BiologyQ1–47 of 47 questions
Page 1 of 1 · English
1
BiologyEasyMCQKCET · 2024
The enzyme $Recombinase$ is required at which stage of $Meiosis \ I$?
A
Pachytene
B
Zygotene
C
Diplotene
D
Diakinesis
Solution
(A) The correct answer is $A$ (Pachytene).
$Recombinase$ is an enzyme that facilitates the process of crossing over between non-sister chromatids of homologous chromosomes.
This process occurs specifically during the $Pachytene$ stage of $Prophase \ I$ in $Meiosis \ I$.
$Recombinase$ is an enzyme that facilitates the process of crossing over between non-sister chromatids of homologous chromosomes.
This process occurs specifically during the $Pachytene$ stage of $Prophase \ I$ in $Meiosis \ I$.
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2
BiologyEasyMCQKCET · 2024
Match the pigments given in List-Iwith their colour in chromatogram given in List-II.
Choose the correct option from the following:
| List-I(Pigments) | List-II(Colour in chromatogram) |
| 1. Chlorophyll 'b' | p. Yellow orange |
| 2. Carotenoids | q. Orange red |
| 3. Chlorophyll 'a' | s. Blue green |
| 4. Xanthophylls | r. Yellow |
| - | t. Yellow green |
A
$1-s, 2-p, 3-q, 4-r$
B
$1-t, 2-p, 3-s, 4-r$
C
$1-t, 2-q, 3-s, 4-p$
D
$1-s, 2-t, 3-q, 4-r$
Solution
(B) According to the paper chromatography of leaf pigments,the colours observed are as follows:
$1$. Chlorophyll '$b$' appears as Yellow green $(t)$.
$2$. Carotenoids appear as Yellow to Yellow orange $(p)$.
$3$. Chlorophyll '$a$' appears as Blue green $(s)$.
$4$. Xanthophylls appear as Yellow $(r)$.
Therefore,the correct matching is $1-t, 2-p, 3-s, 4-r$.
$1$. Chlorophyll '$b$' appears as Yellow green $(t)$.
$2$. Carotenoids appear as Yellow to Yellow orange $(p)$.
$3$. Chlorophyll '$a$' appears as Blue green $(s)$.
$4$. Xanthophylls appear as Yellow $(r)$.
Therefore,the correct matching is $1-t, 2-p, 3-s, 4-r$.
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3
BiologyEasyMCQKCET · 2024
Which is the intermediate compound that links the end product of Glycolysis with the $TCA$ Cycle?
A
Acetyl CoA
B
Pyruvic Acid
C
$OAA$
D
Citric acid
Solution
(A) The correct answer is $A$.
Glycolysis results in the production of $2$ molecules of $Pyruvic$ $Acid$ from one molecule of glucose.
In the presence of oxygen, $Pyruvic$ $Acid$ undergoes oxidative decarboxylation to form $Acetyl$ $CoA$.
$Acetyl$ $CoA$ acts as the connecting link between glycolysis and the $TCA$ cycle (Krebs cycle) by entering the mitochondrial matrix and initiating the cycle.
Glycolysis results in the production of $2$ molecules of $Pyruvic$ $Acid$ from one molecule of glucose.
In the presence of oxygen, $Pyruvic$ $Acid$ undergoes oxidative decarboxylation to form $Acetyl$ $CoA$.
$Acetyl$ $CoA$ acts as the connecting link between glycolysis and the $TCA$ cycle (Krebs cycle) by entering the mitochondrial matrix and initiating the cycle.
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4
BiologyEasyMCQKCET · 2024
Auxins: Apical dominance:: Gibberellins: . . . . . .
A
Adventitious shoot formation
B
Accelerates abscission
C
Closure of stomata
D
Bolting
Solution
(D) The correct answer is $D$.
Gibberellins promote bolting,which refers to the internode elongation just prior to flowering in plants with a rosette habit (e.g.,beet,cabbage).
Gibberellins promote bolting,which refers to the internode elongation just prior to flowering in plants with a rosette habit (e.g.,beet,cabbage).
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5
BiologyEasyMCQKCET · 2024
The typical 'lub-dub' sounds heard during a heartbeat are produced due to:
A
Closure of semilunar valves
B
Closure of bicuspid and tricuspid valves
C
Closure of bicuspid and tricuspid valves followed by semilunar valves
D
Opening of bicuspid and tricuspid valves followed by semilunar valves
Solution
(C) The correct answer is $C$.
During a cardiac cycle,two prominent heart sounds are produced.
The first heart sound,known as 'lub',is produced due to the closure of the atrioventricular valves (tricuspid and bicuspid valves) at the beginning of ventricular systole.
The second heart sound,known as 'dub',is produced due to the closure of the semilunar valves at the beginning of ventricular diastole.
Therefore,the sequence of sounds is the closure of bicuspid and tricuspid valves followed by the closure of semilunar valves.
During a cardiac cycle,two prominent heart sounds are produced.
The first heart sound,known as 'lub',is produced due to the closure of the atrioventricular valves (tricuspid and bicuspid valves) at the beginning of ventricular systole.
The second heart sound,known as 'dub',is produced due to the closure of the semilunar valves at the beginning of ventricular diastole.
Therefore,the sequence of sounds is the closure of bicuspid and tricuspid valves followed by the closure of semilunar valves.
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6
BiologyEasyMCQKCET · 2024
The term $Uremia$ refers to:
A
Accumulation of urea in blood.
B
Presence of glucose in the urine.
C
Accumulation of uric acid in blood.
D
Accumulation of uric acid in kidneys.
Solution
(A) $Uremia$ refers to the accumulation of urea in the blood.
This condition occurs due to the malfunctioning of the kidneys,which are unable to filter urea effectively from the bloodstream.
This condition occurs due to the malfunctioning of the kidneys,which are unable to filter urea effectively from the bloodstream.
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7
BiologyEasyMCQKCET · 2024
The functional unit of contraction is a
A
Portion of myofibril between two successive $Z$-lines
B
Portion of myofibril between two successive $M$-lines
C
Centre of the $H$-zone
D
Centre of the $I$-band
Solution
(A) The correct answer is $A$. The functional unit of muscle contraction is known as the $sarcomere$.
It is defined as the portion of a myofibril that lies between two successive $Z$-lines.
During muscle contraction,these $sarcomeres$ shorten as the actin and myosin filaments slide past each other.
It is defined as the portion of a myofibril that lies between two successive $Z$-lines.
During muscle contraction,these $sarcomeres$ shorten as the actin and myosin filaments slide past each other.
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8
BiologyEasyMCQKCET · 2024
Match the parts of the brain given in List-$I$ with their functions given in List-$II$.
Choose the correct option from the following:
| List-$I$ (Parts of the brain) | List-$II$ (Functions) |
|---|---|
| $1$. Medulla oblongata | $p$. Body temperature |
| $2$. Hypothalamus | $q$. Olfaction |
| $3$. Cerebral cortex | $r$. Respiration |
| $4$. Limbic system | $s$. Motor function |
Choose the correct option from the following:
A
$1-p, 2-r, 3-s, 4-q$
B
$1-q, 2-s, 3-r, 4-p$
C
$1-s, 2-p, 3-q, 4-r$
D
$1-r, 2-p, 3-s, 4-q$
Solution
(D) The correct matching is as follows:
$1$. Medulla oblongata: It contains centers which control respiration,cardiovascular reflexes,and gastric secretions. Thus,$1-r$.
$2$. Hypothalamus: It contains a number of centers which control body temperature,urge for eating and drinking. Thus,$2-p$.
$3$. Cerebral cortex: It is responsible for motor functions,sensory perceptions,and complex cognitive processes. Thus,$3-s$.
$4$. Limbic system: It is involved in the regulation of sexual behavior,expression of emotional reactions (e.g.,excitement,pleasure,rage,and fear),and motivation. It is also associated with the sense of smell (olfaction). Thus,$4-q$.
Therefore,the correct sequence is $1-r, 2-p, 3-s, 4-q$.
$1$. Medulla oblongata: It contains centers which control respiration,cardiovascular reflexes,and gastric secretions. Thus,$1-r$.
$2$. Hypothalamus: It contains a number of centers which control body temperature,urge for eating and drinking. Thus,$2-p$.
$3$. Cerebral cortex: It is responsible for motor functions,sensory perceptions,and complex cognitive processes. Thus,$3-s$.
$4$. Limbic system: It is involved in the regulation of sexual behavior,expression of emotional reactions (e.g.,excitement,pleasure,rage,and fear),and motivation. It is also associated with the sense of smell (olfaction). Thus,$4-q$.
Therefore,the correct sequence is $1-r, 2-p, 3-s, 4-q$.
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9
BiologyEasyMCQKCET · 2024
Identify the phylum which shows the following characteristics:
$1$. Animals are exclusively marine,radially symmetrical,and diploblastic.
$2$. Body bears eight external rows of ciliated comb plates which help in locomotion.
$3$. Digestion is both extracellular and intracellular.
$4$. Reproduction is only by sexual means.
$1$. Animals are exclusively marine,radially symmetrical,and diploblastic.
$2$. Body bears eight external rows of ciliated comb plates which help in locomotion.
$3$. Digestion is both extracellular and intracellular.
$4$. Reproduction is only by sexual means.
A
Coelenterata
B
Mollusca
C
Arthropoda
D
Ctenophora
Solution
(D) The correct answer is $D$.
Phylum $Ctenophora$ exhibits the following characteristics:
$1$. They are exclusively marine,radially symmetrical,and diploblastic organisms.
$2$. Their body bears eight external rows of ciliated comb plates,which help in locomotion.
$3$. Digestion in these organisms is both extracellular and intracellular.
$4$. Reproduction takes place only by sexual means.
Phylum $Ctenophora$ exhibits the following characteristics:
$1$. They are exclusively marine,radially symmetrical,and diploblastic organisms.
$2$. Their body bears eight external rows of ciliated comb plates,which help in locomotion.
$3$. Digestion in these organisms is both extracellular and intracellular.
$4$. Reproduction takes place only by sexual means.
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10
BiologyEasyMCQKCET · 2024
Which of the following statements is correct with reference to vacuoles?
A
It is membrane bound and contains storage proteins and lipids
B
It is membrane bound and contains water and excretory substances.
C
It lacks membrane and contains air
D
It lacks membrane and contains water and excretory substances
Solution
(B) is the correct answer.
Vacuoles are membrane-bound spaces found in the cytoplasm of plant and animal cells.
They contain water,sap,excretory products,and other materials not useful for the cell.
The membrane surrounding the vacuole is called the tonoplast.
Vacuoles are membrane-bound spaces found in the cytoplasm of plant and animal cells.
They contain water,sap,excretory products,and other materials not useful for the cell.
The membrane surrounding the vacuole is called the tonoplast.
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11
BiologyEasyMCQKCET · 2024
Exoskeleton of arthropods is made up of a unique complex polysaccharide known as
A
Hyaluronic Acid
B
Chitin
C
Waxes
D
Cellulose
Solution
(B) The correct answer is $B$.
Arthropods possess an exoskeleton that provides structural support and protection.
This exoskeleton is composed of $Chitin$,which is a complex polysaccharide.
$Chitin$ is a polymer of $N$-acetylglucosamine units linked by $\beta-1,4$-glycosidic bonds.
Arthropods possess an exoskeleton that provides structural support and protection.
This exoskeleton is composed of $Chitin$,which is a complex polysaccharide.
$Chitin$ is a polymer of $N$-acetylglucosamine units linked by $\beta-1,4$-glycosidic bonds.
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12
BiologyEasyMCQKCET · 2024
Which of the following is not a parasitic adaptation?
A
Loss of unnecessary sense organs
B
Absence of adhesive organs or suckers
C
Loss of digestive system
D
High reproductive capacity
Solution
(B) The correct answer is $B$. Parasites have evolved various adaptations to survive within or on their hosts. These include the loss of unnecessary sense organs,the development of adhesive organs or suckers for attachment,the loss of a complex digestive system (as they absorb predigested food),and a high reproductive capacity to ensure the continuation of their species. Therefore,the absence of adhesive organs or suckers is not a parasitic adaptation; rather,their presence is a key adaptation.
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13
BiologyEasyMCQKCET · 2024
When a flower has both stamens and carpels,it is described as:
A
Asexual
B
Unisexual
C
Bisexual
D
Dioecious
Solution
(C) The correct answer is $C$.
Bisexual flowers,also known as hermaphrodite flowers,are those that contain both male reproductive organs (stamens) and female reproductive organs (carpels or pistils) within the same flower.
Bisexual flowers,also known as hermaphrodite flowers,are those that contain both male reproductive organs (stamens) and female reproductive organs (carpels or pistils) within the same flower.
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14
BiologyEasyMCQKCET · 2024
Match the content of List-$I$ with List-$II$:
Choose the correct option from the following:
| List-$I$ | List-$II$ |
|---|---|
| $1$. Polyembryony | $p$. Black pepper |
| $2$. Perisperm | $q$. Banana |
| $3$. False fruit | $r$. Lemon |
| $4$. Parthenocarpy | $s$. Apple |
Choose the correct option from the following:
A
$1-r, 2-p, 3-s, 4-q$
B
$1-p, 2-r, 3-s, 4-q$
C
$1-q, 2-p, 3-s, 4-r$
D
$1-r, 2-s, 3-p, 4-q$
Solution
(A) $1$. Polyembryony: The occurrence of more than one embryo in a seed is known as polyembryony, commonly seen in Citrus and Lemon $(1-r)$.
$2$. Perisperm: The residual persistent nucellus in a seed is called perisperm, found in Black \text{ } pepper and beet $(2-p)$.
$3$. False fruit: Fruits that develop from the thalamus or other floral parts instead of the ovary are called false fruits, e.g., Apple $(3-s)$.
$4$. Parthenocarpy: The development of fruit without fertilization is called parthenocarpy, commonly seen in Banana $(4-q)$.
Therefore, the correct matching is $1-r, 2-p, 3-s, 4-q$.
$2$. Perisperm: The residual persistent nucellus in a seed is called perisperm, found in Black \text{ } pepper and beet $(2-p)$.
$3$. False fruit: Fruits that develop from the thalamus or other floral parts instead of the ovary are called false fruits, e.g., Apple $(3-s)$.
$4$. Parthenocarpy: The development of fruit without fertilization is called parthenocarpy, commonly seen in Banana $(4-q)$.
Therefore, the correct matching is $1-r, 2-p, 3-s, 4-q$.
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15
BiologyEasyMCQKCET · 2024
Transfer of pollen grains from the anther to the stigma of another flower of the same plant is called:
A
Xenogamy
B
Autogamy
C
Cleistogamy
D
Geitonogamy
Solution
(D) $Geitonogamy$ is the transfer of pollen grains from the anther to the stigma of another flower of the same plant.
Although it is functionally a type of cross-pollination because it involves a pollinating agent,genetically it is similar to self-pollination because the pollen grains come from the same plant.
Although it is functionally a type of cross-pollination because it involves a pollinating agent,genetically it is similar to self-pollination because the pollen grains come from the same plant.
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16
BiologyEasyMCQKCET · 2024
The role of the filiform apparatus in synergids is to
A
Protect the egg apparatus
B
Endosperm formation
C
Guide the entry of the pollen tube
D
Prevent the entry of gametes
Solution
(C) The correct answer is $C$.
The filiform apparatus consists of specialized cellular thickenings at the micropylar end of the synergids.
Its primary function is to secrete chemical substances that guide the entry of the pollen tube into the synergids during the process of fertilization.
The filiform apparatus consists of specialized cellular thickenings at the micropylar end of the synergids.
Its primary function is to secrete chemical substances that guide the entry of the pollen tube into the synergids during the process of fertilization.
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17
BiologyEasyMCQKCET · 2024
Which of the following is based upon the principle of antigen-antibody interaction?
A
$PCR$
B
$ELISA$
C
rDNA technology
D
Gel Electrophoresis
Solution
(B) $ELISA$ (Enzyme-Linked Immunosorbent Assay) is a diagnostic technique based on the principle of antigen-antibody interaction.
In this technique,specific antibodies are used to detect the presence of antigens in a sample.
This interaction is highly specific,allowing for the accurate identification of pathogens or proteins.
In this technique,specific antibodies are used to detect the presence of antigens in a sample.
This interaction is highly specific,allowing for the accurate identification of pathogens or proteins.
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18
BiologyEasyMCQKCET · 2024
Which among the following is used to treat Emphysema?
A
Human Hormone - $\alpha$-Antitrypsin
B
Human $\alpha$-Interferon
C
Human protein - $\alpha$-Antitrypsin
D
Human $\alpha$-Lactalbumin
Solution
(C) The correct answer is $C$.
Human protein $\alpha$-antitrypsin is used in the treatment of emphysema.
It is produced by $DNA$ manipulation in transgenic animals (specifically transgenic sheep).
This protein inhibits the enzyme elastase,which otherwise causes the breakdown of alveolar walls in the lungs,leading to emphysema.
Human protein $\alpha$-antitrypsin is used in the treatment of emphysema.
It is produced by $DNA$ manipulation in transgenic animals (specifically transgenic sheep).
This protein inhibits the enzyme elastase,which otherwise causes the breakdown of alveolar walls in the lungs,leading to emphysema.
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19
BiologyEasyMCQKCET · 2024
The annual net primary productivity of the biosphere is approximately
A
$170$ billion tons
B
$55$ billion tons
C
$170$ million tons
D
$55$ million tons
Solution
(A) The correct answer is $A$.
The annual net primary productivity of the entire biosphere is approximately $170$ billion tons of organic matter (dry weight).
This value represents the biomass available for consumption by heterotrophs.
The annual net primary productivity of the entire biosphere is approximately $170$ billion tons of organic matter (dry weight).
This value represents the biomass available for consumption by heterotrophs.
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20
BiologyEasyMCQKCET · 2024
According to the hypothesis proposed by environmental biologists,a relatively constant environment in the tropics promotes:
A
Niche specialization and lesser species diversity
B
Niche specialization and greater species diversity
C
Niche diversity and lesser species specialization
D
Niche diversity and greater species specialization
Solution
(B) The correct answer is $B$.
Environmental biologists suggest that the relatively constant and stable climate in tropical regions allows for $Niche$ specialization.
Because the environment is stable and provides consistent favorable conditions along with abundant solar energy,it supports the evolution of a wider variety of species,leading to greater species diversity.
Environmental biologists suggest that the relatively constant and stable climate in tropical regions allows for $Niche$ specialization.
Because the environment is stable and provides consistent favorable conditions along with abundant solar energy,it supports the evolution of a wider variety of species,leading to greater species diversity.
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21
BiologyEasyMCQKCET · 2024
Identify the incorrect match with respect to recently extinct animals and their place of extinction according to the $IUCN$ Red List.
A
Dodo - Mauritius
B
Quagga - Africa
C
Thylacine - Australia
D
Steller's Sea Cow - North America
Solution
(D) is the incorrect match.
Steller's Sea Cow ($Hydrodamalis$ $gigas$) became extinct from the Commander Islands in Russia (Bering Sea) due to overexploitation by humans within $27$ years of its discovery in $1741$.
It was not native to North America.
Steller's Sea Cow ($Hydrodamalis$ $gigas$) became extinct from the Commander Islands in Russia (Bering Sea) due to overexploitation by humans within $27$ years of its discovery in $1741$.
It was not native to North America.
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22
BiologyEasyMCQKCET · 2024
$A$ strict protection of biodiversity hotspots could reduce the ongoing mass extinction by almost (in $\%$)
A
$20$
B
$25$
C
$30$
D
$35$
Solution
(C) The correct answer is $C$.
Biodiversity hotspots are regions with very high levels of species richness and high degree of endemism.
These regions are also under constant threat.
Ecologists have estimated that strict protection of these biodiversity hotspots could reduce the ongoing mass extinction by almost $30\%$.
Biodiversity hotspots are regions with very high levels of species richness and high degree of endemism.
These regions are also under constant threat.
Ecologists have estimated that strict protection of these biodiversity hotspots could reduce the ongoing mass extinction by almost $30\%$.
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23
BiologyEasyMCQKCET · 2024
With reference to human sperm,match the List-$I$ with List-$II$.
Choose the correct option from the following:
| List-$I$ | List-$II$ |
|---|---|
| $1$. Head | $s$. Contains haploid nucleus |
| $2$. Acrosome | $p$. Filled with enzymes |
| $3$. Middle piece | $q$. Contains mitochondria |
| $4$. Tail | $r$. Sperm motility |
Choose the correct option from the following:
A
$1-r, 2-q, 3-s, 4-p$
B
$1-s, 2-p, 3-q, 4-r$
C
$1-s, 2-r, 3-p, 4-q$
D
$1-q, 2-s, 3-r, 4-p$
Solution
(B) The correct match is $1-s, 2-p, 3-q, 4-r$.
$1$. Head $(s)$: The sperm head contains an elongated haploid nucleus.
$2$. Acrosome $(p)$: The anterior portion of the head is covered by a cap-like structure called the acrosome,which is filled with enzymes that help in the fertilization of the ovum.
$3$. Middle piece $(q)$: The middle piece possesses numerous mitochondria,which produce energy for the movement of the tail.
$4$. Tail $(r)$: The tail facilitates sperm motility,which is essential for fertilization.
$1$. Head $(s)$: The sperm head contains an elongated haploid nucleus.
$2$. Acrosome $(p)$: The anterior portion of the head is covered by a cap-like structure called the acrosome,which is filled with enzymes that help in the fertilization of the ovum.
$3$. Middle piece $(q)$: The middle piece possesses numerous mitochondria,which produce energy for the movement of the tail.
$4$. Tail $(r)$: The tail facilitates sperm motility,which is essential for fertilization.
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24
BiologyEasyMCQKCET · 2024
If two primary spermatocytes and two primary oocytes undergo meiosis simultaneously,what will be the ratio of spermatozoa and ova produced at the end of the gametogenesis?
A
$2$:$1$
B
$4$:$1$
C
$6$:$2$
D
$1$:$2$
Solution
(B) During spermatogenesis,one primary spermatocyte undergoes meiosis to produce $4$ functional spermatozoa.
Therefore,$2$ primary spermatocytes will produce $2 \times 4 = 8$ spermatozoa.
During oogenesis,one primary oocyte undergoes meiosis to produce only $1$ functional ovum (and $2-3$ polar bodies which degenerate).
Therefore,$2$ primary oocytes will produce $2 \times 1 = 2$ ova.
The ratio of spermatozoa to ova produced is $8:2$,which simplifies to $4:1$.
Therefore,$2$ primary spermatocytes will produce $2 \times 4 = 8$ spermatozoa.
During oogenesis,one primary oocyte undergoes meiosis to produce only $1$ functional ovum (and $2-3$ polar bodies which degenerate).
Therefore,$2$ primary oocytes will produce $2 \times 1 = 2$ ova.
The ratio of spermatozoa to ova produced is $8:2$,which simplifies to $4:1$.
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25
BiologyEasyMCQKCET · 2024
Which of the following hormones is not secreted by the human placenta?
A
Progestogen
B
$HCG$
C
Estrogen
D
$LH$
Solution
(D) The correct answer is $D$.
$LH$ (Luteinizing Hormone) is secreted by the anterior pituitary gland,not the placenta.
The human placenta acts as an endocrine tissue and produces several hormones such as Human Chorionic Gonadotropin $(HCG)$,Human Placental Lactogen $(HPL)$,estrogens,and progestogens to maintain pregnancy.
$LH$ (Luteinizing Hormone) is secreted by the anterior pituitary gland,not the placenta.
The human placenta acts as an endocrine tissue and produces several hormones such as Human Chorionic Gonadotropin $(HCG)$,Human Placental Lactogen $(HPL)$,estrogens,and progestogens to maintain pregnancy.
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26
BiologyEasyMCQKCET · 2024
In human females,the endometrium of the uterus consists of:
A
Smooth muscle
B
Glandular layer
C
Adipose layer
D
Cartilaginous layer
Solution
(B) The wall of the human uterus is made up of three layers of tissue: the perimetrium (outer thin layer),the myometrium (middle thick layer of smooth muscle),and the endometrium (inner glandular layer).
The endometrium is the innermost layer that lines the uterine cavity.
It is a highly vascular and glandular layer that undergoes cyclic changes during the menstrual cycle in preparation for implantation of the embryo.
Therefore,the correct option is $B$.
The endometrium is the innermost layer that lines the uterine cavity.
It is a highly vascular and glandular layer that undergoes cyclic changes during the menstrual cycle in preparation for implantation of the embryo.
Therefore,the correct option is $B$.
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27
BiologyEasyMCQKCET · 2024
The Government of India legalised $MTP$ with some strict regulations in the year
A
$1951$
B
$1961$
C
$1971$
D
$2001$
Solution
(C) The Government of India legalised $Medical$ $Termination$ $of$ $Pregnancy$ $(MTP)$ in the year $1971$.
This was done with some strict regulations to prevent the misuse of $MTP$ and to control the indiscriminate and illegal female foeticides which are reported to be high in India.
Therefore,the correct option is $C$.
This was done with some strict regulations to prevent the misuse of $MTP$ and to control the indiscriminate and illegal female foeticides which are reported to be high in India.
Therefore,the correct option is $C$.
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28
BiologyEasyMCQKCET · 2024
The process in which a small part of the $vas \ deferens$ is removed or tied up through a small incision is called:
A
$MTP$
B
Vasectomy
C
Tubectomy
D
$GIFT$
Solution
(B) Vasectomy.
Vasectomy is a surgical method of sterilization in males.
In this procedure,a small part of the $vas \ deferens$ is removed or tied up through a small incision on the scrotum.
This prevents the transport of sperm,thereby acting as a permanent contraceptive method.
Vasectomy is a surgical method of sterilization in males.
In this procedure,a small part of the $vas \ deferens$ is removed or tied up through a small incision on the scrotum.
This prevents the transport of sperm,thereby acting as a permanent contraceptive method.
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29
BiologyEasyMCQKCET · 2024
Find the incorrect statement among the following:
A
In sex-linked recessive traits,the gene is transmitted from an unaffected carrier female to some of the male progeny.
B
Accumulation of phenylpyruvic acid in the brain results in mental retardation.
C
Individuals affected by Down's Syndrome will have congenital heart defects and are more intelligent.
D
Turner's Syndrome is caused due to the absence of one $X$-chromosome.
Solution
(C) The incorrect statement is $C$. Individuals affected by Down's Syndrome do not have higher intelligence; rather,they typically exhibit mental retardation,short stature,and physical abnormalities such as congenital heart defects. Therefore,the claim that they are 'more intelligent' is factually incorrect.
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30
BiologyEasyMCQKCET · 2024
The genotype ratio of incomplete dominance is
A
$3$:$1$
B
$1$:$2$:$1$
C
$1$:$1$:$2$
D
$9$:$3$:$3$:$1$
Solution
(B) The correct answer is $B$ $(1:2:1)$.
In incomplete dominance,the phenotypic and genotypic ratios are identical.
For a cross between two heterozygous individuals (e.g.,$Rr \times Rr$),the resulting offspring follow the ratio $1$ (homozygous dominant) : $2$ (heterozygous) : $1$ (homozygous recessive).
Thus,the genotypic ratio is $1:2:1$.
In incomplete dominance,the phenotypic and genotypic ratios are identical.
For a cross between two heterozygous individuals (e.g.,$Rr \times Rr$),the resulting offspring follow the ratio $1$ (homozygous dominant) : $2$ (heterozygous) : $1$ (homozygous recessive).
Thus,the genotypic ratio is $1:2:1$.
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31
BiologyEasyMCQKCET · 2024
$A$ test cross in a pea plant is defined as:
A
$A$ cross between an $F_2$ tall plant and a recessive parent.
B
$A$ cross between an $F_2$ dwarf plant and a recessive parent.
C
$A$ cross between an $F_2$ tall plant and a dominant parent.
D
$A$ cross between two $F_1$ plants.
Solution
(A) The correct answer is $A$.
$A$ test cross is a genetic cross between an individual showing a dominant phenotype (whose genotype is unknown, such as an $F_2$ tall plant) and a homozygous recessive parent.
This cross is performed to determine the genotype of the dominant individual.
If the $F_2$ tall plant is homozygous dominant $(TT)$, all offspring will be tall.
If the $F_2$ tall plant is heterozygous $(Tt)$, the offspring will be $50\%$ tall and $50\%$ dwarf.
$A$ test cross is a genetic cross between an individual showing a dominant phenotype (whose genotype is unknown, such as an $F_2$ tall plant) and a homozygous recessive parent.
This cross is performed to determine the genotype of the dominant individual.
If the $F_2$ tall plant is homozygous dominant $(TT)$, all offspring will be tall.
If the $F_2$ tall plant is heterozygous $(Tt)$, the offspring will be $50\%$ tall and $50\%$ dwarf.
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32
BiologyEasyMCQKCET · 2024
In a dihybrid cross between a true-breeding round yellow-seeded and a true-breeding wrinkled green-seeded pea plant,the ratio of segregation of round and wrinkled seed traits in $F_2$ is:
A
$9$:$1$
B
$3$:$1$
C
$9$:$3$
D
$3$:$3$
Solution
(B) In a dihybrid cross,the inheritance of two traits is independent of each other according to the Law of Independent Assortment.
For the seed shape trait,the cross is between round $(RR)$ and wrinkled $(rr)$.
The $F_1$ generation is heterozygous round $(Rr)$.
Selfing the $F_1$ generation $(Rr \times Rr)$ results in an $F_2$ phenotypic ratio of $3$ round : $1$ wrinkled.
Therefore,the segregation ratio for the seed shape trait is $3:1$.
For the seed shape trait,the cross is between round $(RR)$ and wrinkled $(rr)$.
The $F_1$ generation is heterozygous round $(Rr)$.
Selfing the $F_1$ generation $(Rr \times Rr)$ results in an $F_2$ phenotypic ratio of $3$ round : $1$ wrinkled.
Therefore,the segregation ratio for the seed shape trait is $3:1$.
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33
BiologyEasyMCQKCET · 2024
When male and female gametes are morphologically distinct,the condition is known as
A
Homogametes
B
Heterogametes
C
Hermaphrodites
D
Sexual Dimorphism
Solution
(B) The correct answer is $B$.
When male and female gametes are morphologically distinct,they are called $Heterogametes$ (or $Anisogametes$).
$Hetero-$ means different and $gametes$ refers to the reproductive cells.
In this condition,the male gamete is usually smaller and motile (sperm/antherozoid),while the female gamete is larger and non-motile (egg/ovum).
When male and female gametes are morphologically distinct,they are called $Heterogametes$ (or $Anisogametes$).
$Hetero-$ means different and $gametes$ refers to the reproductive cells.
In this condition,the male gamete is usually smaller and motile (sperm/antherozoid),while the female gamete is larger and non-motile (egg/ovum).
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34
BiologyEasyMCQKCET · 2024
Following representations $P$,$Q$ and $R$ denote few steps of the Griffith Experiment. Identify the correct one$(s)$:
$P. R$ strain $\rightarrow$ Inject into mice $\rightarrow$ Mice die
$Q. S$ strain (Heat killed) $\rightarrow$ Inject into mice $\rightarrow$ Mice die
$R. R$ strain $\rightarrow$ Inject into mice $\rightarrow$ Mice live
$P. R$ strain $\rightarrow$ Inject into mice $\rightarrow$ Mice die
$Q. S$ strain (Heat killed) $\rightarrow$ Inject into mice $\rightarrow$ Mice die
$R. R$ strain $\rightarrow$ Inject into mice $\rightarrow$ Mice live
A
$P$ only
B
$R$ only
C
$P$ and $R$
D
$Q$ and $R$
Solution
(B) The correct option is $R$ only.
In the Griffith experiment:
$P$: $R$ strain is non-virulent (rough),so mice injected with it survive. The statement says they die,which is incorrect.
$Q$: $S$ strain (Heat-killed) is non-virulent because the heat kills the bacteria,so mice injected with it survive. The statement says they die,which is incorrect.
$R$: $R$ strain is non-virulent,so mice injected with it survive. This statement is correct.
In the Griffith experiment:
$P$: $R$ strain is non-virulent (rough),so mice injected with it survive. The statement says they die,which is incorrect.
$Q$: $S$ strain (Heat-killed) is non-virulent because the heat kills the bacteria,so mice injected with it survive. The statement says they die,which is incorrect.
$R$: $R$ strain is non-virulent,so mice injected with it survive. This statement is correct.
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35
BiologyEasyMCQKCET · 2024
The $mRNA$ has Untranslated Regions $(UTRs)$
A
At $3'$-end beyond Terminator codon
B
At $5'$-end before $AUG$
C
At both $3'$-end and $5'$-end beyond Terminator codon and before $AUG$ respectively
D
$AUG$ and Terminator codon flanks the $UTR$
Solution
(C) $mRNA$ contains Untranslated Regions $(UTRs)$ which are sequences of $RNA$ that are not translated into proteins.
These $UTRs$ are present at both the $5'$-end (before the start codon $AUG$) and the $3'$-end (after the stop/terminator codon).
They are not translated but are required for the efficient translation process and stability of the $mRNA$ molecule.
These $UTRs$ are present at both the $5'$-end (before the start codon $AUG$) and the $3'$-end (after the stop/terminator codon).
They are not translated but are required for the efficient translation process and stability of the $mRNA$ molecule.
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36
BiologyEasyMCQKCET · 2024
In a structural gene,the template $DNA$ strand has the nucleotide sequence $3'-ATGCATGCATGCATGC-5'$. Find the correct and complementary nucleotide sequence on the coding strand.
A
$5'-ATGCATGCATGCATGC-3'$
B
$3'-GCATGCATGCATGCAT-5'$
C
$5'-TACGTACGTACGTACG-3'$
D
$3'-TACGTACGTACGTACG-5'$
Solution
(C) The correct option is $(C)$.
In a $DNA$ double helix,the two strands are antiparallel and complementary to each other.
The template strand runs in the $3' \rightarrow 5'$ direction,while the coding strand runs in the $5' \rightarrow 3'$ direction.
According to the base-pairing rules,$A$ pairs with $T$ and $G$ pairs with $C$.
Given template strand: $3'-ATGCATGCATGCATGC-5'$
Complementary coding strand: $5'-TACGTACGTACGTACG-3'$
In a $DNA$ double helix,the two strands are antiparallel and complementary to each other.
The template strand runs in the $3' \rightarrow 5'$ direction,while the coding strand runs in the $5' \rightarrow 3'$ direction.
According to the base-pairing rules,$A$ pairs with $T$ and $G$ pairs with $C$.
Given template strand: $3'-ATGCATGCATGCATGC-5'$
Complementary coding strand: $5'-TACGTACGTACGTACG-3'$
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37
BiologyEasyMCQKCET · 2024
In $tRNA$,the region that binds with $mRNA$ is:
A
Anticodon loop of $tRNA$
B
Amino acid acceptor end of $tRNA$
C
Amino acyl synthetase loop of $tRNA$
D
Ribosomal binding loop of $tRNA$
Solution
(A) $tRNA$ (transfer $RNA$) acts as an adapter molecule during protein synthesis.
It contains an anticodon loop,which consists of a sequence of three nitrogenous bases that are complementary to the specific codon present on the $mRNA$ strand.
During translation,the anticodon of the $tRNA$ base-pairs with the codon of the $mRNA$,ensuring that the correct amino acid is incorporated into the growing polypeptide chain.
Therefore,the correct region is the anticodon loop.
It contains an anticodon loop,which consists of a sequence of three nitrogenous bases that are complementary to the specific codon present on the $mRNA$ strand.
During translation,the anticodon of the $tRNA$ base-pairs with the codon of the $mRNA$,ensuring that the correct amino acid is incorporated into the growing polypeptide chain.
Therefore,the correct region is the anticodon loop.
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38
BiologyEasyMCQKCET · 2024
Read the following statements:
Statement $I$: All vertebrates develop a row of vestigial gill slits during the embryonic stage.
Statement $II$: Embryos always pass through the adult stages of other animals.
Which of the following options is correct with reference to these statements?
Statement $I$: All vertebrates develop a row of vestigial gill slits during the embryonic stage.
Statement $II$: Embryos always pass through the adult stages of other animals.
Which of the following options is correct with reference to these statements?
A
Statement $I$ is correct, Statement $II$ is incorrect.
B
Statement $I$ is incorrect, Statement $II$ is correct.
C
Both Statements $I$ and $II$ are correct.
D
Both Statements $I$ and $II$ are incorrect.
Solution
(A) Statement $I$ is correct because all vertebrate embryos, including humans, exhibit pharyngeal gill slits at some stage of their development, which are considered vestigial in terrestrial vertebrates.
Statement $II$ is incorrect because the theory of recapitulation (von Baer's law) states that embryos do not pass through the adult stages of other animals; rather, they share common embryonic features that diverge as development progresses.
Statement $II$ is incorrect because the theory of recapitulation (von Baer's law) states that embryos do not pass through the adult stages of other animals; rather, they share common embryonic features that diverge as development progresses.
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39
BiologyEasyMCQKCET · 2024
The most ape-like ancestral primate was:
A
Dryopithecus
B
Ramapithecus
C
Australopithecus
D
Neanderthal man
Solution
(A) $Dryopithecus$ was the most ape-like ancestral primate.
It existed about $15$ million years ago.
$Dryopithecus$ was more ape-like,while $Ramapithecus$ was more man-like in its morphological features.
It existed about $15$ million years ago.
$Dryopithecus$ was more ape-like,while $Ramapithecus$ was more man-like in its morphological features.
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40
BiologyEasyMCQKCET · 2024
Stanley Miller simulated the conditions of pre-biotic Earth using a spark-discharge apparatus. Which organic compounds were observed by him upon analyzing the end product of his experiment?
A
Pigments
B
Fats
C
Nitrogen bases
D
Amino acids
Solution
(D) Amino acids.
Stanley Miller,in his experiment,simulated the conditions of the primitive Earth by using a mixture of $CH_4$,$H_2$,$NH_3$,and water vapor $(H_2O)$ in a closed flask.
He provided energy through electric discharges (sparks) to mimic lightning.
Upon analyzing the end product of his experiment,he observed the formation of simple organic compounds,specifically amino acids such as glycine,alanine,and aspartic acid.
Stanley Miller,in his experiment,simulated the conditions of the primitive Earth by using a mixture of $CH_4$,$H_2$,$NH_3$,and water vapor $(H_2O)$ in a closed flask.
He provided energy through electric discharges (sparks) to mimic lightning.
Upon analyzing the end product of his experiment,he observed the formation of simple organic compounds,specifically amino acids such as glycine,alanine,and aspartic acid.
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41
BiologyEasyMCQKCET · 2024
The genome of $HIV$ replicates in the macrophages with the help of an enzyme called:
A
$DNA$ Polymerase
B
$RNA$ Polymerase
C
Reverse Transcriptase
D
$DNA$ Ligase
Solution
(C) $HIV$ is a retrovirus that contains $RNA$ as its genetic material.
Upon entering the host cell (macrophage),the viral $RNA$ genome is converted into $DNA$ by the enzyme $Reverse \ Transcriptase$.
This $DNA$ then integrates into the host cell genome,allowing the virus to replicate.
Therefore,the correct enzyme is $Reverse \ Transcriptase$.
Upon entering the host cell (macrophage),the viral $RNA$ genome is converted into $DNA$ by the enzyme $Reverse \ Transcriptase$.
This $DNA$ then integrates into the host cell genome,allowing the virus to replicate.
Therefore,the correct enzyme is $Reverse \ Transcriptase$.
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42
BiologyEasyMCQKCET · 2024
The principle of vaccination is based on which property of the immune system?
A
Memory
B
Specificity
C
Diversity
D
Plasticity
Solution
(A) The principle of vaccination is based on the property of 'Memory' of the immune system.
During vaccination,an inactivated or weakened pathogen (vaccine) is introduced into the body.
The immune system recognizes the antigen and produces specific antibodies against it.
Crucially,the immune system also generates memory $B$ and $T$ cells.
When the body encounters the actual pathogenic agent later,these memory cells recognize the pathogen immediately and produce a massive amount of antibodies to neutralize the infection.
Therefore,the correct option is $A$.
During vaccination,an inactivated or weakened pathogen (vaccine) is introduced into the body.
The immune system recognizes the antigen and produces specific antibodies against it.
Crucially,the immune system also generates memory $B$ and $T$ cells.
When the body encounters the actual pathogenic agent later,these memory cells recognize the pathogen immediately and produce a massive amount of antibodies to neutralize the infection.
Therefore,the correct option is $A$.
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43
BiologyEasyMCQKCET · 2024
The chemical substances which are produced by some microbes and can kill or retard the growth of other microbes are known as:
A
Statins
B
Streptokinases
C
Cyclosporins
D
Antibiotics
Solution
(D) Antibiotics.
Antibiotics are chemical substances produced by some microbes that can kill or retard the growth of other disease-causing microbes.
The term 'antibiotic' comes from the Greek words 'anti' meaning 'against' and 'bios' meaning 'life',together meaning 'against life' (in the context of pathogens).
Antibiotics are chemical substances produced by some microbes that can kill or retard the growth of other disease-causing microbes.
The term 'antibiotic' comes from the Greek words 'anti' meaning 'against' and 'bios' meaning 'life',together meaning 'against life' (in the context of pathogens).
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44
BiologyEasyMCQKCET · 2024
Select the correct statement from the following:
A
Methanobacterium is an aerobic bacteria found in the rumen of cattle.
B
Biogas is produced by the activity of aerobic bacteria.
C
Biogas is pure methane.
D
Activated sludge in sediment tanks is a rich source of aerobic bacteria.
Solution
(D) The correct statement is $D$. Activated sludge in sediment tanks is a rich source of aerobic bacteria.
$A$ is incorrect because $Methanobacterium$ is an anaerobic bacterium.
$B$ is incorrect because biogas is produced by the activity of anaerobic bacteria,not aerobic bacteria.
$C$ is incorrect because biogas is a mixture of gases (primarily $CH_4$,$CO_2$,and $H_2S$),not pure methane.
$A$ is incorrect because $Methanobacterium$ is an anaerobic bacterium.
$B$ is incorrect because biogas is produced by the activity of anaerobic bacteria,not aerobic bacteria.
$C$ is incorrect because biogas is a mixture of gases (primarily $CH_4$,$CO_2$,and $H_2S$),not pure methane.
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45
BiologyEasyMCQKCET · 2024
The $DNA$ polymerase of $Thermus$ $aquaticus$ is:
A
Thermolabile
B
Thermophobic
C
Exonuclease
D
Thermostable
Solution
(D) The correct answer is $D$.
$Thermus$ $aquaticus$ is a bacterium that lives in hot springs.
Its $DNA$ polymerase enzyme,known as $Taq$ polymerase,is thermostable.
This means it can withstand high temperatures without denaturing.
This property is crucial in the Polymerase Chain Reaction $(PCR)$ technique,as it remains active during the high-temperature denaturation step of the $DNA$ double strands.
$Thermus$ $aquaticus$ is a bacterium that lives in hot springs.
Its $DNA$ polymerase enzyme,known as $Taq$ polymerase,is thermostable.
This means it can withstand high temperatures without denaturing.
This property is crucial in the Polymerase Chain Reaction $(PCR)$ technique,as it remains active during the high-temperature denaturation step of the $DNA$ double strands.
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46
BiologyEasyMCQKCET · 2024
Which of these enzymes is required to cleave a plasmid?
A
Ligase
B
Endonuclease
C
Exonuclease
D
Polymerase
Solution
(B) The correct answer is $B$.
Endonucleases are enzymes that cleave the phosphodiester bonds within a polynucleotide chain.
Specifically,restriction endonucleases are used in biotechnology to cut plasmid $DNA$ at specific recognition sequences,which is a crucial step in recombinant $DNA$ technology.
Endonucleases are enzymes that cleave the phosphodiester bonds within a polynucleotide chain.
Specifically,restriction endonucleases are used in biotechnology to cut plasmid $DNA$ at specific recognition sequences,which is a crucial step in recombinant $DNA$ technology.
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47
BiologyEasyMCQKCET · 2024
If a recombinant $DNA$ bearing a gene for resistance to Ampicillin is transferred into $E. coli$ cells,the host cells become transformed into Ampicillin-resistant cells. What happens when these $E. coli$ are grown on a medium containing Ampicillin?
A
Non-transformants will grow and transformants will die
B
Non-transformants will die and transformants will grow
C
Both non-transformants and transformants will die
D
Both non-transformants and transformants will grow
Solution
(B) The correct answer is $B$.
When $E. coli$ cells are transformed with a recombinant $DNA$ molecule containing an Ampicillin-resistance gene,they acquire the ability to survive in the presence of the antibiotic Ampicillin.
Non-transformants are cells that have not taken up the recombinant $DNA$ and therefore lack the resistance gene.
When these cells are cultured on a medium containing Ampicillin,the non-transformants will die due to the antibiotic's effect,while the transformants will survive and grow to form colonies.
When $E. coli$ cells are transformed with a recombinant $DNA$ molecule containing an Ampicillin-resistance gene,they acquire the ability to survive in the presence of the antibiotic Ampicillin.
Non-transformants are cells that have not taken up the recombinant $DNA$ and therefore lack the resistance gene.
When these cells are cultured on a medium containing Ampicillin,the non-transformants will die due to the antibiotic's effect,while the transformants will survive and grow to form colonies.
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