Endocrinology and Reproduction MCQ Quiz in मराठी - Objective Question with Answer for Endocrinology and Reproduction - मोफत PDF डाउनलोड करा

The correct answer is Option 3 i.e. B and D

Concept:

• A decreased sense of smell and delayed or missing puberty are two features of Kallmann syndrome.
• This condition is a type of hypogonadotropic hypogonadism, which is a disorder brought on by a deficiency in the production of specific hormones that control sexual development.
• The hypothalamus, a region of the brain, is where these hormones are often produced.
• Men with hypogonadotropic hypogonadism are born with undescended testes (cryptorchidism) and an extremely small penis (micropenis).
• The majority of afflicted people do not experience secondary sex characteristics at puberty, including

1. Growth spurt in both sexes
2. Development of breast tissue in females,
3. Beginning of monthly periods (menstruation),
4. Growth of facial hair growth, and voice deepening in males.

• Most affected men and women remain sterile without treatment.
• The sense of smell is either impaired (hyposmia) or nonexistent in Kallmann syndrome (anosmia).
• The majority of other types of hypogonadotropic hypogonadism do not damage the sense of smell, which sets Kallmann syndrome apart from them.
• Other signs and symptoms include:
  • unilateral renal agenesis - it is a failure of the development of one kidney.
  • abnormalities in bones of toes and fingers
  • cleft lip with/without opening in the roof.
  • abnormal eye movement
  • hearing loss
  • abnormal tooth development. 

Explanation:

• Hypogonadism is a condition where the sex organ produces little or no sex hormones.
• In the absence of sex hormones, sexual development in the individual is impaired and the individual is mainly sterile.
• Kallmann syndrome is a hypogonadism caused by a decrease in the production of sex hormones by gonads.
• Hence, Statement A is incorrect and statement B is correct.
• Hypothalamus secretes gonadotropin-releasing hormone (GnRH) which in turn stimulates the pituitary gland to secrete hormones (FSH and LH) that in turn stimulate the secretion of sex hormones by the gonads.
• In the case of Kallmann syndrome, GnRH is not secreted and gonadotrophins (LH and FSH ) are not secreted.
• Hence, there is a very low level of circulating gonadotrophin in a person suffering from Kallmann syndrome.
• Hence, Statement C is incorrect and statement D is correct.

Hence, the correct answer is Option 3.

The correct answer is B and C

Concept:

• Erythroblastosis Fetalis is a process relating to a blood incompatibility disorder. 
• This disorder is characterized  by  the mother rejecting the fetus  due to an incompatibility of blood group types caused by previous pregnancy or blood transfusions.
• Gamma globulin’s are transported across the placenta and attach themselves to antigenic sites on the fetal red blood cells. 
• Fetal erythrocytes are thus sensitized and removed from the circulation by the fetal  reticuloendothelial system thus creating an anemia.
• Erythroblastosis fetalis originates from the fact that during this removal process, the hemoglobin concentration in the patient falls due to the accelerated red cell destruction.  New red blood cells are produced by the fetus in an attempt  to correct the anemia that has been created.
• Excessive production of erythroblasts (immature cells), a large portion of the circulating volume is compromised and the infant is born with severe anemia, compensatory erythropoiesis, severe hyperbilirubinemia, and congestive heart failure.

Fig 1: Mechanism of Erythroblastosis Fetalis

Explanation:

• ​Treatment for Erythroblastosis Fetalis is if a baby experiences erythroblastosis fetalis in the womb, they may be given intrauterine blood transfusions to reduce anemia. When the baby’s lungs and heart mature enough for delivery, a doctor may recommend delivering the baby early.
• After a baby is born, further blood transfusions may be necessary. Giving the baby fluids intravenously can improve low blood pressure. The baby may also need temporary breathing support from a ventilator or mechanical breathing machine.

A. Mother Rh+ and father Rh: This scenario generally does not cause erythroblastosis fetalis, as the mother is Rh positive and thus does not produce anti Rh antibodies.

B. Mother Rh and father Rh+: Correct cause leading to erythroblastosis fetalis.

C. Replacement of neonate's blood with Rh blood:This treatment is relevant when the newborn is Rh positive and affected by erythroblastosis fetalis, as replacing the neonate's blood with Rh negative blood helps remove the antibodies.

D. Replacement of neonate's blood with Rh+ blood: Since the issue arises due to the presence of Rh+ blood in the neonate already being attacked by antibodies, replacing it with Rh+ blood would not be an effective treatment.

The correct answer is Option 2 i.e. A,C and E.

Concept:

• hCG is a glycoprotein with the molecular weight of 36,000 -40,000 daltons.
• It is produced by the placental trophoblastic cells and it is made up of two subunits- α and β.
• α-subunit of the hormone is produced by the cytotrophoblasts and it is structurally similar to the LH, FSH, and TSH.
• β-subunit of the hormone is produced by the syncytiotrophoblasts and can it has a unique structure. 
• Because of its unique structure, β-subunit is the earliest characteristic clinical marker for detecting pregnancy. 
• The following are the function of hCG:
  • It rescues and maintains the corpus luteum for 6-8 weeks of pregnancy.
  • It stimulates the secretion of progesterone by the corpus luteum.
  • In males, it stimulates, Leydig cells to secrete testosterone hormone.
  • It stimulates placental and adrenal steroidogenesis.
  • It stimulates the maternal thyroid.
• A low level of hCG is detected in the case of early pregnancy and ectopic pregnancy.

Explanation:

• Human chorionic gonadotropin (hCG), is a glycosylated protein, it has 8 chains of carbohydrates.
• The α-subunit contains two N-glycosylation sites while the β-subunit contains 6 glycosylated sites. 
• Galactose, mannose, fructose, GluNAC and GalNAC are present in the glycosylated chain of the hCG.
• So, Statement A is correct.
• hCG is a dimeric protein consisting of one α and one β-subunit.
• α-subunit consists of 92 amino acid residues while the β-subunit that consists of 145 amino acid residues. 
• So, the β-subunit is the larger subunit while the α-subunit is the smaller subunit. 
• So, Statement B is incorrect and Statement C is correct.
• α-subunit of the hCG is identical to the alpha subunit of LH, FSH and TSH.
• So, Statement D is incorrect.
• hCG is secreted right after implantation occurs in the uterus, since, implantation occurs after 6-7 days after conception, hence, traces of hCG can be detected in the blood 6 days after conception and 14 days after conception in urine.
• So, Statement E is correct. 

Hence, the correct answer is Option 2.

The correct answer is Only C

Explanation:

• Calcitropic hormones, such as calcitriol (active vitamin D), parathyroid hormone (PTH), and calcitonin, play a key role in maintaining calcium homeostasis in the body. These hormones regulate calcium levels in the blood by influencing calcium absorption, bone remodeling, and renal calcium reabsorption or excretion.
• Calcium homeostasis is crucial for various physiological processes, including bone health, muscle contraction, nerve signaling, and blood clotting.
  • Option A:  Calcium-binding proteins (CaBP) enhance the movement of calcium from the intestinal brush border into the cytoplasm of enterocytes. This process is essential for intestinal calcium absorption, which is regulated by calcitriol. Therefore, this statement is correct.
  • Option B: Calcitonin acts on receptors present in osteoclasts, where it increases cyclic AMP (cAMP) production. This action inhibits osteoclast activity, reducing bone resorption and contributing to calcium homeostasis. Hence, this statement is also correct.
  • Option C: Parathyroid hormone (PTH) does not work independently to mobilize bone mineral. Instead, it often works in concert with vitamin D (calcitriol) to regulate calcium homeostasis. PTH stimulates the production of calcitriol in the kidneys, which enhances calcium absorption in the intestines and works synergistically to maintain calcium levels. Therefore, this statement is incorrect.
  • Option D: Calcitriol, the active form of vitamin D, is the major calcitropic hormone responsible for regulating intestinal calcium absorption. It increases the synthesis of calcium-binding proteins (CaBP) in enterocytes, thereby facilitating calcium uptake from the diet. This statement is correct.

The correct answer is C and D

Explanation:

• The thyroid gland produces thyroid hormones, primarily thyroxine (T4) and triiodothyronine (T3), which are critical for regulating metabolism, growth, and development.
• The biosynthesis of thyroid hormones involves several steps, including the uptake of iodide (I^-), its transport into the colloid, iodination of tyrosine residues on thyroglobulin, and coupling reactions to form T3 and T4.
• Key transporters and enzymes, such as the Na+/I- symporter (NIS), pendrin, thyroid peroxidase (TPO), and thyroglobulin, play vital roles in this process.

Statement C: "Pendrin, a Cl^-/I^- exchanger, helps I^- entry into the colloid" is correct.

• Pendrin, a Cl^-/I^- exchanger located on the apical membrane of thyroid follicular cells, facilitates the transport of iodide (I^-) from the cytoplasm into the colloid of the thyroid follicle.
• This step is essential for the subsequent iodination of tyrosine residues on thyroglobulin.

Statement D: "Iodination of tyrosine residue takes place first on the 3rd position in the thyroglobulin protein" is correct.

• In the colloid, iodide is oxidized by thyroid peroxidase (TPO) and incorporated into the tyrosine residues of thyroglobulin.
• The iodination occurs at the 3rd position of the tyrosine ring, forming monoiodotyrosine (MIT). A subsequent iodination at the 5th position leads to the formation of diiodotyrosine (DIT).

Other Options:

Statement A: "An antiporter transports two Na⁺ ions and one I^- ion across the thyroid follicular cells" is incorrect.

• The Na⁺/I^- symporter (NIS) is not an antiporter. It is a symporter that transports two Na⁺ ions and one I^- ion into the thyroid follicular cells from the bloodstream.
• This process is driven by the sodium gradient established by the Na⁺/K⁺ ATPase pump, not through antiport mechanisms.

Statement B: "Pendrin, a Cl^-/I^- symporter, helps I^- entry into the colloid" is incorrect.

• Pendrin is not a Cl^-/I^- symporter. It is a Cl^-/I^- exchanger, meaning it facilitates the exchange of chloride (Cl^-) and iodide (I^-) ions across the apical membrane.

The correct answer is A and C

Concept:

• The androgen receptor (AR) plays a critical role in the development and progression of prostate diseases, including benign prostatic hyperplasia (BPH) and prostate cancer.
• AR is a nuclear receptor that binds to androgens (male hormones like testosterone and dihydrotestosterone) to regulate gene expression. It consists of three key functional domains:
  • N-terminal domain (NTD): Responsible for transcriptional activity.
  • Nuclear localization signal (NLS) domain: Facilitates movement of AR into the cell nucleus.
  • Ligand-binding domain (LBD): Binds to androgens and mediates conformational changes in the receptor.
• In the context of prostate disease, targeting specific AR domains can help inhibit the receptor's activity and alleviate symptoms of an enlarged prostate.

Explanation:

​A. Drugs should target the N-terminal domain of the AR.

• The N-terminal domain (NTD) is essential for the transactivation function of the AR. Targeting this domain can effectively inhibit AR activity because it contains activation function-1 (AF-1) which is crucial for AR-mediated transcription. Therapeutics targeting this region could impact the AR's ability to stimulate the transcription of genes involved in prostate growth.

B. Drugs should not target the NLS domain of the AR.

• Targeting the Nuclear Localization Signal (NLS) domain could prevent AR from entering the nucleus, effectively inhibiting its function.
• Some AR-targeting therapies work by blocking nuclear translocation, so avoiding the NLS domain is not necessary..

C. The drug should bind to the ligand-binding domain of the AR.

• The ligand-binding domain (LBD) is where androgens such as testosterone and dihydrotestosterone (DHT) bind. Drugs that target this domain can effectively block androgen binding, diminishing the AR's ability to activate androgen-responsive genes that contribute to prostate enlargement. Most current anti-androgens already use this mechanism.

D. The drug should activate CYP17A1 to facilitate conversion of pregnenolone to DHEA.

• CYP17A1 (17α-hydroxylase/17,20-lyase) is an enzyme involved in steroidogenesis
• Activating CYP17A1 increases the production of dehydroepiandrosterone (DHEA), which is a precursor to testosterone and other androgens.
• This could potentially be counterproductive, since promoting androgen synthesis may worsen prostate growth, contrary to the goal of the treatment.

The correct answer is A, B, and C.

Explanation:

A. Oogenesis starts with the formation of primary oocytes during fetal development, which remain in prophase I until puberty.

• True: Oogenesis begins during fetal development when oogonia (the stem cells) divide by mitosis to form primary oocytes. These primary oocytes enter meiosis I but get arrested in prophase I until puberty, at which point they resume development in response to hormonal signals.

B. Each month, a single primary oocyte resumes meiosis, completing meiosis I and forming a secondary oocyte.

• True: Each menstrual cycle, under the influence of hormones like FSH, one primary oocyte resumes meiosis I and divides to form two cells: a larger secondary oocyte and a smaller polar body. The secondary oocyte is arrested in metaphase II and will only complete meiosis if fertilization occurs.

C. The secondary oocyte is arrested in metaphase II and only completes meiosis II if fertilization occurs.

• True: After completing meiosis I, the secondary oocyte enters meiosis II and is arrested in metaphase II. It will only complete meiosis II if it is fertilized by a sperm cell. Upon fertilization, the secondary oocyte completes meiosis II, forming a mature ovum and another polar body.

D. At the time of ovulation, the secondary oocyte is the product of meiosis I and is immediately ready to undergo fertilization.

• False: At ovulation, the secondary oocyte is indeed released, but it is arrested in metaphase II of meiosis II, not the final product of meiosis I. It is not immediately ready to undergo fertilization until it completes meiosis II after being fertilized by a sperm.

Summary:

• Statements A, B, and C correctly describe the process of oogenesis, including the arrest points and resumption of meiosis.
• Statement D is incorrect because the secondary oocyte is arrested in metaphase II at ovulation, not the end product of meiosis I, and it needs fertilization to complete meiosis II.

The correct answer is A,C and D

Explanation:

A. The theca interna cells develop the ability to secrete steroid sex hormones after stimulation by LH. - Correct.

• Theca interna cells, upon stimulation by Luteinizing Hormone (LH), gain the ability to synthesize androgens, which are then converted into estrogens by the granulosa cells through the action of the enzyme aromatase. This process is a critical part of ovarian follicle development and estrogen production.

B. The granulosa cells are responsible for the secretion of GnRH. - Incorrect.

• Gonadotropin-Releasing Hormone (GnRH) is secreted by the hypothalamus, not by granulosa cells. Granulosa cells are involved in the conversion of androgens to estrogens, the secretion of inhibin (which inhibits FSH production), and other supportive functions related to the maturation of the ovarian follicle and egg.

C. Estrogen secretion from the follicle diminishes immediately after ovulation. - Correct.

• After ovulation, the ruptured follicle transforms into the corpus luteum, which then primarily secretes progesterone and, to a lesser extent, estrogen. The shift toward progesterone production signifies a decrease in estrogen secretion from the follicle following ovulation. This change in hormone secretion prepares the uterus for potential pregnancy during the luteal phase.

D. FSH and LH cause rapid swelling and subsequent ovulation of the follicle. - Correct.

• Follicle-Stimulating Hormone (FSH) and Luteinizing Hormone (LH) are critical for the development and maturation of the ovarian follicles, leading up to ovulation. The rapid increase in LH (LH surge) triggers ovulation, the process where the mature egg is released from the ovar

The correct answer is B and D

Explanation:

A. Sperm become motile immediately after formation in the seminiferous tubules. - Incorrect.

• Sperm produced in the seminiferous tubules are initially non-motile. They gain motility and the capability to fertilize an egg as they mature during their transit through the epididymis, not immediately after their formation.

B. Sperm acquire motility after residing in the epididymis for 18 to 24 hours, even though inhibitors in the epididymal fluid prevent final motility until after ejaculation. - Correct.

• Sperm begin to acquire motility during their time in the epididymis, a process that typically takes several days rather than just 18 to 24 hours. However, it is correct that factors in the epididymal fluid prevent them from exhibiting full motility until after ejaculation, which helps conserve energy for the journey through the female reproductive tract.

C. The primary storage location for sperm is the vas deferens, where they can remain fertile for over a month. - Partially correct.

• While the epididymis is primarily recognized as the site where sperm mature and are initially stored, the vas deferens also serves as a conduit and storage site for sperm, playing a role in their transport during ejaculation. Sperm can survive in the male reproductive system for some time, but describing the vas deferens as the primary storage location might be somewhat misleading; the epididymis is more commonly noted for this. The idea that sperm can remain viable for a prolonged period within the male reproductive system is accurate.

D. Growth hormone deficiency, as seen in pituitary dwarfs, can lead to a significant reduction or absence of spermatogenesis. - Correct.

• Growth hormone (GH), along with other pituitary hormones such as LH and FSH, can affect reproductive function. GH deficiency, especially when part of broader pituitary dysfunction, can interfere with the normal development of sexual organs and the process of spermatogenesis, potentially leading to reduced sperm production or infertility.

The correct answer is B and C

Explanation:

A. Spermatozoa become fully motile within the seminiferous tubules immediately after their formation. - Incorrect. Spermatozoa initially formed in the seminiferous tubules are not fully motile. They gain motility and the ability to fertilize an egg as they pass through and mature in the epididymis.

B. Spermatozoa are stored in the epididymis, where they undergo maturation over several days. - Correct. After being produced in the seminiferous tubules, spermatozoa move to the epididymis, where they are stored and undergo further maturation over several days. This process is crucial for acquiring motility and fertilization capability.

C. Maturation in the epididymis involves the development of motility and the capability to fertilize an ovum. - Correct. During their time in the epididymis, sperm undergo important changes, including the development of motility and the ability to fertilize an ovum. Changes in the sperm membrane and the acquisition of enzymes critical for penetrating the outer layers of the ovum are part of this maturation process.

D. The acrosome reaction occurs within the epididymis before ejaculation. - Incorrect. The acrosome reaction does not occur within the epididymis. Instead, this reaction happens after ejaculation, when the sperm encounters the egg's zona pellucida. The acrosome reaction involves the release of enzymes that help the sperm penetrate the outer layers of the ovum to achieve fertilization.

Conclusion:

The correct statements that describe the maturation of sperm are B and C.