Kpk Biology 2018 — Solved MCQs

Question 1

Yak, Gorilla and Chimpanzee are the examples of animals

Accepted Answer

C. Tamed — C) Tamed: Taming refers to the process of training and conditioning wild animals to be comfortable around humans and to follow certain commands. However, even though they may be tamed, they are still considered wild animals. Examples of tamed animals include yak , chipanzee , gorilla etc.

Question 2

Human insulin gene was inserted into the bacterium Escherichia coli by:

Accepted Answer

C. Herbert Boyer — C) Herbert Boyer: Herbert Boyer, along with Stanley Cohen, conducted groundbreaking research in genetic engineering. They successfully inserted the human insulin gene into the bacterium Escherichia coli in 1973. This achievement marked a milestone in biotechnology, leading to the production of synthetic human insulin in bacteria for medical use.Therefore, the correct answer is C) Herbert Boyer, who played a pivotal role in the development of recombinant DNA technology by inserting the human insulin gene into Escherichia coli.

Question 3

The most widely recognized damage by ultra violet rays occurs to the _

Accepted Answer

B. Skin — B) Skin: UV rays are well-known for causing damage to the skin. They can lead to sunburn, premature aging, and an increased risk of skin cancer, including melanoma.

Question 4

One Dobson Unit (DU) is equal to _

Accepted Answer

C. 0.01 mm — C) 0.01 mm. Dobson Unit (DU) is a unit used to measure the thickness of the ozone layer. It represents the thickness of ozone that would be required to create a layer of pure ozone with a thickness of 0.01 millimeters at standard temperature and pressure.

Question 5

In Rhizobium _ binds to oxygen

Accepted Answer

C. Leg - Haemoglobin — C) Leg hemoglobin: Leg hemoglobin is a protein found in the root nodules of leguminous plants, such as peas and beans. It binds to oxygen, creating a low-oxygen environment that allows Rhizobium bacteria to perform nitrogen fixation.In the case of Rhizobium, it is leg hemoglobin that binds to oxygen, creating the necessary conditions for nitrogen fixation.

Question 6

There are 10% of AA individuals and 15% BB individuals in the population what the function of individuals are heterozygotes?

Accepted Answer

A. 75% — To determine the frequency of heterozygotes in the population, we first need to find the frequency of the homozygous individuals (AA and BB) and then subtract it from 100% to find the frequency of the heterozygotes.Given:- 10% of individuals are AA (homozygous dominant).- 15% of individuals are BB (homozygous recessive).Let's find the frequency of homozygous individuals:Frequency of AA = 10%Frequency of BB = 15%Total frequency of homozygous individuals = Frequency of AA + Frequency of BBTotal frequency of homozygous individuals = 10% + 15%Total frequency of homozygous individuals = 25%Now, to find the frequency of heterozygotes:Frequency of heterozygotes = 100% - Total frequency of homozygous individualsFrequency of heterozygotes = 100% - 25%Frequency of heterozygotes = 75%So, the correct answer is A) 75. The function of individuals who are heterozygotes comprises 75% of the population.

Question 7

_ that occurs in more than one percent of the population are called polymorphism.

Accepted Answer

A. Mutations — Polymorphisms are genetic variations that occur in more than one percent of the population. A) Mutation: Mutations are changes in the DNA sequence that can lead to genetic variations, but not all mutations are considered polymorphisms.So, the correct answer is A) mutation. Polymorphisms are a type of genetic variation resulting from mutations that are present in a significant portion of the population.

Question 8

Down's syndrome and Klinefelter's syndrome are the examples of _

Accepted Answer

C. Trisomy — Down's syndrome and Klinefelter's syndrome are examples of trisomy.C) Trisomy: Trisomy is a condition where an individual has an extra copy of a particular chromosome, resulting in three copies instead of the usual pair. Down's syndrome, characterized by an extra copy of chromosome 21, and Klinefelter's syndrome, characterized by an extra X chromosome in males, are both examples of trisomy.

Question 9

Crossing a heterozygous individual for two traits (F1) with it recessive parent (P1) tells us about _

Accepted Answer

B. F2 generation — F2 generation: Yes, crossing the F1 individual with its recessive parent helps us understand the traits and genetic combinations that will appear in the F2 generation.

Question 10

Blastopore is found in _

Accepted Answer

B. Gastrula — The correct answer is B) gastrula.B) Gastrula: During gastrulation, which follows the morula stage, the embryo undergoes significant morphological changes, forming three germ layers: ectoderm, mesoderm, and endoderm. The blastopore is the invagination or indentation that appears during gastrulation, marking the beginning of the formation of the digestive tract. It is the site where cells ingress and migrate, leading to the formation of the embryonic gut.

Question 11

The process of spermatogenesis is controlled by hormonal from_

Accepted Answer

D. Both pituitary gland and hypothalamaus — D) Both B and C: This option is correct because spermatogenesis is controlled by hormones from both the pituitary gland and the hypothalamus. The hypothalamus signals the pituitary gland to release hormones that directly influence spermatogenesis.Therefore, the correct answer is D) both B and C.

Question 12

The tropic movement of curvature included in plant organs in response to unilateral effect of light is called _

Accepted Answer

C. Phototropism — The tropic movement of curvature in plant organs in response to the unilateral effect of light is called phototropism (option C).Therefore, the correct answer is C) phototropism, which specifically describes the tropic movement of curvature in plant organs in response to the unilateral effect of light.

Question 13

In male LH is also known as

Accepted Answer

B. ICSH — B) ICSH: ICSH stands for interstitial cell-stimulating hormone. As I mentioned earlier, it is another name for LH when referring to its role in stimulating the interstitial cells (Leydig cells) in the testes to produce testosterone. Testosterone is an important hormone involved in male reproductive functions.So, in summary, LH in males is also known as ICSH (option B).

Question 14

Speed of nerve impulses is faster in

Accepted Answer

B. Myelinated neutron — B) Myelinated neuron: A myelinated neuron is a nerve cell that has a protective covering called myelin sheath around its axon. This myelin sheath acts as an insulator and helps to speed up the conduction of nerve impulses. The myelin sheath allows the nerve impulses to "jump" from one node of Ranvier to another, which is known as saltatory conduction. This process significantly increases the speed of nerve impulses in myelinated neurons.

Question 15

Bone fracture is said to be _ fracture if this bone ends penetrate the skin and form a wound.

Accepted Answer

D. Both A and B — If option D is true, which states that a bone fracture is both an open fracture (option A) and a compound fracture (option B),Therefore, if option D is true, the correct answer would be D) both A and B, as both "open" and "compound" fractures refer to the situation where the bone ends penetrate the skin and form a wound.

Question 16

Which portion of the Nephron active pumps our Na+ with Cl- passively to create a salty medulla?

Accepted Answer

B. Ascending loop of Henle — The correct answer is B) ascending loop of Henle.B) Ascending loop of Henle: This segment of the nephron actively pumps sodium (Na⁺) out of the tubular fluid into the interstitial space of the renal medulla. This active transport of sodium creates a concentration gradient in the medulla, making it salty. Chloride ions (Cl⁻) are passively reabsorbed along with sodium ions, contributing to the osmotic gradient in the medulla. The ascending loop of Henle is crucial for establishing the hypertonic environment necessary for water reabsorption in the collecting ducts.

Question 17

Sharks is _

Accepted Answer

A. Ureotelic — Sharks are ureotelic because their primary nitrogenous waste product is urea, which is less toxic and requires less water for excretion compared to ammonia. This adaptation is essential for sharks, which inhabit environments with varying levels of salt concentration, allowing them to efficiently regulate their internal osmolarity while conserving water.

Question 18

Tuberculosis (TB) is the bacterial infection of _

Accepted Answer

A. Lungs

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