← Combined Technical (NG)
Past Questions — 2025
_______ is an example for potassium sparing diuretic.
A)
Spiranalactone
B)
Frusemide
C)
Torsemide
D)
Mannitol
Diuretics are classified based on their mechanism of action and their effect on potassium levels.
1. **Spironolactone (misspelled as Spiranalactone in the options)**: This is an aldosterone antagonist that acts on the distal convoluted tubule and collecting duct of the nephron. By blocking aldosterone, it prevents the reabsorption of sodium and the secretion of potassium, thereby 'sparing' potassium from being excreted in the urine. Thus, it is a potassium-sparing diuretic.
2. **Frusemide (Furosemide)**: This is a loop diuretic that acts on the thick ascending limb of the loop of Henle. It inhibits the Na+/K+/2Cl- cotransporter, leading to significant excretion of sodium, chloride, and potassium. It is a potassium-wasting diuretic.
3. **Torsemide**: Like Frusemide, Torsemide is a loop diuretic. It also causes the loss of potassium in the urine.
4. **Mannitol**: This is an osmotic diuretic. It increases the osmolarity of the glomerular filtrate, pulling water into the tubules. While its primary effect is on water, it typically leads to an increase in the excretion of electrolytes, including potassium.
Therefore, Spironolactone is the only potassium-sparing diuretic among the given choices.
_______ is considered as gold standard treatment for urethral obstruction in Veterinary case.
A)
Perineal urethrostomy
B)
Tube cystotomy
C)
Urethrotomy
D)
Urinary bladder marsupialisation
In veterinary medicine, particularly in male cats and dogs, urethral obstruction (often due to urethroliths or plug formation) requires a permanent solution if medical management fails or if the obstruction is recurrent.
1. **Perineal Urethrostomy (PU)**: This is considered the gold standard surgical treatment for chronic or recurrent urethral obstruction. It involves creating a permanent wide opening between the urethra and the perineum, effectively bypassing the narrowest part of the urethra (the penile urethra) to prevent future blockages.
2. **Tube Cystotomy**: This is a temporary measure where a tube is placed in the bladder to allow drainage. It is not a definitive treatment for urethral obstruction.
3. **Urethrotomy**: This involves an incision into the urethra to remove a stone. While it solves the immediate problem, it does not prevent recurrence and often leads to strictures, making it inferior to PU for long-term management.
4. **Urinary Bladder Marsupialisation**: This is the creation of a permanent opening of the bladder to the skin (usually on the ventral abdomen). This is a salvage procedure used only when PU is not possible or has failed, as it carries a high risk of ascending urinary tract infections and leakage.
A cholinergic drug that is used for the management of ruminal atony is _______.
A)
Methacholine
B)
Acetylcholine
C)
Carbachol
D)
Nor-adrenaline
Ruminal atony (stasis) is a condition where the motility of the rumen is diminished or absent, often requiring the administration of parasympathomimetics (cholinergic drugs) to stimulate muscle contraction and restore motility.
1. **Carbachol (Option c):** This is a potent, synthetic cholinergic agonist that is resistant to hydrolysis by acetylcholinesterase. Because of its stability and ability to stimulate muscarinic receptors in the gastrointestinal tract, it is clinically used in veterinary medicine to treat ruminal atony and other forms of GI stasis.
2. **Methacholine (Option a):** While a cholinergic agonist, it is primarily used in diagnostic tests (like the bronchial challenge test) rather than for the therapeutic management of ruminal atony.
3. **Acetylcholine (Option b):** This is the natural neurotransmitter. It is not used therapeutically for this condition because it is extremely rapidly degraded by acetylcholinesterase in the blood and tissues, making it ineffective as a systemic drug.
4. **Nor-adrenaline (Option d):** This is a sympathomimetic (adrenergic) drug. Activation of the sympathetic nervous system generally inhibits gastrointestinal motility, which would worsen ruminal atony rather than treat it.
The most appropriate relationship of the following intermediate host and a parasite is:
A)
Land snail, Angiostrongylus cantonensis
B)
Mosquito, Spirocerca lupi
C)
Water snail, Dicrocoelium dentriticum
D)
Fresh water fish, Paragonimus westermani
Step-by-step analysis of the options:
1. **Option (a): Land snail, Angiostrongylus cantonensis** - This is correct. Angiostrongylus cantonensis (the rat lungworm) utilizes land snails and slugs as its essential intermediate hosts. The larvae develop from the first stage (L1) to the infective third stage (L3) within these gastropods.
2. **Option (b): Mosquito, Spirocerca lupi** - This is incorrect. Spirocerca lupi (the esophageal worm of dogs) uses dung beetles (Coleoptera) as its primary intermediate host, not mosquitoes.
3. **Option (c): Water snail, Dicrocoelium dendriticum** - This is incorrect. Dicrocoelium dendriticum (the lancet liver fluke) is unique among flukes because it does not use water snails; it uses land snails (e.g., Zebrina spp.) as its first intermediate host and ants as its second intermediate host.
4. **Option (d): Fresh water fish, Paragonimus westermani** - This is incorrect. Paragonimus westermani (the lung fluke) uses a water snail as the first intermediate host and freshwater crustaceans (crabs or crayfish), not fish, as the second intermediate host.
Therefore, the only accurate pairing of an intermediate host and its corresponding parasite is the land snail and Angiostrongylus cantonensis.
Which among the statements is true for Sarcoptes?
A)
The female parasite lay up to 3000 eggs (approx.) in a burrow and dies at the end of a tunnel
B)
The female parasite lay up to 30 eggs (approx.) in a burrow and dies at the end of a tunnel
C)
The female parasite lay up to 3000 eggs (approx.) in a burrow dig another tunnel
D)
The female parasite lay up to 30 eggs (approx.) in a burrow and dig another tunnel
Sarcoptes scabiei is the mite responsible for scabies. The life cycle involves the female mite burrowing into the stratum corneum of the skin. During her lifespan, the female mite creates a burrow (tunnel) where she deposits a small number of eggs—typically ranging from 2 to 3 eggs per day, totaling approximately 20 to 30 eggs over her entire life cycle. Once the female reaches the end of her burrowing activity and completes her egg-laying process, she typically dies within the tunnel. Option (a) and (c) are incorrect because 3000 eggs is an impossibly high number for a single Sarcoptes mite. Option (d) is incorrect because the female typically dies at the end of her primary burrow rather than continuously digging new tunnels indefinitely.
Under microscopic examination of stained blood smear, various sizes of RBCs are seen which is indicative of:
A)
Erythropoiesis
B)
Hyperthyroidism
C)
Acute viral disease condition
D)
All of the above
The presence of various sizes of Red Blood Cells (RBCs) in a stained blood smear is known as 'anisocytosis'. Anisocytosis is a non-specific finding that occurs whenever there is a disruption in the normal production or destruction of RBCs.
1. Erythropoiesis: During active or accelerated erythropoiesis (such as after blood loss or in response to anemia), the bone marrow releases reticulocytes (immature RBCs), which are larger than mature erythrocytes, leading to a variation in cell size.
2. Hyperthyroidism: Thyroid hormones influence the rate of erythropoiesis. Hyperthyroidism can lead to an increase in RBC turnover and production, often resulting in anisocytosis.
3. Acute viral disease condition: Certain viral infections can cause bone marrow suppression or trigger hemolytic processes, leading to the release of abnormal or varying sizes of RBCs into the peripheral blood.
Since all three conditions can lead to the observation of varying RBC sizes (anisocytosis) under a microscope, the correct option is 'All of the above'.
A dog suffering from high fever, anaemia/pale mucous membrane, ascites with tick infestation is indicative of _______ and treatment may be given with _______.
A)
Neospora caninum, Imidocarb
B)
Trypanosoma evansi, Diaminazine aceturate
C)
Babesia gibsoni, Imidocarb
D)
Hepatozoon canis, Diaminazine aceturate
Step-by-step reasoning: 1. Clinical Signs: The dog presents with high fever, anemia (pale mucous membranes), and ascites (fluid accumulation in the abdomen). These are classic signs of a hemoprotozoan infection causing hemolytic anemia and hypoalbuminemia. 2. Vector: The mention of 'tick infestation' strongly points toward tick-borne diseases such as Babesiosis or Hepatozoonosis. 3. Pathogen Identification: Babesia gibsoni is a common tick-borne parasite in dogs that causes severe hemolytic anemia, fever, and often leads to secondary complications like ascites due to severe protein loss or organ failure. 4. Treatment: Imidocarb dipropionate is the drug of choice for treating Babesiosis (specifically B. gibsoni and B. canis). 5. Evaluating other options: (a) Neospora caninum is primarily associated with neuromuscular signs and abortion, not typically acute hemolytic anemia with ascites. (b) Trypanosoma evansi (Surra) is more common in livestock; while it causes fever and anemia, it is transmitted by tabanids/ticks but is not the primary suspect for this specific canine clinical triad. (d) Hepatozoon canis typically causes musculoskeletal pain and fever, but not the severe hemolytic anemia/pale membranes characteristic of Babesia, and Diaminazine is not the primary treatment for it.
Lactose intolerance happen where:
A)
The small intestine does not produce enough lactase
B)
The liver does not produce enough lactase
C)
The small intestine does not produce enough lactocase
D)
The liver does not enough lactocase
Lactose intolerance is a digestive disorder caused by the body's inability to digest lactose, which is the primary sugar found in milk. The digestion of lactose requires an enzyme called lactase. This enzyme is produced by the enterocytes (specialized cells) lining the villi of the small intestine. When the small intestine does not produce sufficient amounts of lactase, undigested lactose passes into the colon, where it is fermented by bacteria, leading to symptoms like bloating, gas, and diarrhea.
Analysis of options:
- Option (a) is correct because lactase is produced in the small intestine.
- Option (b) is incorrect because the liver does not produce lactase; it primarily produces bile.
- Option (c) and (d) are incorrect because 'lactocase' is a misspelling or a non-existent term in this biological context; the correct enzyme is 'lactase'.
The RBC of _______ is biconcave in nature.
A)
goat
B)
dog
C)
camel
D)
pig
In mammals, the morphology of Red Blood Cells (RBCs) varies between species. Most mammals, including dogs, goats, and pigs, possess biconcave, enucleated (non-nucleated) erythrocytes. However, the question asks to identify which of the provided options has biconcave RBCs. While goats, dogs, and pigs all generally have biconcave RBCs, the camel (Option c) is the distinct exception. Camels possess oval-shaped (elliptical) RBCs, which is an adaptation to prevent osmotic lysis and allow the blood to flow more easily during severe dehydration. In the context of comparative physiology exams, this question is typically designed to test the student's knowledge of the unique oval RBCs of the camel versus the standard biconcave shape of other mammals. Since the question asks which is biconcave, and the camel is definitely NOT biconcave, the other options are technically correct; however, in standard veterinary physiology textbooks, the dog (canine) is often used as the primary example of the classic mammalian biconcave disc. Note: If this is a single-choice question where only one can be correct, it is often a poorly phrased question because a, b, and d are all biconcave. But based on the physiological distinction, the camel is the only one that is NOT biconcave.
The etiology of alveolar hydatidosis is_______.
A)
Echinococcus granulosus
B)
Echinococcus multilocularis
C)
Echinococcus vogeli
D)
E. oligarthrus
Alveolar hydatidosis (or Alveolar Echinococcosis) is a parasitic disease caused by the larval stage of the tapeworm Echinococcus multilocularis. Unlike cystic echinococcosis, which forms slow-growing, well-defined cysts, alveolar hydatidosis is characterized by an infiltrative, multicystic growth pattern that resembles a malignant tumor, typically affecting the liver.
Breakdown of options:
- a. Echinococcus granulosus: This is the causative agent of Cystic Echinococcosis (Cystic Hydatidosis), which forms single, large fluid-filled cysts.
- b. Echinococcus multilocularis: The correct agent for Alveolar Hydatidosis.
- c. Echinococcus vogeli: This causes Polycystic Echinococcosis, primarily found in Central and South America.
- d. E. oligarthrus: A less common species of Echinococcus that does not typically cause alveolar hydatidosis in humans.
Cat is the only definitive infection for Toxoplasma gondii which is due to:
A)
Presence of delta-6-desaturase
B)
Absence of delta-6-desaturase
C)
Presence of alpha-6-desaturase
D)
Absence of alpha-6-desaturase
Toxoplasma gondii is a coccidian parasite that can infect almost all warm-blooded animals, but it can only undergo sexual reproduction (forming oocysts) in members of the Felidae family (cats). This unique requirement is linked to the metabolism of linoleic acid. Cats are deficient in the enzyme delta-6-desaturase, which means they cannot convert linoleic acid into arachidonic acid efficiently. Consequently, cats have high levels of linoleic acid in their intestinal epithelium. T. gondii requires high concentrations of linoleic acid to trigger the sexual cycle and the formation of oocysts. In other animals, the presence of delta-6-desaturase converts linoleic acid into other metabolites, preventing the parasite from completing its sexual stage. Therefore, the absence of delta-6-desaturase in cats is the biochemical reason why they are the only definitive hosts.
Consumption of measly pork leads to an infection known as:
A)
Cysticercosis
B)
Taeniasis
C)
Hydatidosis
D)
All of the above
Step-by-step reasoning: 1. 'Measly pork' refers to pork containing the larval stage (cysticerci) of the tapeworm Taenia solium. 2. When a human consumes undercooked 'measly pork', the cysticerci attach to the wall of the small intestine and develop into adult tapeworms. This intestinal infection is called Taeniasis. 3. Cysticercosis (Option a) occurs when a human ingests the eggs of Taenia solium (usually via fecal-oral route), causing the larvae to migrate to tissues like the brain or muscles; it is not caused by eating the pork itself, but by ingesting the eggs. 4. Hydatidosis (Option c) is caused by the larval stage of Echinococcus granulosus, typically transmitted from dogs to humans, not from pork. Therefore, the specific infection resulting from the consumption of measly pork is Taeniasis.
During skeletal muscle contraction, _______ binds to troponin.
A)
calcium
B)
magnesium
C)
sodium
D)
potassium
The process of skeletal muscle contraction is governed by the sliding filament theory and the excitation-contraction coupling mechanism.
1. **Mechanism**: When an action potential reaches the muscle fiber, it triggers the release of calcium ions (Ca²⁺) from the sarcoplasmic reticulum into the sarcoplasm.
2. **Binding**: These calcium ions bind specifically to **troponin C**, a protein complex located on the thin (actin) filaments.
3. **Conformational Change**: The binding of calcium causes a conformational change in the troponin complex, which in turn pulls **tropomyosin** away from the myosin-binding sites on the actin filament.
4. **Contraction**: Once these sites are exposed, myosin heads can bind to actin, forming cross-bridges and initiating the power stroke that shortens the muscle fiber.
**Why other options are incorrect:**
- **Magnesium (b)**: While magnesium is a cofactor for ATP and helps in muscle relaxation by competing with calcium, it does not trigger contraction by binding to troponin.
- **Sodium (c)**: Sodium ions are crucial for the depolarization of the sarcolemma (action potential), but they do not bind to troponin.
- **Potassium (d)**: Potassium ions are involved in repolarizing the cell membrane to its resting state, not in the activation of the contractile apparatus.
The following statement about Marek’s disease is true:
A)
It is caused by avian herpesvirus (MDV) which induce B-cell lymphomas and immunosuppression.
B)
It is caused by avian adenovirus serotype 4 (FAdV) which induce B-cell lymphomas and immunosuppression.
C)
It is caused by avian herpesvirus (MDV) which induce T-cell lymphomas and immunosuppression.
D)
It is caused by avian adenovirus serotype 4 (FAdV) which induce T-cell lymphomas and immunosuppression.
Step-by-step reasoning: 1. Etiology: Marek's Disease (MD) is caused by the Gallid alphaherpesvirus 2, commonly known as Marek's Disease Virus (MDV), which is a member of the herpesvirus family. This eliminates options (b) and (d), which incorrectly attribute the disease to an adenovirus. 2. Pathogenesis: MDV is characterized by its ability to transform lymphocytes. Specifically, it induces the neoplastic transformation of T-cells, leading to the development of T-cell lymphomas in various organs (such as the nerves, skin, and viscera). 3. Immunosuppression: The virus also targets the B-cells and T-cells in the bursa of Fabricius and thymus, leading to severe immunosuppression, which makes the birds more susceptible to other opportunistic infections. Therefore, the combination of 'avian herpesvirus', 'T-cell lymphomas', and 'immunosuppression' makes option (c) the correct statement. Option (a) is incorrect because MDV induces T-cell lymphomas, not B-cell lymphomas (B-cell lymphomas are more characteristic of avian leukosis).
Humoral type of immunity is stimulated when the antigen induced:
A)
Th1 type of immune response
B)
Th2 type of immune response
C)
Cytotoxic type of immune response
D)
All of the above
Humoral immunity refers to the aspect of the adaptive immune system mediated by antibodies produced by B-lymphocytes. The differentiation of naive CD4+ T cells into specific helper T cell subsets determines the type of immune response:
1. **Th2 (T-helper 2) cells**: These cells secrete cytokines such as IL-4, IL-5, and IL-13. These cytokines stimulate B-cells to proliferate, undergo isotype switching, and differentiate into plasma cells that secrete antibodies. Therefore, the Th2 response is the primary driver of humoral immunity.
2. **Th1 (T-helper 1) cells**: These cells secrete cytokines like IFN-gamma and IL-2, which activate macrophages and cytotoxic T-cells. This pathway drives 'Cell-Mediated Immunity' (CMI), not humoral immunity.
3. **Cytotoxic T-cells (CD8+)**: These are the effectors of cell-mediated immunity, responsible for killing virally infected or cancerous cells directly. They do not produce antibodies.
Since the question asks specifically for the stimulation of humoral immunity, the Th2 response is the correct mechanism.
In apicomplexan parasite, recognition and penetration of a target cell is taken by the following organelles such as _______ & _______, respectively.
A)
Conoid, microneme
B)
Dense granule, rhoptry
C)
Sub-pellicular microtubule, conoid
D)
Microneme, Conoid
Apicomplexan parasites (such as Toxoplasma gondii and Plasmodium) possess a specialized 'apical complex' designed for host cell entry.
1. Recognition: The conoid is a spiral-shaped organelle of tubulin filaments that helps in the mechanical probing and initial recognition/attachment to the target host cell membrane.
2. Penetration: The micronemes are small, secretory organelles that release adhesins and enzymes upon contact with the host cell. These secretions are critical for the gliding motility and the actual penetration/invasion process into the host cell.
Analysis of other options:
- Dense granules (Option b) are involved in the modification of the parasitophorous vacuole after entry, not the initial penetration.
- Rhoptries (Option b) are involved in creating the parasitophorous vacuole membrane during entry, but the initial penetration is triggered by micronemes.
- Sub-pellicular microtubules (Option c) provide structural integrity and shape to the parasite but do not actively perform recognition or penetration.
- Option d reverses the order or misidentifies the primary sequence of recognition and penetration.
Eschar is developed in the skin of human which is the hallmark lesion of:
A)
Scrub typhus
B)
Q-fever
C)
Black death
D)
Black leg
Step-by-step reasoning: 1. Definition of Eschar: An eschar is a dry, dark, necrotic scab that forms at the site of a bite or inoculation of certain pathogens. 2. Scrub Typhus: Caused by Orientia tsutsugamushi and transmitted by larval mites (chiggers). The hallmark clinical sign of scrub typhus is the formation of a painless eschar at the site of the mite bite, which is highly diagnostic. 3. Q-fever: Caused by Coxiella burnetii. It typically presents as a flu-like illness or pneumonia and does not produce a characteristic eschar. 4. Black Death (Bubonic Plague): Caused by Yersinia pestis. Its hallmark is the 'bubo' (swollen lymph node), not a primary eschar at the bite site. 5. Black Leg: This is a bacterial disease (Clostridium chauvoei) affecting livestock (cattle/sheep), not a human skin lesion disease. Therefore, the eschar is the hallmark lesion of Scrub typhus.
The nature of ASF virus is:
A)
Double stranded which multiply in the cytoplasm of the infected cell
B)
Single stranded which multiply in the cytoplasm of the infected cell
C)
Retro virus which multiply in the cytoplasm of the infected cell
D)
It can transform in any form which makes the virus extremely lethal for pigs
African Swine Fever Virus (ASFV) is a large, complex virus belonging to the family Asfarviridae. Its genome consists of a linear, double-stranded DNA (dsDNA). Unlike most DNA viruses that replicate in the nucleus of the host cell, ASFV is unique because it replicates and assembles exclusively within the cytoplasm of the infected cell (specifically in specialized 'viral factories').
Analysis of options:
- Option (a) is correct: It is double-stranded DNA and multiplies in the cytoplasm.
- Option (b) is incorrect: ASFV is not single-stranded.
- Option (c) is incorrect: ASFV is not a retrovirus (retroviruses are RNA viruses that use reverse transcriptase to integrate into the host genome).
- Option (d) is incorrect: While ASFV is extremely lethal, it does not 'transform in any form' as a biological mechanism for its lethality; its virulence is due to its ability to target macrophages and cause systemic hemorrhagic fever.
The component of vaccine against Newcastle in poultry is:
A)
Live virus attenuated vaccine- La Sota strain
B)
Killed virus vaccine- La Sota strain
C)
Live virus attenuated vaccine- S19 strain
D)
Killed virus vaccine- S19 strain
Newcastle Disease (ND) is caused by avian orthoavuloviruses. To prevent this highly contagious disease in poultry, vaccines are used. The most common vaccines used globally are live attenuated vaccines. Two of the most widely used strains for live attenuated vaccines are the La Sota strain and the B1 strain. The La Sota strain is highly effective for providing immunity and is typically administered via spray, drinking water, or injection. While killed vaccines exist, they are less common for primary immunization compared to the live attenuated La Sota strain. The S19 strain is not a standard recognized strain for Newcastle Disease vaccines (it is more commonly associated with other veterinary vaccines or specific research contexts, but not the primary ND vaccine component). Therefore, the combination of 'Live virus attenuated' and 'La Sota strain' is the correct component.
RNA is more labile to hydrolysis than DNA because of the presence of _______.
A)
2'-hydroxyl group (2'-OH) on its ribose sugar in RNA
B)
2'-hydroxyl group (2'-OH) on its fructose sugar in RNA
C)
5'-hydroxyl group (2'-OH) on its ribose sugar in RNA
D)
5'-hydroxyl group (2'-OH) on its fructose sugar in RNA
The fundamental difference between RNA (Ribonucleic Acid) and DNA (Deoxyribonucleic Acid) lies in the sugar component of their nucleotides. RNA contains ribose sugar, which has a hydroxyl group (-OH) at the 2' carbon position. DNA contains 2-deoxyribose, where this hydroxyl group is replaced by a hydrogen atom (-H).
This 2'-OH group in RNA makes the molecule significantly more chemically reactive and susceptible to hydrolysis. Under alkaline (basic) conditions, the 2'-OH group can act as a nucleophile, attacking the adjacent phosphodiester bond that links the nucleotides together. This leads to the cleavage of the RNA backbone (autocatalytic cleavage). Because DNA lacks this 2'-OH group, it is much more stable and resistant to hydrolysis, which is why DNA is the ideal molecule for long-term genetic storage.
Analysis of other options:
- Option b and d are incorrect because RNA contains ribose sugar, not fructose sugar.
- Option c is incorrect because the 5'-hydroxyl group is involved in the phosphodiester linkage in both DNA and RNA and is not the reason for the specific instability of RNA compared to DNA.
The first step in disease investigation:
A)
Analytical epidemiology
B)
Horizontal epidemiology
C)
Descriptive epidemiology
D)
Predictive epidemiology
In the field of epidemiology, the investigation of a disease typically follows a logical sequence. The first step is Descriptive Epidemiology, which focuses on characterizing the disease by observing the distribution of the health event. It answers the questions: 'Who is affected?' (Person), 'Where is it occurring?' (Place), and 'When is it occurring?' (Time). Once the descriptive patterns are established, researchers can formulate hypotheses, which are then tested using Analytical Epidemiology (the second step) to determine the cause or risk factors.
Breakdown of options:
- a. Analytical epidemiology: This is the second stage, used to test hypotheses generated during the descriptive phase to establish causal relationships.
- b. Horizontal epidemiology: This is not a standard primary step in disease investigation; it generally refers to the study of disease prevalence across a population at a single point in time (cross-sectional).
- d. Predictive epidemiology: This involves forecasting future outbreaks based on existing data, which can only happen after descriptive and analytical phases are complete.
During slaughtering of pig, the captive bolt should be placed _______.
A)
on top of head
B)
middle of forehead
C)
2.5 centimeters above eye level
D)
13 mm running down the centre of forehead
In animal welfare and humane slaughter practices, the goal of stunning (using a captive bolt) is to induce immediate unconsciousness by destroying the cerebral cortex. For pigs, the anatomical target is a specific point on the forehead. According to standard veterinary and slaughter guidelines (such as those provided by the AVMA or EFSA), the bolt should be placed in the center of the forehead, approximately 2-2.5 cm above the eyes, but the precise anatomical target for the bolt's trajectory is often described as being centered on the forehead. Option (d) refers to the specific alignment/placement along the midline of the forehead to ensure the bolt penetrates the skull at the correct angle to reach the brainstem/cerebrum. While option (c) mentions 'above eye level' (which is a general guideline for height), option (d) specifies the 'centre of forehead' which is the critical axis for a successful stun. In many technical manuals for pig stunning, the focus is on the midline (centre) of the forehead to avoid the thickest parts of the frontal bone or missing the brain entirely.
High titre of IgM is seen during serological examination that indicates:
A)
Primary infection or initial stage of infection
B)
Secondary infection or chronic stage of infection
C)
Tumor formation
D)
All of the above
In serology, Immunoglobulin M (IgM) is the first antibody produced by the immune system in response to a new infection. Because it appears early and typically disappears after the acute phase of the illness, a high titre of IgM is a diagnostic marker for a primary infection or the initial stage of an infection. In contrast, Immunoglobulin G ( IgG) is produced later and persists longer, serving as a marker for secondary infections, chronic stages, or long-term immunity. Tumor formation (Option c) is generally not diagnosed via IgM titres, as that would involve specific tumor markers or different immunological assays. Therefore, Option a is the only correct answer.
Recording of events occurring in a particular point of time is called _______.
A)
Prospective survey
B)
Retrospective survey
C)
Cross-sectional survey
D)
Longitudinal survey
In epidemiology, the timing of data collection is critical for defining the study design.
1. **Cross-sectional survey (Correct):** This type of study analyzes data from a population, or a representative subset, at a specific point in time. It is often described as a 'snapshot' of the population, recording the prevalence of a disease or the occurrence of events at that particular moment.
2. **Prospective survey (Incorrect):** This is a longitudinal study that follows a group of individuals forward in time (from the present into the future) to see if certain outcomes develop.
3. **Retrospective survey (Incorrect):** This looks backward in time, examining existing records or interviewing subjects about past events to determine the cause of a current condition.
4. **Longitudinal survey (Incorrect):** This is a general term for studies that involve repeated observations of the same variables over a long period of time (including both prospective and retrospective designs), rather than a single point in time.
The normal pH of milk is _______.
A)
4.6
B)
5.6
C)
6.6
D)
7.6
The normal pH of fresh cow's milk typically ranges from 6.6 to 6.8, making it slightly acidic. This pH level is maintained by a complex buffering system involving phosphates, citrates, and proteins (caseins).
Analysis of options:
- Option (a) 4.6: This is the isoelectric point of casein. When milk pH drops to 4.6 (due to lactic acid production by bacteria), the casein precipitates, causing the milk to curdle.
- Option (b) 5.6: This is too acidic for fresh milk and would indicate significant spoilage or fermentation.
- Option (c) 6.6: This falls within the standard physiological range (6.6–6.8) for fresh bovine milk.
- Option (d) 7.6: This is alkaline. Fresh milk is not alkaline; an alkaline pH in milk usually indicates mastitis (inflammation of the udder) or contamination.