Biology Test

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Biology Test - 55

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Weekly Quiz Competition

Question 1
1 / -0

The stage in cell division that consists of the G1 phase, S phase, G2 phase is:

A
Metaphase

B
Prophase

C
Anaphase

D
Interphase
Solution
The stage in cell division that consists of the G1 phase, S phase, G2 phase is the Interphase.
Interphaseis the period between the end of one cell division to the beginning of the next cell division. Theinterphase takes almost 95% of the total timeof the cell cycle. During interphase the cell prepares itself for the next division, it grows in size. So thecell is the most active metabolically in Interphase.
Interphase is divided into 3 phases:
1. G1 phase/Post mitotic/Pre-DNA synthetic phase/Ist gap phase
2. S-phase/Synthetic phase
3. G2-phase/Pre mitotic/Post synthetic phase/IInd gap phase.
Cell growth:Itis the period of synthesis and duplication of various components of the cell.
Cell Divisionis the process by which a mature cell divides and forms 2 nearly equal daughter cells which resemble the parental cell in a number of characteristics. 2 processes take place during cell reproduction. It is the process when a mature cell divides into two cells.The Cell cycle completes in 2 steps:
1. Interphase
2. M-phase/Dividing phase
Question 2
1 / -0

A given double stranded DNA molecule is 1,00,000 base pairs long. What will be the length of this DNA molecule?

A
10,000 nm

B

C

D
2,00, 000 nm

Solution

The length of this DNA molecule will be since one complete spiral of the helix is 3.4 nm (34 A° long and has 10 base pairs.

Question 3

1 / -0

The table below gives the populations (in thousands) of ten species (A-J) in four areas (p-s) consisting of the number of habitats given within brackets against each. Study the table and answer the question which follows.

Which area out of p- s shows maximum species diversity?

Area and No. of habitats	Species, and their populations (in thousands) in the areas
Area and No. of habitats	A	B	C	D	E	F	G	H	I	J
p(11)	2.3	1.2	0.52	6.0	-	3.1	1.1	9.0	-	10.3
q(11)	10.2	-	0.62	-	1.5	3.0	-	8.2	1.1	11.2
r(13)	11.3	0.9	0.48	2.4	1.4	4.2	0.8	8.4	2.2	4.1
s(12)	3.2	10.2	11.1	4.8	0.4	3.3	0.8	7.3	11.3	2.1

s

p

q

r

Solution

Species diversity is related to the variety in the number and richness of the species within a region and is measured at the level of 'species'. Thus, it is the product of species richness and species evenness. Species richness refers to the number of species per unit area. As the area of the site increases, the number of species also increases due to more availability of natural resources. Species evenness is the relative abundance with which each species is represented in an area. Thus, variation in the number of species, kinds of species as well as the number of individuals per species lead to greater diversity. In the given table, the area which shows maximum species diversity is 's'.

Question 4
1 / -0

Aestivation of petals in the flower of cotton is correctly shown in

A
(a)

B
(b)

C
(c)

D
(d)

Solution

In cotton, china rose and lady's finger margins of sepals or petals overlap that of the next one this mode of arrangement (aestivation) is called twisted.
Question 5
1 / -0

With increase in age which of the following is true for lungs?

A
Pulmonary compliance increases

B
Residual volume decreases

C
Mucocillary clearance increases

D
Fibrous of interstitium decreases

Solution

Pulmonary compliance increases with increase in age is true for lungs.

Changes in lung compliance can indicate issues within the lungs. These generally cause, or are caused by, injuries, illnesses, or impairments. Increased compliance can be present where there is degeneration of lung tissue. Degenerative lung tissue diseases (eg emphysema) make it harder for the lungs to expand, and harder to exhale as there is less elastic recoil.
Question 6
1 / -0

Match the pteridophytes with their common names:

A B

1. Psilotum a. Ground pine

2. Marsilea b. Spike moss

3. Selaginella c. Whisk fern

4. Lycopsida d. Water fern

A
1 - a, 2 - d, 3 - b, 4 - c

B
1 - c, 2 - b, 3 - d, 4 - a

C
1 - c, 2 - d, 3 - b, 4 - a

D
1 - d, 2 - c, 3 - b, 4 - a
Solution
The correct match is: 1 - c, 2 - d, 3 - b, 4 - a.
- Psilotum - Psilotum is a genus of fern-like vascular plants. It is one of two genera in the family Psilotaceae commonly known as whisk ferns, the other being Tmesipteris
- Marsilea -Marsilea is a genus of approximately 65 species of aquatic ferns of the family Marsileaceae. The name honours Italian naturalist Luigi Ferdinando Marsili (1656–1730).
- Selaginella - Selaginella is a pteridophyte. It is also called spikemoss or club moss. It is the largest and the only living genus of the family Selaginellaceae.
- Lycopsida - a subdivision of Tracheophyta coextensive with the class Lycopodineae comprising vascular plants (as the club mosses and related forms) with small leaves, sessile and adaxial sporangia, and no leaf gaps in the primary vascular cylinder — compare psilopsida , pteropsida , sphenopsida.
Question 7
1 / -0

Viroids have

A
Single-stranded RNA not enclosed by a protein coat

B
Single-stranded DNA and enclosed by a protein coat

C
Double-stranded DNA enclosed by a protein coat

D
Double-stranded RNA enclosed by a protein coat

Solution

Viroids consist of short strands of circular, single-stranded RNA without protein coats. The genome is extremely small in size, ranging from 246 to 467 nucleobases. These are mostly pathogenic to the plant. On infection, it enters the cell through plasmodesmata.

Question 8

1 / -0

Study the following lists and match them to the appropriate answers.

List - I	List - II
(A) Dieback in citrus	(I)Urease
(B)Mottled leaf	(II)Hexokinase
(C)Mouse-ear in pecan	(III)Nitrogenase
(D)Whiptail in cauliflower	(IV)Cytochrome C oxidase
	(V) Carboxypeptidase

A- III, B- II, C- IV, D- I

A- IV, B- V, C- I, D- III

A- IV, B- I, C- III, D- V

A- V, B- III, C- II, D- IV

Solution

Copper is concerned with the enzyme cytochrome-C and the deficiency of copper in citrus plants is called 'dieback'.

Mottled leaves are caused by zinc deficiency and in the enzyme carboxypeptidase, Zn acts as a prosthetic group. Mouse-ear in pecan is a growth abnormality resulting from a deficiency of nickel in pecan trees and the enzyme urease requires nickel. Whiptail in cauliflower is caused due to deficiency of molybdenum which is required by the enzyme nitrogenase.

Question 9
1 / -0

After fertilization the ovule develops into;

A
Endosperm

B
Seed

C
Embryo sac

D
Fruit

Solution

After fertilization the ovule develops into seed.
Fertilization occurs when one of the sperm cells fuses with the egg inside of an ovule. After fertilization occurs, each ovule develops into a seed. Each seed contains a tiny, undeveloped plant called an embryo. The ovary surrounding the ovules develops into a fruit that contains one or more seeds.
Question 10
1 / -0

Hypodermis in monocotyledonous stem is?

A
Parenchymatous

B
Chlorenchymatous

C
Collenchymatous

D
Sclerenchymatous

Solution

Hypodermis in monocotyledonous stem is sclerenchymatous.
The region present below the epidermis in monocot stem is hypodermis. It is composed with sclerenchyma. Sclerenchyma is dead mechanical tissue. It provides mechanical strength to the stem.

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A	B
1. Psilotum	a. Ground pine
2. Marsilea	b. Spike moss
3. Selaginella	c. Whisk fern
4. Lycopsida	d. Water fern

Biology Test - 55

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Biology Test - 55

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Question 1

Metaphase

Prophase

Anaphase

Interphase

Solution

Question 2

10,000 nm

2,00, 000 nm

Solution

Question 3

s

p

q

r

Solution

Question 4

(a)

(b)

(c)

(d)

Solution

Question 5

Pulmonary compliance increases

Residual volume decreases

Mucocillary clearance increases

Fibrous of interstitium decreases

Solution

Question 6

1 - a, 2 - d, 3 - b, 4 - c

1 - c, 2 - b, 3 - d, 4 - a

1 - c, 2 - d, 3 - b, 4 - a

1 - d, 2 - c, 3 - b, 4 - a

Solution

Question 7

Single-stranded RNA not enclosed by a protein coat

Single-stranded DNA and enclosed by a protein coat

Double-stranded DNA enclosed by a protein coat

Double-stranded RNA enclosed by a protein coat

Solution

Question 8

A- III, B- II, C- IV, D- I

A- IV, B- V, C- I, D- III

A- IV, B- I, C- III, D- V

A- V, B- III, C- II, D- IV

Solution

Question 9

Endosperm

Seed

Embryo sac

Fruit

Solution

Question 10

Parenchymatous

Chlorenchymatous

Collenchymatous

Sclerenchymatous

Solution

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