CSIR UGC NET MATHS IN LIFE SCIENCES.pptx

CSIR UGC NET- MATHEMATICAL QUESTIONS AND
SOLUTIONS IN LIFE SCIENCES
Dr. D. Charumathi CSIR NET, TN SET
Assistant Professor,
Department of Biotechnology
D.K.M College for Women (Autonomous), Vellore-1

• The genome of bacterium composed of a single DNA molecule which is
10 9 bp long. How many moles of genomic DNA is present in the
bacterium? (consider Avagadro no.= 6x 1023)
• Options a) (1/ 6 )x 10 -23 b) (1/ 6.023) x 10 -14 c) 6x 10 14 d) 6.023) x 10 23
• Answer:a)
• 1 mole contains 6.023 x 10 23 molecules
• Therefore 10 9 bp long single dna molecule will be present in how many
moles
• = 1 x 1/ 6.023 x 10 23 = (1/ 6.023) x 10 -23 moles

• The phosphoglucomutase catalyses
Glucose 6-P Glucose 1-P. The
standard free energy Change ΔG° of
this is +1.3 Kcal mol-1 at 25 ° C. If
initial concentration of Glucose 6-P
was used to carry out the reaction,
then the concentration of Glucose 6-
P and Glucose 1-P at equilibrium
would be
• A)50 mM and 50 mM respectively
• B)45 mM and 96 mM respectively
• C)96mM and 45 mM respectively
• D)100 mM and 10 mM respectively.
Answer:
Given: ΔG° = 1.3
Temperature= 25C= 25+273=298 Kelvin
We know ΔG° = -RTlnKeq; R= gas constant=
8.314 joule;
1 joule =0.00024 Kcal; there fore, R=8.314x
0.00024 Kcal,
Therefore, 1.3= -8.314x0.00024x298xln Keq
1.3= -0.5946x ln Keq;
1.3/-0.5946=lnKeq
-2.186=ln Keq
e-2.186 = Keq= 0.11= 11/100= Keq
Keq= (product concentration)/(reactant
concentration)
Keq= Glucose 1-P /Glucose 6-P
therefore, 11 mM and 100 mM
e0 =1; e-1 =0.367; e-2 = 0.13; e-10 =0; e1=2.7183; e100
=infinity

• The apparent pH of a fluid is 7.45 where bicarbonate buffer is
involved for maintaining its pH. Values of pKa of carbonic acid are
6.15 and 10.45. The molar rate of [conjugate base] is to acid is
• A) 1: 20; b) 20:1; c) 1:1000 d) 1000: 1
Answer
Given:
pH= 7.45; pKa= 6.15
pH=pKa+ log base/acid
7.45=6.15+log base/acid
7.45-6.15=log base/acid
1.3=log base/acid
Antilog 1.3= base/acid= 19.9/1
So, b) 20: 1

• 100 ml of 0.1 M sodium acetate solution has pH 8.0. To this solution 1000µl
of 1 M acetic acid (pKa= 4.76) of pH 2.80is added. The pH of this mixture
will be
• a) 8.90 b) 4.76 c) 2.80 d) 5.76
Answer: d) 5.76
Given pKa= 4.76;
1000µl= 1ml; 1ml of 1 M acetic acid added to 100 mlof 0. 1M sodium
acetate…So the mixture contains , 1M acetic acid is diluted with 0.1 M
sodium acetate so, the concentration of acetic acid in the mixture becomes
1ml /100ml x 1M = 0. 01 M and the concentration of base that is sodium
acetate is 0.1M.
pH = pKa + log base/acid
?= 4.76+ log 0.1/0.01
?= 4.76+ log 10
?=4.76+1
=5.76

• The internal energy of a gas increases by 1J when it is compressed by
a force of 1 Newton through 2 metres. The heat change of the system
is
• A) 1 j b) -1 J c) 2 J d) -2 J
• Given
F= 1 newton ; Distance = 2m
Work = F x d = 1 x 2= 2
ΔU=Q + W, where Change in internal energy
Q= heat change; W= work.
So, 1= Q+2; Q=1-2= -1
So answer b) -1J

The hydrolysis of pyrophosphate to orthophosphate is important for several biosynthetic
reactions. In E. coli the molecular mass of the enzyme pyrophosphatase is 120 KD, and it consists
of six identical subunits. The enzyme activity is defined as the amount of enzyme that hydrolyses
10µmol of pyrophosphate in 15 min at 37C under standard assay condition. The purified enzyme
has a Vmax of 2800 units per milligram of the enzyme.
How many moles of the substrate are hydrolyzed per second per milligram of the enzyme when
the substrate concentration much greater than Km?
a) 0.05µmol b) 62µmol c)31.1 µmol d) 1 µmol
Answer: c)
Given:
Enzyme activity= µmole of substrate utilized or product formed / (incubation time X Volume of
enzyme used)
Vmax = µmole of substrate utilized or product formed / (incubation time)
When [S]>> km; V= Vmax.
1 U= 10µmol of substrate hydrolysed/ 15 x 60; since time given in min convert it to sec, because
in question they have asked moles of substrate per sec.
Therefore 2800 U= 10 X 2800/(15X 60)= 31.1 µmol

• The following small peptide substrates are used for determining elastase activity and the following data have
been recorded.
• Substrate Km (mM) Kcat (s-1)
PAPA G 4.02 26
PAPA A 1.51 37
PAPA F 0.64 18
The arrow indicates the cleavage site, From the above observations, it appears that
A. PAPAF is digested most rapidly
B. PAPAG is digested most rapidly
C. A hydrophobic residue at the C terminus seems to be favoured.
D. A smaller residue at the C- terminus seems to be favoure
E. Elastase always requires a smaller residue at the N terminus of the cleavage site
Which of the following I true?
a) A, C and E b) B, D and E C) E only d) D and E only
Answer:
Kcat= turn over number= Vmax/ Et
Catalytic efficiency =Km/Kcat. Find the highest catalytic efficiency from given data. It shows highest catalytic
efficiency for 3rd peptide…So its is rapily digesed than others….Cleavage sites is followed by hydrophobic
aminoacids , Elastase always requires a smaller residue at the N terminus of the cleavage site
So answer option a) A, C and E

• A 1% (W/v) solution of a sugar polymer is digested by an enzyme
(20µg, MW= 200,000). The rate of monomer sugar (MW= 00)
liberated was determined to have a maximal initial velocity of 10 mg
formed /min. The turn over number (min-1) will be
• A) 5 x 104 b) 2.5 x 10-2 C) 4.0 x 10-4 D)2.5 x 105
• Answer: Kcat= turn over number = Vmax/Et
• To find Vmax ie amount of product formed or substrate utilized/
incubation time in moles, we should, convert amount of substrate
which is given in mg to moles, by converting mg into g and then
dividing it by molecular weight of the substrate.
• Therefore vmax= 10 x 10-3 /400 = 2.5 x 10-5 moles/min
Et= 20 x 10-6 /200000 = 1 x 10-10 moles
Therefore, Kcat= 2.5 x 10-5 / 1 x 10-10 =2.5 x105 /min
Answer d)

• The turn over number and specific activity of an enzyme (molecular weight 40,000D) in a reaction
(Vmax =4µmol of substrate reacted/min, enzyme amount = 2µg are
• A) 80,000/min , 2 x 10 3µ mol substrate/min
• B) 80,000/min, 2 x 10 3µ mol substrate/sec
• C) 40,000/min, 1 x 10 3µ mol substrate/min
• D)40,000/min, 2 x 10 3µ mol substrate/min
Answer:
Turn over number =Vmax/ Et
Vmax= 4µ mol /min = 4 x 10 -6 moles / min
Enzyme concentration (total)= 2 x 10 -6 / 40,000 moles
Vmax/Et= 4 x 10 -6 moles / 2 x 10 -6 / 40,000 = 80,000 /min
Specifc activity= Enzyme activity / mg of protein
Enzyme activity = 4µmol of substrate reacted
protein = 2µg= 0.002 mg
Therefore specific activity = = 4µmol of substrate reacted/ 0.002 mg
=2000 µ mol substrate
Answer:
A) 80,000/min , 2 x 10 3 µ mol substrate/min

• t1/2 of an irreversible first order reaction S P is 1 hr . The time required to
reach 75% completion is
• A) 2 h b)3 h c)1 h d) 4 h
• Answer: a)
• K= 0.693/t1/2
• K= 0.693/1
• We know, ln (No/Nt)=KX t
• Therefore let No the initial concentration of substrate= 100 % and at
completion of 75% of reaction, the remaining concentration Nt will be 100-
75%= 25%
• Sub in eqn,
• Ln (100/25)=0.693 x t
• Ln 4 =0.693 x t
• Ln 4/ 0.693= t = 2

A segment of B-DNA encodes an enzyme of molecular mass 50 KDa. The
estimated length of this segment in µm would be (CSIR 2013 June Part C)
a) 0.1547 b) 0.1547 x 10-3 C) 0.4641 d) 0.4641 x 10-3
Answer:
Molecular mass = 50 KDa = 50,000Da
Molecular mass of amino acid= 110Da
Therefore no. of amino acid present in 50,000Da= 50,000/110= 454.5 ≈ 455
3 nucleotides code for 1 amino acid.. There fore how many nucleotides codes
for 455 amino acid.= 455 x 3= 1365 bases
The distance between two bases in B DNA is 0.34 nm
Therefore distance between first base and the last base i.e the length of the
fragment= 1365x 0.34 = 464.1 nm
1nm= 1/1000µm
464.1 nm= 0.4641 µm

• A rapidly growing bacterial species such as E.coli exhibits typical
phase of growth cycle in liquid nutrient broth (lag phase – log phase-
stationary phase- death phase). If a bacterial culture has a starting
density of 103/ ml has lag time 0f 10 min of a generation time of 10
minutes, what will the cell denstiy be at (cells/ml) 30 minutes.
• Answer:
• Nt = No x 2n
• Nt = cell density at time t
• No= initial cell density
• n= no. of generation; n= total duration- lag duration/ generation time
• = 30-10/10=20/10=2
• Nt = No x 2n
• Nt= 10 3 x 2 2 = 4 x 103 cells/ ml

• The frequency of cells in a population that are undergoing mitosis (the mitotic
index) is a convenient way to estimate the length of the cell cycle. In order to
measure the cell cycle in the liver of he adult mouse by measuring the mitotic
index , liver slices are prepared and stained to easily identify cells undergoing
mitosis. It was observed that only 3 out of 25,000 cells are formed o be
undergoing mitoss. Assuming that M phase lasts for 30 min., then calculate the
approximate length of the cell cycle in the liver of an adult mouse?
• Answer:
• Mitotic Index= (total no of cells in mitosis/ total no of cells observed) X 100
• MI= 3/25000 X 100=3/250;
• duration of M phase = 30 min= ½ h
• Cell cycle length =duration of M phase / Mitotic index
• = (½)/(3/250)
• =41.6 h

Lipid rafts are involved in signal transduction in cells. Rafts have composition different from
rest of the membrane. Rafts were isolated and found to have cholesterol to sphingolipids
ratio 2:1. The estimated size of the raft is 35 nm2. If the surface area of the cholesterol is
40Å2 and sphingolipid is 60Å2 , how many cholesterol and sphingolipids are present in one
raft.
a) 50 cholesterol : 25 sphingolipid b) 200 cholesterol :100 sphingolipid
c) 40 cholesterol : 20 sphingolipid d) 20 cholesterol : 10 sphingolipid
Given: the surface area of raft = 35 nm2, Cholesterol to sphingolipid ratio= 2:1; Surface area
of one cholesterol molecule = 40Å2 ; Surface area of one sphingolipid = 60Å2
let residues of sphingolipids by y. for every sphingolipid molecule 2 molecule of
cholesterol present therefore, number of cholesterol=2y.
The equation becomes, surface area of all cholesterol molecules + surface area ofall
sphingolipid molecules = total surface area
1 nm = 10 Å; 1 nm2 = 100 Å2; 35 nm2 =3500 Å2
Total surface area=(2y x 40) + (y x 60) = 3500
80y + 60 y= 3500
140y=3500; y=3500/140= 25/1
Y=no .of sphingolipids= 25; 2y= no. of cholesterol= 2x 25= 50
So answer a) 50 cholesterol : 25 sphingolipid

• If the diameter of cylindrical histone octamer is 9nm and height 5 nm, which are 32 million in
nucleus of diameter 6µm.What fraction nucleus is occupied by histone ? CSIR June 2012 Part C
• a) 1/11 b) 1/21 c) 10/11 d) 10/21
• Answer:
• Given:
• Diameter of cylindrical histone octamer = 9nm ie radius= 9/2 nm
• Height of cylindrical histone octamer= 5nm; diameter of nucleus = 6µm = 6 x 103 nm , radius= 3x 103
nm
• No of histone octamer = 32 million= 32 x106
• To Find the fraction of histone occupied in nucleus.
• Volume of histone cylindrical octamers/ volume of nucleus
• Volume of a cyclinder= π x r2x h
• Volume of total histone cylindrical octamers= π x r2x h x 32 x106
• = π x (9/2)2 x 5x 32 x106
• Volume of nucleus (spherical shape)= (4/3) x π x r3 =(4/3) x π x 3x3x3x 109
• Fraction= Volume of total histone cylindrical octamers/ Volume of nucleus (spherical shape )
• = π x (9/2)2 x 5x 32 x106 /(4/3) x π x 27x 109 = 0.09= 9/100= 1/11

• The time taken for atrial systole and diastole in a normal heart are tas
and tad seconds respectively. If ventricular systole takes tvs seconds
calculate the ventricular diastolic time (seconds) (CSIR 2011 JUNE
PARTC)
• A.(tas + tad)- tvs b) (tas-tad)+tvs c) (tad-tas)+ tvs d) (tas +tad) X tvs
• Answer:
• tvs+tvd = tas+tad;
• So , tvd=(tas+tad)-tvs

• Mendel crossed tall pea plants with dwarf ones. The F1 plants were all tall. When theses F1 plants were
selfed to produce F2 generation, he got a 3:1 tall to dwarf ratio in the offspring. What is the probability that
out of three plants (of F2 generation ) picked up at random two would be dwarf and one would be tall?
(PART C CSIR JUNE 2011).
• A) ¾ b)3/8 c) 9/64 d) 9/32
• Answer: c)
• Given : n of off spring picked=3; no. of dwarf to be picked (x)= 2; no of tall to be picked (n-x)=1;
• off spring tall : dwarf ratio= 3:1
• For monohybrid cross F2 generation,
binomial equation= only two possible outcomes,
• P (x)= (n! /(x! (n-x)!) x px x qn-x
• x: The number of successes that result from the binomial experiment. n: The number of trials in
the binomial experiment. P: The probability of success on an individual trial.
• The letter p denotes the probability of a success on one trial, and q denotes the probability of a failure on
one trial.
• probability of success ie dwarf (p) = 1/4
• Probability of failure ie. tall (q) = 3/4
• probability of picking 2 dwarf and one tall=
• = (3!/(2!(3-2)!) x (1/4)2 x (3/4) 3-2 = 9/64

• Two plant species having cob length of 9 cm and 3 cm were crossed. Assume trait for cob
length is polygenic and additive property, what would be the cob length of progenies ?
(CSIR June 2012 PART B).
• A) 3 cm b) 6cm ) 9 cm d)12 cm
• Answer : b)
• Polygenic inheritance = Polygenic inheritance occurs when one characteristic is
controlled by two or more genes.
• Cob length 9cm due to dominant alleles, cob length 3cm due to recessive allele.
• Let two genes control cob length A, B, therefore plant with cob length 9cm is having
dominant alleles AABB whereas the cob length 3cm plant has recessive alleles aabb.
• The cross between these plants will give AaBb offspring.
• AABB alleles (4 dominant) give 9cm cob length there fore each dominant allele gives=
9/4= 2.2 cm cob length.
• Similarly, aabb (4 recessive alleles) give 3cm cob length therefor each recessive allele
gives = ¾=0.75.
• Therefore AaBb give cob length can be got from 2 dominant allele contribution +2
recessive alleles contribution = 2X2.2 + 2X0.75= 5.9 = 6cm.
•

• How many gametes would be formed from genotype AaBBccDdEe?
CSIR JUNE 2012 part B.
• A) 8 b) 16 c) 32 d) 64
• Answer:
• No of gametes = 2n
• Where n= no of. Heterozygous genes
• Here n= 3
• Therefore no .of gametes= 23 = 8

• Two plants with white flowers are crossed . White flowers arise due to recessive mutation. All F1
progeny have red flowers. When the F1 plants are selfed, both red and white flowered progeny
are observed . In what ratio will red flowered plants and white flowered plants occur? CSIR 2014
Dec, Part B.
• A) 1:1 b) 3:1 c) 9:7 d) 15:1 Answer: c)
• If it is mono hydrid cross, white flower is due to homozygous recessive aa. Red will be due to Aa
or AA ..if white flower plant crosses with other recessive white plant then its f1 progeny will be
all white. aa x aa = aa ..but in question all F1 progeny is having red flower. therefore more than
one gene code for the red color. Let us consider two genes contribute for red color.
• Let two genes A, and B be involved so the chance for genotype of red flower plant would be
AABB, AaBb, AaBB, white flowers bearing parent would be Aabb, aabb, aaBB, aaBb, AAbb.
• When we cross white flower plant, for example self cross these genotypes and crosses like
Aabb x aabb, Aabb xAAbb , they will not result in red flower plant because not two non allelic
dominant genes present together in F1 progeny.
• …on the other hand, crosses like Aabb x aaBb, Aabb x aaBB , Aabb x aaBB , will produce plants
with red flower and white flower. so only one cross AAbb x aaBB will give f1 progeny with
genotype AaBb (red flower) all progeny will be red fowering in this cross. Therefore AAbb and
aaBB are parents. So all F1 Progeny has AaBb genotype
• Therefore on self crossing f1 progeny AaBb we get both red and white flowers in the ratio 9:7

aB aB aB aB
Ab AaBb AaBb AaBb AaBb
AAbb x aaBB

• Genes A,B and C control three phenotypes whih assort
independently. A plant with the genotype AaBbCc is selfed. What is
the probability for progeny which shows the dominant phenotype
for ATLEAST ONE of the phenotype controlled by genes A,B, C (CSIR
2015 JUNE PART B)
• A) 1/64 b) 27/64 c) 63/64 d) cannot be predicted
• Answer c
• This is trihybrid cross so the ratio is 27:9:9:9:3:3:3:1
• out of 64 plant , Only one 1 plant will have all recessive genes aabbcc
and other plants 64-1=63, will have at least one dominant allele
which will result in expression of the phenotype.
• Therefore the probability for progeny which shows the dominant
phenotype for ATLEAST ONE of the phenotype controlled by genes
A,B, C is 63/64.

• A cross was made between pure wild type males and brown eyed curled winged females of D. melanogaster. The
F1 progeny obtained was as follows(CSIR 2012DEc pArt B)
• Wild type 200,
• Brown eyes, curled wings 150,
• brown eyes, normal wings 30,
• normal eyes, curled wings 20 ,
• total 400.
• The genetic distance (cM) between brown eye and curled wing loci is (CSIR 2012 DEC PART B)
• A) 12.5 b) 50 c) 150 d) 25
• Answer: a) 12.5
• Genetic distance
• between two genes in MAP (cM)
• =No. of recombinants/ total number of offsprings X 100
• Recombinants are brown eyes, normal wings 30,
• normal eyes, curled wings 20
• Therefore, Genetic distance = 30+20/400 X 100= 12.5 cM
• 1 cM= 1x 106 bp

• Poplar is a dioecious plant . A wild plant with 3 genes AABBCC was crossed
with a triple recessive mutant aabbcc. The F1 male hydrid AaBbCc was then
back crossed with the triple mutant and the phenotype recorded are as
follows
• AaBbCc – 300 aaBbCc- 100 aaBbcc- 16 AabbCc- 14 AaBbcc- 65
• aabbCc- 75 aabbcc- 310 Aabbcc- 120
• The distance in Map unit (mu) between A to B and B to C is (CSIR 2016 June
PART C).
• Answer:
Highest frequency parents here, aabbcc- 310, AaBbCc- 300
Total progeny =1000
Lowest frequency double cross over for example if the order of genes A, B, C in a
chromosome is F1,
A------B-------C a-----B------c gamete of f1 X a b c of parent =aaBbcc = 16
X x =
a------b-------c A-----b-------C gamete of f1 x abc of parent = AabbCc=14

• In double cross over the middle gene moves from one chromatid to
other
• Single cross over, F1
• Between A and B genes,
• A------B-------C a----B-----C gamete of f1 x abc = aaBbCc =100
• X =
• a------b--------c A----b-----c gamete of f1 x abc =Aabbcc=120
• Cross between B and C genes
A-----B-------C A------B-------c gamete of f1 x abc = AaBbcc=65
X =
a------b-------c a------b-------C gamete of f1 x abc =aabbCc=75

No cross over
A-----B------C A-----B------C gamete of f1 x abc =AaBbCc = 300
(F1 )
=
a------b-------c a------b-------c gamete of f1 x abc =aabbcc = 310.
recessive parent
The linkage distance between two genes= No. of recombinants/ total number of offsprings X 100
Crossing overs will result in recombinants therefore both single cross over and double cross resulted in
recombinants. Distance between A and B
=100+120+16+14/1000 * 100 = 25 mu
Distance between B and C =17 mu
Distance between A and C= 25 +17= 42
A--25---B----17—C = 42.

• A three point test cross was carried out in Drosophila melanogaster involving three adjacent
genes X,Y,Z arranged in the same order. The distance between X to Y is 32.5 map unit and Z to Y is
20.5 map. The coefficient of coincidence 0.886 .What is the percentage of double recombinants
in the progeny obtained from test cross? CSIR 2016 JUNE PART C.
• Answer:
• A) 6% b) 8% c) 12%d)16%
• Coefficient of coincidence= Observed double cross/expected double cross = 0.886
• Expected double cross = frequency of single cross in 1st region X frequency of single cross in 2nd
region x no of progeny=
• Frequeny of cross over= map distance /100
• Expected = 0.325 x 0.205 x1000=66.625
• COC = observed/expected= 0.886=observed/66.625
• Observed =0.886 x 66.625 = 59 out of 1000 are observed double cross, therefore fore 100,
59/1000 X100 =5.9 = 6%
• Answer = a)6%
• Coincidence + interference =1
• Interference is formation of one chiasma reduces the probability of another chiasma formation.
• When COC= 0, there is no crossover, complete interference,
• When it is between 0 to 1 indicated partial interference, if it is 1, no interference all expected
double crosses are observed.

• In a transformation experiment , donor DNA from an E. coli strain with the
genotype Z+ Y+ was used to transform strain of genotype Z-Y-. The
frequencies of transformed classes were: CSIRR 2012 Dec PART B
• Z+Y+ = 200
• Z+Y- = 400
• Z-Y+ = 400
• Total 1000
• Where is the frequency (%) with which Y locus is cotransformed with Z
locus?
• A) 1 b) 20 c) 33.3 d) 40
• Answer b)
• Cotransformation frequency = Number of transformed cells with two
genes/ total number of transformed cells x 100.
• =200/1000 x 100 =20

• DNA from a strain of bacteria with genotype a+ b+ c+ d+ e+ was
isolated and used to transform a strain of bacteria that was a- b- c- d- e
. The transformed cells were tested for the presence of donated
genes . The following genes are found to be co transformed
• i) a+ and d+ ii) b+ and e+ iii) c+ and d+ iv) c+ and e+
• The order of genes on the bacterial chromosome is (CSIR 2013 DEC
PART C)
• A) a-b-c-d-e b) a-d-c-e-b c) a-c-d-e-b d)a-d-b-e-c
• Answer b)
• Co transformed genes will present in closer proximity than non co
transformed genes

• An interrupted mating experiment was performed between Hfr Strs
a+b+c+ strains. Th genotype of majority of streptomycin resistant
(Strr) exconjurait after 10, 20 and 30 min of interrupted mating is
given below.
• 10 min a+ b- c-
• 20 min a+ b- c+
• 30 min a+ b+ c+
• The most probable gene order would be ( CSIR 2014 Dec Part B)
A) abc b) cab c) bac d) acb
Answer d) acb
•

• The ABO blood type in human is under the control of autosomal multiple alleles .
Color blindness is recessive Xlinked trait. A male with blood type A and Normal
vision marries a female who also has blood type A and normal vision . The
couple’s first child is a male who is color blind and has O blood group. What is the
probability that their next female child has normal vision and O blood group.(CSiR
2011 De Part C)
• A) ¼ b) ¾ c) 1/8 d) 1
• Answer : c) 1/8
• Male with blood type A and normal vision= AAXY, AOXY
• Female with blood type A and normal vision=AAXX, AOXx
• On mating, AAXY x AAXX, AOXY x AAXX, AOXY x AAXX, AAXY x AOXx
• = A blood group and normal male and female,
• AOXY x AOXx provides 1st child as male color blind and blood group O.
• = OOxY genotype
• Next Female = OOXX, OOXx
• = 2/16= 1/8

AX Ax OX Ox
AX AAXX AAXx AOXX AOXx
AY AAXY AAxY AOXY AOxY
OX AOXX AOXx OOXX OOXx
OY AOXY AOxY OOXY OOxY
AAXX, AOXX=Normal vision female A blood group
AAXx, AOXx=Normal vision female Carrier recessive gene A blood group
AAXY, AOXY= Normal vision male with A blood group
AAxY, AOxY= Color blind male with Ablood group
OOXX, = normalfemal vision blood group O
OOXx= nomal female carrier blood group O
OOXY = normal male with blood group O
OOxY= colorblind male with blood group O
Next Female normal blood group O = OOXX, OOXx
Out of 16 genotypes, 2 is of OOXX, OOXx .. Therefore , 2/16 = 1/8
AOXY x AOXx
Normal vision male A blood group x normal vision female A blood group

Answer: b)
If it is X linked disorder then , male parent would be affected because only one X chromosome is parent in it and
that is recessive but it is not affected. So not due to X chromosome, If it si Y inked then male would be affected
because only one Y chromosome is present in them. So not due to Y chromosome. So the genetic disorder due
to autosomes. Both the four parents must be carrier. Let the dominant allele be A and recessive allele be a,
which is the cause for genetic disorder. There fore Aa x Aa will give AA, Aa, Aa and aa, which could be either
male or female. That is out of 4, 3 carries recessive allele. The probability of parenst to carry recessive allele is ¾.
In II2, the genotype could be AA, Aa dominant or carrier, same way, II 3, could be AA, Aa. Therefore but in
question, child will show only if parents carry recessive allele. Therefore, AA is not the genptype of II 2,3 .
Therefore mating, Aa x Aa will results in AA, Aa, Aa, aa that is out of four, 1 will be affected. ¼.

Answer.
AbC is due to double cross over, A-----B-----C A-----b-------C
x x =
a-----b-----c a-----B--------c
Expected double cross over= frequency of recombinants in 1st region x frequency of recombination in 2nd region x total
no. of progeny
Frequency of recombination = total no of recombinants /total number of progeny= map distance /100
Expected cross over = 10/100 X 20/100 x1000 = 20double cross overs.
20 double cross overs will be having AbC or aBc … therefore, only AbC is present in 20/2= 10 progeny
10 out of 1000 progeny are AbC double cross overs , there % = 10/1000 x100 = 1%.

• Assuming a 1: 1 sex ratio, what is the probablility that three children from the same
parents will consist of two daughters and one son?
• A) 0.375 b) 0.125 c) 0.675 d) 0.75
• Answer: A)
• Two laws of probability,
• First law multiplicative or productive law. Used when probability of this and that . Over
all probability = probability of this X probability of that
• Second law : additive law. used when probability of either this or that
• Over all probability = probability of this + probability of that
• Therefore here, and is used for productive law,
• Probability of daughter = ½
• Probability of son= ½
• Over all probability of two daughters and one son in any of the following order GGB or
GBG or BGG = (½ x ½ x ½ ) + (½ x ½ x ½ ) + (½ x ½ x ½ ) = 1/8 + 1/8 +1/8 = 3/8 = 0.375.
• If the in question the order of children is given, for example first two daughters and last is
son then the probability could be ½ x ½ x ½ = 1/8 =0.125.. So read question carefully.

An analysis of four microsatellite markers was carried out in a family showing a genetic disorder. Th results are
summarized here
Based on the above which of the markers show linkage to the disorder? CSIR 2014 Dec Part C
a) M1 b)M2 c) M3 d) M4
Answer: b) M2
b) The microsatellite markers that cause genetic disorder in the family must be present in all affected individuals.
M2 is present in P1, 1,2, 5,6 the affected individuals.

Three somatic hybrid cell lines, designated as X,Y,Z have been scored for the
presence or absence of chromosomes 1 through 8, as well as for their ability to
produce the hypothetical gene product A,B, C and D as shown in the following table.
Which of the following option has most appropriately assigned chromosomes for each of the given genes? (CSIR 2015
Dec Part C)
a) Gene A on chromosome 5, Gene B on chromosome 3, Gene C on chromosome 8 and Gene D on chromosome 1
b) Gene A on chromosome 5, Gene B on chromosome B only
c) Gene D on chromosom 8, Gene C on chromosome 1, Gene B on chromosome 5 and Gene A on chromosome 4
d) Gene A on chromosome 5, Gene B on chromosome 3 and Gene D on chromosome
Answer D:

• What kind of aneuploid gametes will be generated if meiotic non dis
junction occurs at first division (n represents the haploid number of
chromosomes) CSIR 2011 Dec Part b
• A) only n+1 and n b) only n-1 and n c) both n+1 and n-1 d) either n+1
or n-1
• Answer c)
• Meiotic nondisjunction: Failure of two members of a chromosome pair to separate from one another
during meiosis, causing both chromosomes to go to a single daughter cell.
• Aneuploidy is the presence of an abnormal number of chromosomes in a cell. A cell with any number of
complete chromosome sets is called a euploid cell.
•

• The total variance in a phenotypic character can be split into two components-
genetic VG and environmental VE. The heritability of a phenotypi trait can be
expressed quantitatively as heritability coefficient h2 = (CSIR 2011 JUN PART C)
• a) VG - VE. B) VE/ VG c) VG /(VG + VE) d) VG /(VG - VE)
• Answer: c)
• Heritability is how much of the variation seen in a certain trait within a
population can be attributed to genetic variation, as opposed to environment. A
high heritability means that for all the variation of a certain trait in the
population, a large portion is caused by genetic differences.
•
• h2 = VG (genetic variance ) / Vp phenotypic variance
• VP= VE+ VG
• Where VE= environmental variance, VG= genotypic variance
• h2 = VG / VE+ VG

Diversity of life forms
• The following table shows the summary of characters between two taxa
based on presence (1) and absence of (0) data
• Which of the following represents Jaccard’s coefficient and simple
matching coefficient respectively? CSIR 2012 Dec PARTC
• A) 0.8 , 0.5 b) 0.6, 0.5 c) 0.8, 0.6 d) 0.5, 0.6
• Answer: d)
Taxon A
Taxon B 1 0
1 40 18
0 22 20

1= presence of characters, 0= absence of characters,
a= number of characters coded as present (1) for both
organisms A and B,
b= number of characters present only in B not in A
c= number of characters present only in A and absent in B
d= number of characters absent in both the organisms A and B
a+b+c+d= total number of characters compared.
Sj= a/ a+b+c= 40/ 40+18+22 =0.5
SSm= a+d/a+b+c+d =40+20/40+18+22+20= 0.6
Taxon A/1
Taxon B/2 1 0
1 a b
0 c d

• Ecological principles
• What will be the approximate effective population size in a panmictic
population of 240 with 200 females and 40 polygamous males? CSIR 2011
JUNE PART C
• A)160 b) 133 c) 63 d) 67
• Answer b)
• Ne = 4Nm xNf / Nm + Nf
• Where Ne = effective population size, Nm = number of males,
• Nf= number of females
• Ne= 4 x 40x200 / 240 = 133.33
• Panmixia (or panmixis) means random mating. A panmictic population is
one where all individuals are potential partners. This assumes that there
are no mating restrictions, neither genetic nor behavioural, upon
the population and that therefore all recombination is possible.

• The population size of a bird increased from 600 to 645 in one year . If
the per capita birth rate of this population is 0.125 what is its per
capita death rate ? CSIR 2014 June Part B
• A) 0.25 b) 0.15 c) 0.05 d) 0.02
• Answer: c
• dN/dt = bN-dN, where b is birth rate capita, d is death rate capita, N is
population size.
• dN/dt = difference in population size in one year = 645-600
• b= 0.125
• dN/dt = bN-dN = 45 =0.125x600 – dx 600
• = 45+d x600 =75
• =d x 600=75-45=30
• =d=30/600 = 0.05

• A population is growing logistically with a growth rate (r) of 0.15/week, in an
environment with carrying capacity of 400. What is the maximum growth rate
(No of individuals / week) that this population can achieve (CSIR 2015 DEC PARTC)
• A) 15 b) 30 c) 22.5 d) 60
• Answer a)
• Exponential growth model, dN/dt = bN-dN = rN where r = b-d, r = intrinsic rate of
increase.
• Expo growth is continuous population growth in an envt where resources are
unlimited. It is density independent growth. Whereas logistic growth is
continuous population growth in an envt where the resources are limited. S
shaped growth curve.
• Logistic growth model , dN/dt = rmax N (K-N/ K)
• On simplification,
• Maximum Growth rate logistically (r max) = -r/K x N2 + r N
• Where r= growth rate, carrying capacity (K) = maximum number of individuals
that a population can support, at rmax, N= population size = K/2
• Rmax= - 0.15/400 x 200 x 200 + 0.15 X 200 = -15 + 30 = 15

• Complete the following hypothetical life table of a species to calculate the
net reproductive rate Ro.
•
• The calculated Ro will be (CSIR 2014 Dec Part C) a) 0.75 b) 1.00 c) 0.65 d)
0.5
• Answer b)
• Age specific survivor ship (lx) = number surviving(Sx)/ number present
initially (nx)
• Age specific fertility rate (ASFR) = mx
• Net reproductive rate Ro = Ʃlxmx
Age class (x) Number
alive (nx)
Number of
dying (dx)
Age specific
survivor
ship
Age specific
fertility
Lx mx
0-1 1000 0
1-2 800 0
2-3 200 0.5
3-4 300 100 1.0
4-5 200 1.0

Age class (x) Number alive
(nx)
Number of
dying (dx)
Number of
surviving
Sx = nx -dx
Age specific
survivor ship
(lx)= Sx/nx
Age specific
fertility (mx)
Lx mx
0-1 1000 200 800 800/1000=
0.8
0 0
1-2 800 300 500 500/800 =
0.625
0 0
2-3 500 200 300 300/500=0.6 0.5 0.6 x0.5=0.3
3-4 300 100 200 200/300=
0.66
1.0 0.6 x 1.0=
0.66
4-5 200 200 0 0/200=0 1.0 0x1.0=0
Ʃlxmx =Ro = 0.96
Age specific survivor ship (lx) = number surviving(Sx)/ number present initially (nx),
Age specific fertility rate (ASFR) = mx
Net reproductive rate Ro = Ʃlxmx
Number surviving (sx) = number alive (nx) - number dying (dx)
Age class x1 alive = 1000
And x2= 800; therefore 1000- 800 = 200 is dying d1= 200; s1 =d2= 800 number of surviving in first row is number of alive
in second row,
L1= s1/n1 = 800/1000= 0.8 Answer Ro= 0.96 = 1.00. When R0 = 1; population neither increasing nor decreasing, Ro> 1
then increase in population, Ro< 1 decrease in population

• A gypsy moth egg density is 160 at time t and 200 at t+1 . What will
be its value at t+3, assuming that egg density continues to increase at
constant rate.? CSIR 2016 June PART C
• A) 250 b) 280 c) 312 d) 390
• Answer: c)
• Nt+1 =Nt x R where Nt = density at time t, R rate of increase
• Nt = 160, Nt+1 = 200, therefore R= ?
• 200 =160 x R
• R=200/160 = 1.25
• Therefore, Nt+2 =Nt+1 X R
• Nt+2 = 200 x 1.25 = 250
Nt+3 =Nt+2 X R = 250 x 1.25= 312.5

• Lotka Volterra model explains competition between species
• Logistic model, dN/dt = rmax N (K-N/ K) so, if two species present 1 and 2 then,
and interspecific competition included in the model
• dN1 /dt = r1N1 (K1-N1)/K1
• dN1 /dt = r1 N1 ((K1-N1-α12N2))/K1
• dN2 /dt = r2N2(K2-N2)/K2
• dN2 /dt = r2 N2 ((K2-N2-α21N1))/K2
• N1 , N2 = population size of species 1 and 2 respectively, K1 and K2 = carrying capacity of species 1 and 2 resp. α 12 , α 21
= interspecific competition coefficients.
• α 12 is denoted as α, α 21 is also denoted as β
• Population growth of the two species will stop when
• dN1 /dt = r1 N1 ((K1-N1-α12N2))/K1 = 0
• dN2 /dt = r2 N2 ((K2-N2-α21N1))/K2 = 0
• Therefore,
• K1-N1-α12N2 =0
• K2-N2-α21N1 = 0

When isocline for species 1 lies above that of species 2, species 1 will
eventually exclude species 2.
Species 1 wins and species 2 dies, N1= K1,
N2= 0
When isocline for species 2 lies above that of species 1, species 2 will
eventually exclude species 1.
Species 2 wins and species 1 dies, N1= 0, N2= K2

the isoclines of the two species cross one another. Here, the carrying capacity
of species 1 (K1) is higher than the carrying capacity of species 2 divided by
the competition coefficient (K2/a21), and the carrying capacity of species 2 (K2)
is higher than the carrying capacity of species 1 divided by the competition
coefficient (K1/a12). Eventually species 1 or 2 wins.
Coexistence is unstable .
Finally, in the fourth scenario we can see that the isoclines cross one another,
but in this case both species' carrying capacities are lower than the other's
carrying capacity divided by the competition coefficient. Rather than
outcompeting one another, the two species are able to coexist at this stable
equilibrium point

• For two species Aand B in competition, the carrying capacitis nd competition
coefficients are KA= 150, KB= 200, α = 1.0, β= 1.3
• According to the Lotka-Volterra model of interspecific competition, the out come
of competition will be (CSIR 2016, June PART C)
• A) species A wins b) Species B wins c) Both species reach a stable equilibrium d)
Both species reach an unstable equilibrium
• Answer b)
• Find KA/α , KB/ β
• KA/ α =150/ 1.0 = 150; KB/ β = 200/1.3 = 153. 85
• Construct curve
• Which is KB/ β is higher than KA and
• KA/ α is lesser than KB
• Answer b) Species B wins

• Three island have identical habitat characteristics . On the first island rodent
species A is present at a density 325/Km2. Second island has only species B at a
density 0f 179 /Km2. On the third island , both A and B coexist with densities 297
/Km2 and 150 /km2 respectively. Which of the following can be inferred from
this? (CSIR 2013 June Part C)
• A) The two species do not compete with each other b) the intra species
competition is more intense than inter species competition c) the inter species
competition is more intense than intra species competition d) the inter and intra
species competition are of the same intensity.
• Answer b)
• Lets find interspecies competition coefficient α12, α21
• K1-N1-α12N2 =0; 325-297- α12 *150 =0; α12 = 0.186
• K2-N2-α21N1 = 0; 179-150- α21 * 297 = 0; α21 = 0.097
• in both case interspecies competition is < than 1 indicating interspecies
competition is less than the intra species competition.
• If α12 is higher than 1 , then it indicates effect of species 2 is more competitive to
species 1 than species 1 on itself.
• If α21 is higher than 1, then it indicates effect of species 1 on species 2 growth is
more competitive than species 2 on itself.

• Lotka Volterra model explains competition between species
• Logistic model, dN/dt = rmax N (K-N/ K) so, if two species present 1 and 2 then, and
interspecific competition included in the model
• dN1 /dt = r1N1 (K1-N1)/K1
• dN1 /dt = r1 N1 ((K1-N1-α12N2))/K1
• dN2 /dt = r2N2(K2-N2)/K2
• dN2 /dt = r2 N2 ((K2-N2-α21N1))/K2
• N1 , N2 = population size of species 1 and 2 respectively, K1 and K2 = carrying capacity of
species 1 and 2 resp. α 12 , α 21 = interspecific competition coefficients.
• α 12 is denoted as α, α 21 is also denoted as β
• Population growth of the two species will stop when
• dN1 /dt = r1 N1 ((K1-N1-α12N2))/K1 = 0
• dN2 /dt = r2 N2 ((K2-N2-α21N1))/K2 = 0
• Therefore,
• K1-N1-α12N2 =0
• K2-N2-α21N1 = 0

• Species richness
• Species richness is the
number of
different species represented
in an ecological community,
landscape or region.
• Measure of Species richness:
• D= Menhinick’s index
• =S/√N
• S = number of different
species in sample
• N= total number of individuals
in sample
Species evenness
Species evenness refers to how close in
numbers each species in an environment is.
Measure of species eveness
Pielous eveness index= J’ = H’/ H’max
H’ = Shannon diversity index
= -Ʃ (p(i) x ln (p (i))
Where pi = number of species / total
number of all species
H’max = ln S
Where S= number of species
Less variation , then Higher J’ value

• Species abundance is the number of individuals per species, and
relative abundance refers to the evenness of distribution of
individuals among species in a community.
• Simpson's Diversity Index is a measure of diversity which takes into
account the number of species present, as well as the relative
abundance of each species. As species richness and evenness
increase, so diversity increases.
• Simpson Index (Ds) = Ʃ (n/N)2
• Where n = total number of organism of particular species
• N= total number of organisms of all species

• The following table shows the number of individuals
of each species found in 2 communities:
• Community Species
A B C D
C1 25 25 25 25
C2 80 05 05 10
(hint : ln values for 0.5 , 0.10, 0.25, and 0.80 are -3.0, -
2.3, -1.4, and -0.2 respectively)
The calculated Shannon diversity Index (H) values for
communities C1 and C2 respectively are (CSIR 2015,
JUNE PART C)
a) 1.4 and 0.69 b) 1.2 and 0.34 c) 2.1 and 0.43 d) 1.8
and 0.37
Answer: b)
• H’ = Shannon diversity index
• = -Ʃ (p(i) x ln (p (i))
• Where pi = number of species / total number of all
species
• H’max = ln S
• Where S= number of species
p(i)= no. of I th species / total number of species
p(A)= no. of A/ no. of A+B+C+D
P(B) = no.of B/no. of A+B+C+D
P(C) = no. of C/no. of A+B+C+D
P(D) = no. of D/ no. of A+B+C+D
Therefore for C1 and C2 Shannon diversity index =
H’= -Ʃ (p(i) x ln (p (i))
= - (p(A) x ln p(A) + p(B) x ln p(B) + p(C) x ln p(C) + p(D) x ln p(D))
Or C1,
p(A) = 25/100 =0.25 ; ln p(A)= ln 0.25 = -1.4;
here p(A) = p(B) =p(C ) = p(D)
H’ = - ( (0.25 x -1.4) + (0.25 x -1.4) ++ (0.25 x -1.4) + (0.25 x -1.4) )
= - (- 1.40) = 1.4
For C2,
p(A) = 80/100 =0.8 ; ln p(A) = ln 0.8 = -0.2;
p(B)= p(c )= 5/100 = 0.05 ; ln p(B) = ln p(C ) = ln 0.05 =-3.0;
p(D )= 10/100 =0.10 ; ln p(D) = ln (0.10) =-2.3
H’ = -((0.8 x -0.2) + (0.05 x -3.0) + (0.05 x -3.0) + (0.1 x -2.3))
= 0.69

• In a census for a lake fish, 10 individuals were marked and released. In
second sampling after few days 15 individuals were caught , of which 5
individuals were found marked . The estimated population of the fish in the
lake will be (CSIR 2012 Dec PART C)
• A) 20 b) 30 c) 25 d) 35
• Answer B) 30
• Marking and recapture method
• NΛ = M*C/R
• Where M
• NΛ = population size at the time of release of marked fish
• M= number of individuals originally captured and marked
• C= total no. of fish caught in second sampling (marked and unmarked)
• R= number of recaptured fish in second sampling (marked in first sampling)
• Given M= 10, C= 15, R= 5
• NΛ = 10 x 15/ 5 =30.

• Using molecular clock, it was estimated that two species A and B must have
diverged from their common ancestor about 9 x 106 years ago. If the rate of
divergence per base pair is estimated to be 0.0015 per million years, what is the
proportion of base pairs that differ between the two species now ? (CSIR 2011,
part C)
• A) 0.0270 b) 0.0135 c) 0.00017 d) 0.0035
• Answer a) 0.0270
• The molecular clock is a figurative term for a technique that uses the mutation
rate of biomolecules to deduce the time in prehistory when two or more life
forms diverged. It is sometimes called a gene clock or an evolutionary clock.
• D= 2 x r x t where,
• D = proportion of base pairs that differ between two sequences
• r = rate of divergence per base pair per million year
• t = time in million years since species common ancestor
• 2= represents 2 diverging lineages
• D= 2 x 0.0015 x 9 = 0.0270

• The frequencies of alleles A and a in a population at hardy Wein burg equilibrium
are 0.7 and 0.3 respectively. In a random sample of 250 individuals taken from the
population how many are expected to be heterozygous? CSIR 2011 June PART B
• A) 112 b) 81 c) 105 d) 145
• Answer c)
• Hardy Weinberg Equilibrium
• In population genetics, the Hardy–Weinberg principle, also known as the Hardy–
Weinberg equilibrium, model, theorem, or law, states that allele and genotype
frequencies in a population will remain constant from generation to generation in
the absence of other evolutionary influences. These influences include genetic
drift, mate choice, assortative mating, natural selection, sexual
selection, mutation, gene flow, meiotic drive, genetic hitchhiking, population
bottleneck, founder effect and inbreeding.
• p+q = 1 when there is one allele, where p = probability of dominant allele, q =
probability of recessive allele

• (p+q)2= p2 +q2 +2pq = 1 when there is two alleles, where 2 pq =
probability of heterozygous alleles, therefore, n = total number of
individuals in a population, 2 pq x n = number of heterozygous alleles,
P2 x n = number of homozygous dominant alleles, q2 x n= number of
homozygous recessive alleles.
• (p+q)4 =p4 +q4 +4p3q + 4pq3 + 6 p2 q2 = 1 when there are two alleleic
pairs.(4 alleles)
• Here, number of herterozygous asked, 2pq x n.
• P= 0.7, q =0.3 n= 250
• Therefore, 2pq x n= 2x 0.7 x 0.3 x 250 =105.

• The frequencies ot 2 alleles p and q for a gene locus in a population at
Hardy-Weinberg equilibrium are 0.3 and 0.7, respectively. After a few
generation of inbreeding, the heterozygote frequency was found to be
0.28. The inbreeding coefficient in this case is (CSIR 2011 Dec, Part C)
• A) 0.42 b) 0.28 c) 0.33 d) 0.67
• Answer c)
• Inbreeding is the production of offspring from the mating or breeding of
individuals or organisms that are closely related genetically.
• Inbreeding results in homozygosity, which can increase the chances of
offspring being affected by deleterious or recessive traits. (F= 1) all
progenies are homozygotes.
• Inbreeding Coefficient or fixation Index (F) = (He-Ho)/He = 1- (Ho/He)
• Where He = expected heterozygotes frequency, Ho= observed
heterozygotes frequency

• Given p= 0.3; q= 0.7 ;
• Ho = observed heterozygote frequency =0.28 = observed 2pq
• Then calculate expected heterozygote frequency He,
• Since two alleles given, We know p2 +q2 +2pq = 1
• Therefore (0.3) 2 + (0.7)2 + 2pq = 1
• 2pq (expected heterozygote frequency) = He = 1 – ((0.3) 2 + (0.7)2 )
• = 1- (0.09+ 0.49) = 1- 0.58 = 0.42
• F= (He-Ho)/ He = (0.42-0.28)/0.42 = 0.333

• The following genotypes were observed in a population.
• Genotype no.
• HH 90
• Hh 60
• hh 50
• Total 200
• Which of the following is correct frequency of H allele and what is the expected number of HH in
the given population?(CSIR Dec PartC )
• A) 0.60 and 72 b) 0.80 and 96 c) 0.50 and 32 d) 0.30 and 90
• Answer:a)
• P(A allele) = number of A allele in a population/Total number of alleles in a population,
• P(H)= (HH) x2 + (Hh)/ (HH+Hh+hh)x2
• Therefore,
• P(H) = (90x2)+60/(90+60+50)x2= 180+60/400= 240/400 = 0.6
• We know P2 x n = expected of homozygous dominant (HH)= (0.6)2 x 200 = 72
• Answer: p(H) = 0.6 and expected (HH) = 72

• In very small population, genetic variation is often lost through genetic drift. If
the population size of a mammal on an isolated island is 50, what percentage of
its genetic variation is lost every generation? (CSIR 2014 June Part B)
• A) 0.01 b) 0.5 c) 0.1 d) 0.05
• Answer a)
• Genetic variation is a term used to describe the variation in the DNA sequence in
each of our genomes.
• Genetic drift: Genetic drift or allelic drift is the change in the frequency of a gene
variant in a population due to random sampling. It is more likely to happen in
small populations.
• It reduces genetic variation
• % of genetic variation lost in every generation = 1/2N or
• Rate of decrease per generation alleles is lost or fixed.
• N is number of individuals in a population,
• Therefore % of genetic variation lost in every generation = 1/ (2 x50) =1/100
=0.01

• In a random sample of 400 individuals from a population with alleles of a trait in Hardy Weinberg
equilibrium , 36 individuals are homozygous for allele a. How many individuals in the sample are
expected to carry at least one allele A? (CSIR 2014 Dec PART C)
• A) 36 b) 168 c) 364 d) 196
• Answer c)
• Here, total individuals in population n= 400,
• Number of individuals with homozygous a (i.e) aa =36
• Question: to find number of individuals with at least one A.
• That is number of heterozygous Aa and number of homozygous AA. Or p2 x n+2pq x n
• We know, when two alleles are given,
• (p+q)2= p2 +q2 +2pq = 1 when there is two alleles, where 2 pq = probability of heterozygous
alleles, therefore, n = total number of individuals in a population, 2 pq x n = number of
heterozygous alleles, P2 x n = number of homozygous dominant alleles, q2 x n= number of
homozygous recessive alleles.
• So p2 x n +q2 x n+2pq xn = 1 x n
• P2 x n + q2 x n +2pq x n = 400
• P2 x n + 2pq x n = 400-36= 364
• Answer :number of individuals with at least one A = 364

• In a population at Hardy Weinberg equilibrium, the genotype frequencies are
f(A1A1)= 0.59; f(A1A2)= 0.16; f(A2A2)= 0.25. What are the frequencies of the two
alleles at this locus ? )(CSIR 2015, June PART C)
• A) A1= 0.59 ; A2 =0.41 b) A1= 0.75 ; A2= 0.25 c) A1= 0.67; A2= 0.33 d) A1= 0.55 ;
A2= 0.44
• Answer b)A1=0.75 ; A2= 0.25
• Given:
• f(A1A1)= p2= 0.59; f(A1A2)= 2pq= 0.16; f(A2A2)= q2= 0.25.
• Question : to find the frequencies of the two alleles separately
• f(A1) = p =?and f(A2) = q=?
• p2= 0.59, √ p2 = p, therefore, √ 0.59= 0.768 =p
• We know p +q =1
• Therefore., q= 1-p = 1-0.768 = 0.232
• The answer close to b) option
• Answer b) f(A1) = p = 0.75 and f(A2) = q= 0.25

• In several populations, each of size N= 20, if genetic drift results in a
change in the relative frequencies of alleles,
• A. What is the rate of increase per generation in the proportion of the
populations in which the allele is lost or fixed?
• B. What is the rate of decrease per generation in each allele
frequency between 0 and 1?
• The correct answer for A and B is (CSIR 2015 Dec Part C)
• A) A= 0.25 ; B= 0.125 b) A= 0.025 ; B= 0.0125 c) A= 0.0125 ; B= 0.025
d) A= 0.125 ; B= 0.25
• Answer: c) A= 0.0125 ; B= 0.025
• rate of increase per generation in the proportion of the populations
in which the allele is lost or fixed = 1/4N = 1/(4 x20)= 0.0125
• rate of decrease per generation in each allele = 1/2N= 1/(2xN)=
1/(2x20)=0.025

• In a population of effective population size Ne, with rate of neutral
mutation µ0, the frequency of heterozygous per nucleotide site at
equilibrium between mutation and genetic drift is a calculated as
• A) 2Ne µ0/ (4Ne µ0 +1) b) 4Ne µ0/ (4Ne µ0 +1) c) Ne µ0/ (4Ne µ0 +1) d)
2Ne µ0/ (4Ne µ0 -1)
• Answer: b) 4Ne µ0/ (4Ne µ0 +1)
• the frequency of heterozygous per nucleotide site at equilibrium
between mutation and genetic drift (H)= 1-F where H is
heterozygosity and F is homozygosity, F= 1/(4Ne µ +1)
• Therefore H= 1- 1/(4Ne µ +1)
• here µ = µ0 (neutral mutation)
• = (4Ne µ0 +1 -1)/(4Ne µ0 +1) = 4Ne µ0/ 4Ne µ0 +1
So option b is correct answer

• Fruit color of wild Solanum nigrum is controlled by 2 alleles of a gene (A and a). The
frequency of A. p = 0.8 and a q =0.2 . In a neighboring field a tetraploid genotype of S.
nigrum was found; which are AAAA, AAAa, AAaa, Aaaa and aaaa. Following Hardy
Weinberg principle and assuming the same allele frequency as that of diploid population.
The numbers of phenotypes calculated within a population of 1000 plants are close to
one of the following (CSIR 2016, June PARTC)
• AAAA : AAAa : AAaa : Aaaa : aaaa
• A) 409 : 409: 154:26:2 b) 420:420:140:11:2c) 409:409:144:36:2
• D) 409: 420: 144: 25:2
• Answer:
• We know when 2 alleleic pairs (4alleles) are there,
• (p+q)4 =p4 +q4 +4p3q + 4pq3 + 6 p2 q2 = 1;
• P= probability of dominant allele A and q = probability of recessive allele a
• Where p4 = f(AAAA) ; q4 = f(aaaa); 4p3q =f(AAAa); 4pq3 =f(Aaaa); 6 p2 q2 = f(AAaa)
• Given: n= 1000
• P = 0.8 and q= 0.2
• Question find no. of individuals having genotypes:
• AAAA : AAAa : AAaa : Aaaa : aaaa

• P4 x n = number of individuals with AAAA genotype = 0.8 x 0.8 x 0.8 x 0.8 x
1000 = 409.6
• q4 x n = number of individuals with aaaa genotype = 0.2 x 0.2 x 0.2 x 0.2 x
1000 =1.6 approximately 2
• 4p3q x n =number of individuals with AAAa genotype= 4 x 0.8 x 0.8 x 0.8 x
0.2 x 1000 = 409.6
• 4pq3 x n = number of individuals with Aaaa genotype = 4 x 0.8 x 0.2 x 0.2 x
0.2 x 1000 = 25.6 approximately 26
• 6 p2 q2 x n = number of individuals with AAaa genotype = 6 x 0.8 x 0.8 x 0.2
x 0.2 x 1000 =153.6 approxiamtely 154.
• Answer for AAAA: AAAa : AAaa : Aaaa : aaaa
• a) 409 : 409: 154: 26: 2

• Consider an autosomal locus with two alleles A1 and A2 at frequencies of
0.6 and 0.4respectively. Each generation, A1 mutates to A2 at a rate of µ =
1x10-5 while A2 mutates to A1 at a rate of 2x10-5 . Assume that the
population is infinitely large and no other evolutionary force is acting. The
equilibrium frequency of allele A1 is (CSIR 2016 June, Part C)
• A) 1.0 b) 0.5 c) 0.67 d)0.33
• Answer: c) 0.67
• Equilibrium frequency of A1 allele = v / u+v
• Equilibrium frequency of A2 allele = u /u+v
• Where u = forward mutation of allele A1 to A2 at a rate
v= backward mutation of allele A2 to A1 at a rate
Therefore Equilibrium frequency of A1 = 2x10-5 / (1x10-5 + 2x10-5 )
= 2 x10-5 /(10-5 (1+2)) =2/3 = 0.66
Answer : c) 0.67

• In an altruistic act, if a donor sacrifices “C” offsprin g which helps the recipient to gain B offspring
and the donor is related to the recipient by coefficient γ . Under which condition would kin
selection favor this altruistic trait? (Csir 2011 june part B)
• A) B > C b) B> γC c) γB –C =0 d) γB-C > 0
• Answer d) γB-C > 0
• Hamilton’s rule ( γ × B > C) specifies the conditions under which reproductive altruism evolves.
• Altruism: behaviour of an animal that benefits another at its own expense
• Hamilton’s rule = γ × B > C which can be written as γB-C > 0
• B is the benefit (in number of offspring equivalents) gained by the recipient of the altruism, C is
the cost (in number of offspring equivalents) suffered by the donor while undertaking the
altruistic behaviour, and γ is the genetic relatedness of the altruist to the beneficiary.
• Relatedness is the probability that a gene in the potential altruist is shared by the potential
recipient of the altruistic behaviour.

• Assume that in terms of genetic fitness , the benefit of performing an altruistic
act to relative is 500 units and the cost involved is 150 units. Following
Hamilton’s rule , the act should be performed if the relative is a (CSIR 2014 June ,
Part C).
• A) only brother b) nephew or niece c) brother or step sister d) only step sister
• Answer:
• A) only brother
Hamilton’s rule: γB-C > 0
• genetic relatedness coefficient for brother is 50%= 0.5
• (0.5 x 500) – 150 = 250 – 150 =100 > 0
• Genetic relatedness coefficient for half sister, nephew , niece is 0.25
• Whereas Step sister is not blood relative
• (0.25x 500)-150 = 125-150 = -25 which is lesser than zero
• So answer:
• A) only brother

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