Thursday, September 2, 2021

Genetics and Evolution : Biology

click here (For More Questions based on Biology )

 Q.1. Frequency of an allele may change in isolated  population due to

A) genetic drift

B) mutation

C) natural selection 

D gene flow


Q.2. The Hardy-Weinberg law of equilibrium was based on the following

A) random mating, selection, gene flow

B) random mating, genetic drift, mutation

C) non-random mating ,mutation, mutation, gene flow

D) random mating, no mutation, no gene flow and no genetic drift


Q.3. What influenced Darwin's idea of natural selection?

A) Hugo de Vries mutation theory

B) Malthus theory of population 

C) Wallace theory of natural selection

D) Lamarck theory of acquired characters


Q.4. Concept of genetic drift was introduced by

A) Sewall Wright

B) Hardy-Weinberg

C) Julian Huxely

D) GG simpson


Q.5. Mendel was born in

A) 17th century

B) 18th century

C) 19th century

D) 8th century


Q.6. In 1900 A.D., three biologists independently discovered Mendel’s principles .They are 

A) De Vries , Correns and Tschermak

B) Sutton , Morgan and Bridges

C) Avery , McLeod and McCarthy

D) Bateson , Punnet and Bridges


Q.7. Organism of pure line is that which produces individuals of

A) Dominant characters

B) Recessive characters

C) Its own characters 

D) Intermediate type


Q.8. In Mendelism , linkage was not observed due to

A) Mutation

B) Independent assortment

C) Synapsis

D) Crossing over


Q.9.  The genotype of an individual is Rr Bb. How many different types of gametes will it produce based on the law of independent assortment ?

A) 16

B) 9

C) 8

D) 4


Q.10.  If in a dihybrid cross , Mendel had used two such characters which have linked , he would have faced difficulty in explaining the results on the basis of his

A) Law of segregation

B) Law of multiple factor hypothesis

C) Law of independent assortment

D) Law of dominance

Statistic and Probability : Mathematics MCQ

Q.1. Two boys b subscript 1 b subscript 2 and three girls g subscript 1 comma g subscript 2 comma g subscript 3 play a tournament. Those of the same sex have equal probabilities of winning but each boy is twice as likely to win as any girl. The probability of winning the tournament by a girl is  [MP PET 2007]

A) 2/7

B) 3/7

C) 1/7

D) None of these

Explanation-

Let probability of winning of girl be x

therefore Probability of winning of boy = 2 x

and x + 2x = 1 rightwards double arrow x = 1 third

therefore Probability of winning of girl = 3 over 5 cross times 1 third equals 1 fifth .


Q.2. Two athletes A and B participate in a race along with other athletes. If the chance of A winning the race is 1/6 and that of B winning the same race is 1/8, then the chance that neither wins the race, is 

A) 1/4

B) 7/24

C) 17/24

D) 35/48

Explanation-

Chance of winning the race=1 over 6

 and chance of B winning the race=1 over 8

Chance ofwinsthe race=1 over 6 plus 1 over 8 plus 14 over 48=7 over 24 

Hence,chancethatneitherwinstherace=1 minus 7 over 24 equals 17 over 24 


Q.3. A speaks truth in 60% cases and b speak truth in 70% cases. The probability that they will say the same thing while describing a single events is 

A) 0.56

B) 0.54

C) 0.38

D) 0.94

Explanation-

straight P left parenthesis straight A right parenthesis equals 60 over 100 equals 3 over 5 comma space straight P left parenthesis straight A with bar on top right parenthesis equals 1 minus 3 over 5 equals 2 over 5
straight P left parenthesis straight B right parenthesis equals 70 over 100 equals 7 over 10 comma space straight P left parenthesis straight B with bar on top right parenthesis equals 1 minus 7 over 10 equals 3 over 10
Probability space of space single space event
equals straight P left parenthesis straight A right parenthesis. straight P left parenthesis straight B right parenthesis equals straight P left parenthesis straight A with bar on top right parenthesis straight P left parenthesis straight B with bar on top right parenthesis
equals 3 over 5 cross times 7 over 10 plus 2 over 5 cross times 3 over 10
equals 27 over 50 equals 0.54 


Q.4. A second order determinant is written down at random using the numbers 1,-1 as elements. The probability that the value of the determinant is non zero is 

A) 1/2

B) 3/8

C) 5/8

D) 1/3

Explanation-

n(S)=2 to the power of 4=16(because each of the four places can be filled in 2 ways).The zero determinants are 

open vertical bar table row 1 1 row 1 1 end table close vertical bar comma open vertical bar table row 1 1 row cell negative 1 end cell cell negative 1 end cell end table close vertical bar comma open vertical bar table row cell negative 1 end cell cell negative 1 end cell row 1 1 end table close vertical bar comma open vertical bar table row cell negative 1 end cell 1 row cell negative 1 end cell 1 end table close vertical bar comma
open vertical bar table row 1 cell negative 1 end cell row 1 cell negative 1 end cell end table close vertical bar comma space open vertical bar table row cell negative 1 end cell 1 row cell negative 1 end cell 1 end table close vertical bar comma open vertical bar table row 1 cell negative 1 end cell row cell negative 1 end cell 1 end table close vertical bar comma open vertical bar table row cell negative 1 end cell cell negative 1 end cell row cell negative 1 end cell cell negative 1 end cell end table close vertical bar
straight n left parenthesis straight E right parenthesis equals 8
therefore straight P left parenthesis straight E right parenthesis equals fraction numerator straight n left parenthesis straight E right parenthesis over denominator straight n left parenthesis straight S right parenthesis end fraction equals 8 over 16 equals 1 half 


Q.5. If n positive integers taken at random and multiplied together, than the chance last digit of the product would be 1,3,5,7 or 9 is 

A) open parentheses 2 over 5 close parentheses to the power of straight n

B) open parentheses 1 half close parentheses to the power of straight n

C) fraction numerator 2 to the power of straight n minus 1 over denominator 5 to the power of straight n end fraction

D) fraction numerator 5 to the power of straight n minus 4 to the power of straight n over denominator 10 to the power of straight n end fraction 

Explanation-

straight P left parenthesis straight E subscript 1 right parenthesis equals fraction numerator straight n left parenthesis straight E subscript 1 right parenthesis over denominator straight n left parenthesis straight S right parenthesis end fraction equals 5 to the power of straight n over left parenthesis 10 right parenthesis to the power of straight n equals open parentheses 1 half close parentheses to the power of straight n 


Q.6. A is a set containing 10 elements. A subset P is A is chosen at random and the set A  is reconstructed by replacing the elements of P. Another subset Q and A is now chosen at random. Then, the probability that if, P and Q have no common elements, is 

A) open parentheses 2 over 3 close parentheses to the power of 10

B) open parentheses 3 over 4 close parentheses to the power of 10

C) open parentheses 4 over 5 close parentheses to the power of 10

D) open parentheses 5 over 6 close parentheses to the power of 10 

Explanation-

 because straight P intersection straight Q equals straight Ï•
therefore straight n left parenthesis straight E right parenthesis equals 3 to the power of 10 space space space space space space space space space left square bracket because left parenthesis straight i right parenthesis not an element of straight P intersection straight Q right square bracket
rightwards double arrow straight P left parenthesis straight E right parenthesis equals 3 to the power of 10 over 4 to the power of 10 equals open parentheses 3 over 4 close parentheses to the power of 10


Q.7. A triangle  informed with the vertices of n sided regular polygon, Then the probability that the triangle, Have exactly one side common with the side of the polygon is 

A) fraction numerator 6 left parenthesis straight n minus 2 right parenthesis left parenthesis straight n minus 3 right parenthesis over denominator straight n left parenthesis straight n minus 1 right parenthesis end fraction

B) fraction numerator 6 left parenthesis straight n minus 3 right parenthesis over denominator left parenthesis straight n minus 1 right parenthesis left parenthesis straight n minus 2 right parenthesis end fraction

C) fraction numerator 6 left parenthesis straight n minus 4 right parenthesis over denominator left parenthesis straight n minus 1 right parenthesis left parenthesis straight n minus 2 right parenthesis end fraction

D) fraction numerator 6 left parenthesis straight n minus 2 right parenthesis over denominator left parenthesis straight n minus 1 right parenthesis left parenthesis straight n minus 3 right parenthesis end fraction 

Explanation-

Total number of ways=

straight i to the power of straight n straight C subscript 3 equals fraction numerator straight n left parenthesis straight n minus 1 right parenthesis left parenthesis straight n minus 2 right parenthesis over denominator 1.2.3 end fraction
space space space space space space space space equals fraction numerator straight n left parenthesis straight n minus 1 right parenthesis left parenthesis straight n minus 2 right parenthesis over denominator 6 end fraction

Favorable ways=If side AB and take one vertex out of (n-4) vertices

equals to the power of straight n minus 4 end exponent straight C subscript 1 cross times straight n
straight n left parenthesis straight E subscript 1 right parenthesis equals left parenthesis straight n minus 4 right parenthesis straight n
Required space probability space straight p subscript 1 equals fraction numerator begin display style fraction numerator straight n left parenthesis straight n minus 4 right parenthesis over denominator straight n left parenthesis straight n minus 1 right parenthesis left parenthesis straight n minus 2 right parenthesis end fraction end style over denominator 6 end fraction
equals fraction numerator 6 space left parenthesis straight n minus 4 right parenthesis over denominator straight n left parenthesis straight n minus 1 right parenthesis left parenthesis straight n minus 2 right parenthesis end fraction 


Q.8. 5 persons A, B, C, D and E are in a queue of a shop. The probability that A and E always occur together, is

A) 1 fourth

B) 2 over 3

C) 2 over 5

D) 3 over 5 

Explanation-

5 persons A, B, C, D and E can stand in a queue in 5! ways. Persons a and E can occur together in 2! ways. Considering A and E as one person, there are 4 persons who can be queued up in 4! ways. So, A and E can occur together in 4! cross times 2! ways. So, required probability = fraction numerator 4 factorial cross times 2 factorial over denominator 5 factorial end fraction space equals space 2 over 5 


Q.9. If a batsman scored 70 in one innings and remained not out at 50 in the 2 nd inning . His batting average  is...

A) 60

B) 120

C) 70

D) 85


Q.10. The best statistical  measure used for comparing two series is...

A)  mean deviation 

B) range 

C) coefficient of variation 

D) none of these