.
e. incorrect. High levels of estrogen enhance the release of prolactin toward the end of pregnancy. So, low serum estrogen would tend to lower serum prolactin.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
9
Q

A newborn presents with short rib-polydactyly syndrome, which is sometimes associated with a large inversion on a chromosome 4. Which of the following chromosome analyses would be most useful in determining whether or not the child has that inversion?

A.High resolution Giemsa banding.
B.Chromosome painting with a chromosome 4 probe.
C.FISH with a chromosome 4 centromere probe.
D.Spectral karyotyping.
E.Comparative genomic hybridization.

A

A.High resolution Giemsa banding.

a. Correct. A large inversion will change the G-banding pattern of the chromosome.
b. Incorrect. The chromosome probe will paint the entire chromosome regardless of inversion.
c. Incorrect. Centromeric FISH probe will stain the centromere regardless of the inversion. In some cases though, when the inversion is pericentric (i.e., includes the centromere), the distance between the ends of chromosome and centromere might change due to the inversion and can be detected on metaphase spreads. Since only certain cases can be detected this way, there should be a better answer.
d. Incorrect. Since no DNA was actually lost from (or added to) this chromosome the staining will show a regular pattern (similar to b; spectral karyotyping is chromosome painting performed with an individual paint color for each chromosome).
e. Incorrect. Since no DNA was actually lost or added comparative genomic hybridization will not show any deletions or extra DNA material.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
10
Q

In the pedigree below, II-1 died of an autosomal recessive disease that is fatal in childhood. The frequency of the disease in the general population is 1/40000. II-3 is pregnant, and her husband has no family history of the disease. What is the chance that their child (III-1) will have the disease?

A.1/600
B.1/800
C.1/1200
D.1/1600
E.1/40000

A

A.1/600

II-1 is affected, therefore I-1 and I-2 are obligate heterozygotes. What is the chance of their daughter being a carrier (remember, the disease is fatal in childhood, so she cannot be a homozygote): 2/3 (see Pannett square below)

A a
A AA Aa
a aA (aa)
What is the chance of II-2 being a carrier:
2pq = 2 x 1 x square root(1/40000) = 1/100.
What is the chance of both parents (II-2 and II-3) being carriers:
2/3 x 1/100 = 1/150
What is the chance of an affected child if both parents are carriers: 1/4
What is the total risk:
1/150 x 1/4 = 1/600.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly