NEET Test 78

Result

NEET Test 78

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
4 / -1

A mass M is divided into two parts Xm and (1−X)m. For a given separation, the value of X, for which the gravitational attraction between the two parts becomes maximum, is

A
1/2

B
3/5

C
1/7

D
2/5

Solution
Question 2
4 / -1

Two waves of intensities I and 4I superpose. What will be the maximum and minimum intensities?

A
5I, 3I

B
9I, I

C
9I, 3I

D
5I, I

Solution
Question 3
4 / -1

R.Q. (Respiratory Quotient) is infinite

A
If the respiratory substrate is a carbohydrate

B
If the respiratory substrate is fat

C
If the respiratory substrate is an organic acid

D
In anaerobic respiration

Solution

The R.Q. (Respiratory Quotient) of anaerobic respiration is infinite. This is because, in anaerobic respiration, the total volume of oxygen consumed is zero as this process occurs in the absence of oxygen and only carbon dioxide is produced.
Question 4
4 / -1

A body is thrown up with a speed u, at an angle of projection θ. If the speed of the projectile becomes u/√2 on reaching the maximum height, then the maximum vertical height attained by the projectile is -

A
u² / 4g

B
u² / 3g

C
u² / 2g

D
u² / g

Solution
Question 5
4 / -1

Which one of the following pairs is correctly matched?

A
Anaphase I - Homologous chromosomes are separated

B
Metaphase I - Pairing of maternal and paternal homologous chromosomes takes place

C
Interphase - A nuclear envelope encloses each haploid set of chromosomes

D
Prophase I - Non-homologous chromosomes are separated

Solution

Anaphase I involves separartion of homologous chromosomes which start moving opposite poles so each tetrad is divided into two daughter dyads.

In Metaphase, I Centromeres of homologous chromosomes lie equidistant from equator and are directed towards the poles while arms generally lie horizontally on the equator. Each homologous chromosome has two kinetochores and both the kinetochores of a chromosome are joined to the chromosomal or tractile fibre of same side.

Whereas Prophase I is divided in five subphases i.e., leptotene, zygotene, pachytene, diplotene, and diakinesis.

Interphase: It is the period between the end of one cell division to the beginning of the next cell division. It is also called the resting phase or not dividing phase.

Hence, the correct option is (A).
Question 6
4 / -1

0.73 g of organic compound on oxidation gave 1.32 g of carbon dioxide. The percentage of carbon in the given compound will be:

A
49.32

B
59.32

C
29.32

D
98.64

Solution
Question 7
4 / -1

The IUPAC name for

A
3−amino−5−heptenoic acid

B
β−amino−δ

C
5−amino−2−heptenoic acid

D
5−amino-hex−2−enecarboxylic acid

Solution

As - COOH group is the highest priority group, it is numbered one. So, the IUPAC name is 3−amino−5−heptanoic acid.

Hence, the correct option is (D).
Question 8
4 / -1

Which of the following national parks is home to the Asiatic lion?

A
Ranthambore National Park

B
Keoladeo National Park

C
Kaziranga National Park

D
Gir National Park

Solution

Gir National Park, The only place to see Asiatic Lions in the world. The Gir Forest National Park and Wildlife Sanctuary is a forest zone located 43 km north-east of Somnath.

Hence, the correct option is (D).
Question 9
4 / -1

In a DNA strand the nucleotides are linked together by:

A
Glycosidic bonds

B
Phosphodiester bonds

C
Peptide bonds

D
Hydrogen bonds

Solution

A nucleotide has three components - nitrogenous base, a pentose sugar ribose in the case of RNA and deoxyribose for DNA), and a phosphate group. There are two types of nitrogenous bases - Purines (Adenine and Guanine) and Pyrimidines (Cytosine and Thymine). A nitrogenous base is linked to the pentose sugar through a N-glycosidic linkage to form a nucleoside. When a phosphate group is linked to 5 -OH of a nucleoside through phosphodiester linkage, a corresponding nucleotide is formed. The nucleotides are linked through 3-5 phosphodiester linkage to form a dinucleotide.

Hence, the correct option is (B).
Question 10
4 / -1

Which of the following features of genetic code does allow bacteria to produce human insulin by recombinant DNA technology?

A
Genetic code is not ambiguous

B
Genetic code is redundant

C
Genetic code is nearly universal

D
Genetic code is specific

Solution

In recombinant DNA technology bacteria is able to produce human insulin because genetic code is nearly universal, that is all contain same nitrogen bases, that is A, G, C, and T.

Hence, the correct option is (C).