Home/Past Papers/Punjab / UHS/Punjab Biology 2019
Punjab Biology 2019 — Solved Past Paper with Answers
All 17 MCQs from Punjab Biology 2019, solved with the correct answer highlighted and a full explanation for every question. This is a free MDCAT Punjab / UHS past paper — no signup, no ads. Practise it interactively in timed mode, drill more with free MDCAT MCQs, or browse all Punjab / UHS papers.
Attempt this paper interactively →Q1. The floral parts of a flowering plant are
- A. Homologous✓
- B. Analogous
- C. Similar
- D. Different
Explanation: The flower parts of a flowering plant are homologous. They are considered to have evolved from leaves, to form sepals, petals, stamens and carpels.
Why the other options are wrong
- B. The flower parts of a flowering plant are homologous. They are considered to have evolved from leaves, to form sepals, petals, stamens and carpels.
- C. The flower parts of a flowering plant are homologous. They are considered to have evolved from leaves, to form sepals, petals, stamens and carpels.
- D. The flower parts of a flowering plant are homologous. They are considered to have evolved from leaves, to form sepals, petals, stamens and carpels.
Q2. Mutualism is a type of
- A. Symbiosis✓
- B. Commensalism
- C. Parasitism
- D. Predation
Explanation: Mutualism is a type of symbiotic relationship where all species involved benefit from their interactions.
Why the other options are wrong
- B. Commensalism, in biology, a relationship between individuals of two species in which one species obtains food or other benefits from the other without either harming or benefiting the latter. commensalism. Category: Animals & Nature. Related Topics: mimicry host inquilinism interspecific association.
- C. Parasitism is generally defined as a relationship between the two living species in which one organism is benefitted at the expense of the other. The organism that is benefitted is called the parasite, while the one that is harmed is called the host. A few examples of parasites are tapeworms, fleas, and barnacles.
- D. Predation is a biological interaction where one organism, the predator, kills and eats another organism, its prey. It is one of a family of common feeding behaviours that includes parasitism and micropredation (which usually do not kill the host) and parasitoidism (which always does, eventually).
Q3. The average rainfall in temperate deciduous forest is between
- A. 700-2500 m.m
- B. 700-800 m.m
- C. 700-1000 m.m
- D. 700-1500 m.m✓
Explanation: The average rainfall is between 750 - 1500 mm.
Why the other options are wrong
- A. The average rainfall is between 750 - 1500 mm.
- B. The average rainfall is between 750 - 1500 mm.
- C. The average rainfall is between 750 - 1500 mm.
Q4. The two main causes of air pollution are industrialization and
- A. Automobile
- B. Urbanization✓
- C. Deforestation
- D. Overgrazing
Explanation: Air pollution as a result of industrialization and urbanization all over the world.
Why the other options are wrong
- A. Air pollution as a result of industrialization and urbanization all over the world.
- C. Air pollution as a result of industrialization and urbanization all over the world.
- D. Air pollution as a result of industrialization and urbanization all over the world.
Q5. The leaves with a very small surface area, are found in
- A. Hydrophytes
- B. Mesophytes
- C. Xerophytes✓
- D. Sciophytes
Explanation: Many xerophytes possess small, thick leaves to limit water loss by reducing surface area proportional to the volume.
Why the other options are wrong
- A. In this type the surface area of leaves is very large to transpire water excessively. Extensive stomata are present on the upper surface facing the atmosphere to promote loss of water
- B. Mesophytes have moderate water availability. In suicient supply of water stomata are kept open to promote loss of excess water, however, in restricted supply stomata close to prevent the loss e.g. Brassica, rose, mango etc.
- D. Sciophytes are plants that grow in the shade and represent the understory in forest stratification. For example, java moss and club moss.
Q6. The compound which takes part in the urea cycle is
- A. Adenine
- B. Guanine
- C. Citrulline✓
- D. Thymine
Explanation: Two ammonia and one carbon dioxide molecule are shunted into the cycle to generate one molecule of urea. One ammonia molecule combines with carbon dioxide and an already available precursor from the previous cycle ornithine to form citrulline, subsequently, another ammonia combines to form arginine. The arginine is split by arginase to form urea and the precursor ornithine for next cycle.
Why the other options are wrong
- A. Two ammonia and one carbon dioxide molecule are shunted into the cycle to generate one molecule of urea. One ammonia molecule combines with carbon dioxide and an already available precursor from the previous cycle ornithine to form citrulline, subsequently, another ammonia combines to form arginine. The arginine is split by arginase to form urea and the precursor ornithine for next cycle.
- B. Two ammonia and one carbon dioxide molecule are shunted into the cycle to generate one molecule of urea. One ammonia molecule combines with carbon dioxide and an already available precursor from the previous cycle ornithine to form citrulline, subsequently, another ammonia combines to form arginine. The arginine is split by arginase to form urea and the precursor ornithine for next cycle.
- D. Two ammonia and one carbon dioxide molecule are shunted into the cycle to generate one molecule of urea. One ammonia molecule combines with carbon dioxide and an already available precursor from the previous cycle ornithine to form citrulline, subsequently, another ammonia combines to form arginine. The arginine is split by arginase to form urea and the precursor ornithine for next cycle.
Q7. Osteomalacia includes a number of disorders in which bones receive inadequate
- A. Water
- B. Oxygen
- C. Blood
- D. Minerals✓
Explanation: Osteomalacia, often referred to as "soft bone disease," is a metabolic bone disorder characterized by the inadequate mineralization of bone tissue. This condition results from a deficiency in vitamin D, calcium, or phosphate, weakening and softening bones.
Why the other options are wrong
- A. Osteomalacia, often referred to as "soft bone disease," is a metabolic bone disorder characterized by the inadequate mineralization of bone tissue. This condition results from a deficiency in vitamin D, calcium, or phosphate, weakening and softening bones.
- B. Osteomalacia, often referred to as "soft bone disease," is a metabolic bone disorder characterized by the inadequate mineralization of bone tissue. This condition results from a deficiency in vitamin D, calcium, or phosphate, weakening and softening bones.
- C. Osteomalacia, often referred to as "soft bone disease," is a metabolic bone disorder characterized by the inadequate mineralization of bone tissue. This condition results from a deficiency in vitamin D, calcium, or phosphate, weakening and softening bones.
Q8. Each A-band has a lighter stripe in its mid section called
- A. A zone
- B. H zone✓
- C. M line
- D. Z line
Explanation: This central region of the A band looks slightly lighter than the rest of the A band and is called the H zone. The middle of the H zone has a vertical line called the M line, at which accessory proteins hold together thick filaments.
Why the other options are wrong
- A. This central region of the A band looks slightly lighter than the rest of the A band and is called the H zone. The middle of the H zone has a vertical line called the M line, at which accessory proteins hold together thick filaments.
- C. This central region of the A band looks slightly lighter than the rest of the A band and is called the H zone. The middle of the H zone has a vertical line called the M line, at which accessory proteins hold together thick filaments.
- D. This central region of the A band looks slightly lighter than the rest of the A band and is called the H zone. The middle of the H zone has a vertical line called the M line, at which accessory proteins hold together thick filaments.
Q9. The receptor cells of planaria are sensitive to
- A. Light and pressure
- B. Light, pressure and touch
- C. Touch, pressure and chemicals✓
- D. Light, pressure, touch and chemicals
Explanation: The receptor cells sensitive to pressure and touch are present in Planaria. There are no specialized sensory cells in Hydra, but some nerve cells are more sensitive, to a particular stimulus - chemical or mechanical, than others.
Why the other options are wrong
- A. The receptor cells sensitive to pressure and touch are present in Planaria. There are no specialized sensory cells in Hydra, but some nerve cells are more sensitive, to a particular stimulus - chemical or mechanical, than others.
- B. The receptor cells sensitive to pressure and touch are present in Planaria. There are no specialized sensory cells in Hydra, but some nerve cells are more sensitive, to a particular stimulus - chemical or mechanical, than others.
- D. The receptor cells sensitive to pressure and touch are present in Planaria. There are no specialized sensory cells in Hydra, but some nerve cells are more sensitive, to a particular stimulus - chemical or mechanical, than others.
Q10. In nature P730 to P680 Conversion occurs in
- A. Dark✓
- B. Light
- C. Morning
- D. Evening
Explanation: In nature, the P 680 to P 730 conversion takes place in daylight and P 730 to P 680 conversion occurs in the dark.
Why the other options are wrong
- B. In nature, the P 680 to P 730 conversion takes place in daylight and P 730 to P 680 conversion occurs in the dark.
- C. In nature, the P 680 to P 730 conversion takes place in daylight and P 730 to P 680 conversion occurs in the dark.
- D. In nature, the P 680 to P 730 conversion takes place in daylight and P 730 to P 680 conversion occurs in the dark.
Q11. Luteinizing hormone in human females induces
- A. Mensuration
- B. Menopause
- C. Oogenesis
- D. Ovulation✓
Explanation: In women, ovulation of mature follicles on the ovary is induced by a large burst of LH secretion - the preovulatory LH surge. Residual cells within ovulated follicles proliferate to form corpora lutea, which secrete the steroid hormones - progesterone and estradiol.
Why the other options are wrong
- A. In women, ovulation of mature follicles on the ovary is induced by a large burst of LH secretion - the preovulatory LH surge. Residual cells within ovulated follicles proliferate to form corpora lutea, which secrete the steroid hormones - progesterone and estradiol.
- B. In women, ovulation of mature follicles on the ovary is induced by a large burst of LH secretion - the preovulatory LH surge. Residual cells within ovulated follicles proliferate to form corpora lutea, which secrete the steroid hormones - progesterone and estradiol.
- C. In women, ovulation of mature follicles on the ovary is induced by a large burst of LH secretion - the preovulatory LH surge. Residual cells within ovulated follicles proliferate to form corpora lutea, which secrete the steroid hormones - progesterone and estradiol.
Q12. The branch of biology which deals with the study of abnormal development is
- A. Morphology
- B. Embryology
- C. Teratology✓
- D. Perinatology
Explanation: Teratology – the study of abnormalities of physiological development.
Why the other options are wrong
- A. Morphology is the study of the internal structure of words and forms a core part of linguistic study today. The term morphology is Greek and is a makeup of morph- meaning 'shape, form', and -ology which means 'the study of something'.
- B. Embryology, the study of the formation and development of an embryo and fetus.
- D. A subspecialty of obstetrics concerned with the care of the fetus and complicated, high-risk pregnancies. Perinatology is also known as maternal-fetal medicine.
Q13. The genetic code for glycine is
- A. UAG
- B. GAU
- C. GUA
- D. GGU✓
Explanation: In the genetic code, glycine is coded by all codons starting with GG, namely GGU, GGC, GGA and GGG.
Why the other options are wrong
- A. In the genetic code, glycine is coded by all codons starting with GG, namely GGU, GGC, GGA and GGG.
- B. In the genetic code, glycine is coded by all codons starting with GG, namely GGU, GGC, GGA and GGG.
- C. In the genetic code, glycine is coded by all codons starting with GG, namely GGU, GGC, GGA and GGG.
Q14. In turner syndrome the affected person has a set of chromosomes
- A. XO✓
- B. XXY
- C. XYY
- D. XXO
Explanation: To diagnose Turner syndrome, doctors use a special blood test that looks at chromosomes, called a karyotype test (chromosomal analysis). Results that indicate TS show only one X chromosome instead of two X chromosomes, with a total of 45 chromosomes instead of the usual 46.
Why the other options are wrong
- B. To diagnose Turner syndrome, doctors use a special blood test that looks at chromosomes, called a karyotype test (chromosomal analysis). Results that indicate TS show only one X chromosome instead of two X chromosomes, with a total of 45 chromosomes instead of the usual 46.
- C. To diagnose Turner syndrome, doctors use a special blood test that looks at chromosomes, called a karyotype test (chromosomal analysis). Results that indicate TS show only one X chromosome instead of two X chromosomes, with a total of 45 chromosomes instead of the usual 46.
- D. To diagnose Turner syndrome, doctors use a special blood test that looks at chromosomes, called a karyotype test (chromosomal analysis). Results that indicate TS show only one X chromosome instead of two X chromosomes, with a total of 45 chromosomes instead of the usual 46.
Q15. The leptotene and zygotene lasts for
- A. Few hours✓
- B. Few days
- C. Few weeks
- D. Few years
Explanation: Pachytene may lasts for days, weeks or even years, whereas leptotene and zygotene can last only for few hours.
Why the other options are wrong
- B. Pachytene may lasts for days, weeks or even years, whereas leptotene and zygotene can last only for few hours.
- C. Pachytene may lasts for days, weeks or even years, whereas leptotene and zygotene can last only for few hours.
- D. Pachytene may lasts for days, weeks or even years, whereas leptotene and zygotene can last only for few hours.
Q16. The maturity-onset diabetes of the young is
- A. An autosomal Recessive trait
- B. An autosomal dominant trait✓
- C. A sex-linked trait
- D. A sex-influenced trait
Explanation: MODY can be inherited as an autosomal dominant trait. About 50% of cases of MODY are caused by mutations in the glucokinase gene.
Why the other options are wrong
- A. MODY can be inherited as an autosomal dominant trait. About 50% of cases of MODY are caused by mutations in the glucokinase gene.
- C. MODY can be inherited as an autosomal dominant trait. About 50% of cases of MODY are caused by mutations in the glucokinase gene.
- D. MODY can be inherited as an autosomal dominant trait. About 50% of cases of MODY are caused by mutations in the glucokinase gene.
Q17. The organism used as biofilters is
- A. Transgenic plant
- B. Transgenic animal
- C. Transgenic bacteria✓
- D. Transgenic virus
Explanation: Industry has found that bacteria can be used as bioilters to prevent airborne chemical pollutants from being vented into the air.
Why the other options are wrong
- A. Industry has found that bacteria can be used as bioilters to prevent airborne chemical pollutants from being vented into the air.
- B. Industry has found that bacteria can be used as bioilters to prevent airborne chemical pollutants from being vented into the air.
- D. Industry has found that bacteria can be used as bioilters to prevent airborne chemical pollutants from being vented into the air.
More Punjab / UHS Solved Papers
Biology
Punjab Biology 2015 Paper 1
60 solved MCQs
Biology
Punjab Biology 2015 Paper 1
58 solved MCQs
Biology
Punjab Biology 2015 Paper 2
60 solved MCQs
Biology
Punjab Biology 2016
60 solved MCQs
Biology
Punjab Biology 2017 Paper 1
60 solved MCQs
Biology
Punjab Biology 2017 Paper 2
60 solved MCQs
Biology
Punjab Biology 2017 Paper 3
60 solved MCQs
Biology
Punjab Biology 2017 Paper 4
60 solved MCQs
Biology
Punjab Biology 2018 Paper 1
60 solved MCQs