Answer Key
F"Y. B.Sc. Life Sciences Paper
I
Q.P. code 754904
Exam held on 17th November 2016
Q.1. A) Fill in the blanks:
histone.
l. H3 is a
2. Major component of Gram negative cell wall is lipopolysaccharide.
3. Euchromatin is lightly stained in an Interphase nucleus.
4. Microtubules are 2! nrn in diameter.
5. Actin is a elobular protein.
@
Q.l B
a-iv
b-v
c-
ii
d-i
e-iii
Q.1 c
1. Virion- The complete infective form of a virus outside host cell. That
consists of an RNA
or DNA core with a protein coat sometimes with an extemal envelope"
2. Condensor- A condenser is an optical lens which renders a divergent beam from a point
source into a parallel or converging beam to illuminate an object under observation. It is most
commonly used in microscopes.
3. Cryopreservation- It is a process where organelles, cells, tissues, organs or any biological
construct susceptible to damage caused by unregulated chemical kinetics are preserved by
cooling at very low temperatures such as -80 degree using solid carbon di oxide or - I 96
degree centigrade
4.
M phase- It is the phase or period in the cell cycle during which cell division takes
place.
5. Minimal Media-A culture medium for microorganisms that contain the minimal
necessities for growth for the wild type of organisms. Components- Inorganic salts,
carbon source and water.
Q 1D.
1. True
True
False
False
False
2.
3.
4.
5.
Q.2
A
1. Tertlary structure of
o
o
o
o
telns
- lMark
Stnrcture-5Marks
Example(s)-1M
Definition
Forces contributing
of bonds)- 3 Marks
2. Buffers
o
o
o
Definition - 1.5 M
Examples - 1.5
Role for sustenance f life (3-a) with reasons / Explanation
B
t.
2.
o
o
Structure of
Ionic Produc
2 Marks
- 3 Marks
Surface tention
o
o
o
Definition -l
Explanation (
- I Mark
Significance 3 Marks
3. Structure and F
o
o
o
Definition -1
Structure
-2
Functions (
least 2)- 2 Marks
4. Structural Proteins
o
o
o
o
Definition: I
Structure or
Example (s)Functions / S
cs:
Mark
-
I
Mark
2 Marks
- 7 Marks
Q"3
A) Any
one
10 mks
1. Compare and Contrast between Prokaryotic and Eukaryotic cell
Ans: Any 10 points of compare and contrast -1 mark each- Yz mark on either side
point
If
diagrams drawn with labels- 5 mks n 5 points written- 5 mks
Linear IINA (in nuclcrrs)
Cir:culnr Db{A {itt *ytos<rl}
No orsanclles
?"Tuclecirl
{*ot mernhrnne hnrrnd}
$
er,erfl I me,nrbrnne bound CIrFrurclles
li{ue|eus {memhmns bounel)
fiirtgle chrom*seimc
$evernl cluomosonms
Flnsmn memhmn*
tlp icn I ly lacks re*ept+rx
Plnsma membrnne with recePtors
{stsrn ls nncl *orbtrhydrntes}
Chemieally complex eeJl wall
(may contnin peptidogtrvcan)
Chenrically silnple cell rvallt
{cellulose (plnnts) nnd +hitin {furngi))
Ialion osrlus$ mirunlttncously {in $yts$01)
DNAtransctription in nucleus- nnd
tn:uslntion iil cytesol
gellum (i.f trrre**nt)
-$imple* hnilt frctn tx.n proteinn
Flngeltrrun (if presqnt)
Cnmplex- huih frnnr misrot\ttruln*s
May huvu pitri and fimXrriae
Fv{*y hrrve ei.lia
I{*plold gsns{ne
(only one ccpy of eaelr geno)
(rnoro than onc copy of each genu)
nNA
trarrsc.tription sucl UIRNA ttans*
Ftra
Mny hnve plnsnrirtri
(DN;\ nntsicl
e chnrurosom e)
Cornpnct grluome
{tiltl* repetitive Pt'34)
Ndoy lrnr*e a
glycccdyx cevcr
Diploidgflrums
Ftrasmid
VsuallSt
not inCImmsn
luge arntunts of ttnn'cnding
*ud repetitive DNA
#lyeocnlyx only if n* cr:llnrull
l$o hishnes in ehrurnosrune
Large ribnsnrnes iu ctr'tnsn#nucleus
snlaltr ribascrtues in eitstnells*
*'$tt)ilnd" *rcunql hinf*nex
Lncks cytaskeleton
etr4eiskstretor: {acti n, ll,licrotubu}*s}
h{ycerlngirtous capsule
No nryc*lug,intus cnpxule
Small ribos$mes
Cell siz$ {anss 0"5*lSfr uttt
$.srxual rryrnrXuctitttt
(trinary fission)
DNA
eeil
siu$ sans€ l0*15CI pttt
Sexual relwcduritinn
{meio*is mel mito*is)
2. Isolation of Microorgarfisms2 marks each/ techniques mQntioned
i) Broth Cultureii)Serial dilution
iii)Streak plating
iv)Spread Plate
v) Pour plate technique
10 mks
5mks
Q.3 B
1. Principle and working Qf a Electron microscopePrinciple I mk ( ray diagr{m can be considered as supplementary- 1 mk) Working-
3/4 mks
2. Structure of a Bacterioph{ge supported by neat and labelled diagram
Description - 3mks
Diagram-2 mks
Struetur-e of,
**,*.r;il;.
iifrr
hur
/'
sw,tMsrrbcft.qom . lttt5?odd
4
The Cell
Q.4.A) Attempt any one of the following: (10 marks)
What are microtubules? Explain their arrangement and role in the working of a
flagellum.
L.
Answer:
.
Microtubules are apart of the cytoskeletal systems involved in cell motility and the
determination of cell shape. Microfubules are responsible for various cell movements.
Examples include the beating of cilia and flagella, the transport of membrane vesicles in
the cytoplasm, and, in some protists, the capture of prey by spiny extensions of the
surface membrane. These movements result from the polymerization and
depolymerization of microtubules or the actions of microtubule motor proteins. (2
marks)
Arrangement (3 marks)
' Flagella are attached to structures known as basal bodies, which in turn are anchored to
the plasma membrane. From the basal bodies the microtubule "backbone" extends,
pushing the plasma membrane out with it.
' Within the flagella, nine outer doublet microtubules surround a central pair of singlet
microtubules (Figure below).
' This characteristic "9 * 2" arrangement of microtubules is viewed in cross section
' Each doublet microtubule consists of A and B fubules: the A tubule is a complete
microtubule with 13 protofilaments, while the B tubule contains 10 protofilaments.
' The flagellar structures are held together by three sets of protein cross-links:
(a) The central pair of singlet microtubules are connected by bridges.
(b) A set of linkers, composed of the protein nexin, joins adjacent outer doublet
microtubules.
(c) Radial spokes, which radiate from the central singlets to each
doublets, form the third linkage system.
A tubule of the
outer
ttdtdlls*t,\\$}
f-,'r$
Diagrams (2 marks)
Flagellar movement: (3 marlfs)
. The force to create the bending of a flagellum is generated by the dynein arms.
. One end of each dynein is permanently bound to the A tubule of a microtubule doublet.
. Using the energy of ATP hydrolysis, the other end of the dynein reaches out to the
adjacent B tubule and walks along this doublet.
. This dynein activity causes 2 microtubule doublets that are next to each other to slide
relative to one another.
. Because this microtubule sliding is restricted, tension is created.
. This tension causes the miqrotubules to bend leading to flagellar movement.
2. Describe the structure of ( )ram positive and Gram negative bacterial cell
wall.
Answer:
Gram Positive 5 marks (2 for diagram + 3 for description)
Gram negative 5 marks (2 fo r diagram * 3 for description)
The cell wall is the outermost I igid structure of bacterial cell which maintains the form of the
cell and protects the internal ct llular constituents. It is present beneath the capsule in case of
encapsulated organisms and it s the outermost layer in non-capsulated bacteria. Based on the
presence or absence and struct nal characteristics of the cell wall, simple sub-division
of
prokaryotes is:
1.
Gram positive bacterir which synthesize a mono-layer cell wall.
2. Gram negative bacteri r which synthesize a bilayer cell wall.
3. Mvcoolasna which do not synthesize a cell wall and the cell membrane serves as the
outermost protective la fer.
Principle chemical diffe :ences between cell wall of Gram positive and Gram negative
bacteria are as follows:
COMPONENT
GRAM POSITIVE CELL
GRAM NEGATIVE CELL
WALL
l.
WALL
Inner layer
Peptidoglycan
f
2. Techoic acid
3. Polysaccharide
4. Proteins
5. Lipopolysaccharides
6. Lipoproteins
Key:*+Present;--+A
Outer layer
T
+
+
J-l
+
T
+/t
Gram positive Cell wall:
rt
+
Gram negative Cell wall:
Q.4. B) Attempt any two of the following: (5 marks each)
1.
Write
a note on: Structure of Nucleus
Answer:
@iagram 2 marks
* Description 3 marks)
Fodnucledar
chromdiln
Inl16nucfgolit
0hrofllailn
Nuclear ofivolopa
Innermornbrana
Nuclear envelope and pores: The nuclear envelope, otherwise known as nuclear membrane,
consists of two cellular membranes, an inner and an outer membrane, affanged parallel to one
another and separated by 10 to 50 nanometres (nm). The nuclear envelope completely
encloses the nucleus and separates the cell's genetic material from the surrounding cytoplasm,
serving as a barrier to prevent macromolecules from diffusing freely between the
nucleoplasm and the cytoplasm. The outer nuclear membrane is continuous with the
membrane of the rough endoplasmic reticulum (RER), and is similarly studded with
ribosomes.Nuclear pores, which provide aqueous channels through the envelope, are
composed of multiple proteins, collectively referred to as nucleoporins.
Nuclear Lamina: The nuclear lamina is composed mostly of lamin proteins. Like all proteins,
lamins are synthesized in the cytoplasm and later transported to the nucleus interior, where
they are assembled before being incorporated into the existing network of nuclear lamina.
Chromosomes: The cell
contains the majority of the cell's genetic material in the
form of multiple linear DNA
organized into structures called chromosomes.
Nucleolus: The nucleolus is a discrete densely stained structure found in the nucleus.
It is not
sunounded by a membrane,
is sometimes called a suborganelle. It forms around tandem
repeats of rDNA, DNA codi
for ribosomal RNA (rRNA). These regions are called
nucleolar organizer regions
R). The main roles of the nucleolus are to synthesize rRNA
and assemble ribosomes.
Write
on: Interme iate Filaments
Answer:
(1 mark for definition * 4 m rks for fypes)
2.
a note
eukaryotic cells from multicellular organisms,
cellular eukaryotes is controversial. In electron
ediate filaments filling the entire cytosol of a
of microfilaments but similar to that of
laments:
Keratin, respectively. produced by different
in a number
leukocytes;
of glia, an
I
I
es
'
'
Tlpe IV include Neurofilafnent H (heavy), M (medium) and L (low).
Type V are the lamins lvtrictr have a nuclear signal sequence so they can
form a
filamentous_support insidf the inner nuclear rn..btun.. Lamins are
vital to the re-
formation of the nuclear erlvelope after cell division.
3. compare and contrast betfreen Heterochromatin and Euchromatin
Answer:
(Any 5 points
- I mark for e ch point)
chromatin is found in two variBties: Euchromatin and Heterochromatin.
The concept of heterochromati was described by Emil Heiu in 192g.
The two forms
sfled cytologically by how intensely they stained
euchromatin is
w
le
indicating tighter
-
the
afr interphase nucleus, while heterochromatin stains intensely,
Heterochromatin DNA is
NOT condensed in Interphase
Euchromatin DNA is Early replicating
Latereplicating
DNA is NOT methylated
Histones from euchromatin are NOT
I
I
I
Heterochromatin is kanscriptionally
Euchromatin is transcriptionally
INACTIVE
ACTIVE
Heterochromatin DOES NOT
participate in genetic recombination
Euchromatinparticipates in genetic
Heterochromatin has a gregarious
EuchromatinDOES NOT have
gregarious instinct
recombination
instinct
4.
Write
a note on: Fungal Cell
a
wall
Answer:
@iagram 2 marks
* Description 3 marks)
Approximately 80% of the wall consists of polysaccharides. Most fungi have a fibrillar
structure built on chitin, chitosan (Zygomycotina), and B-glucans, and a variety of
heteropolysaccharides. The fibres are contained in a complex gel-like matrix. Proteins
constitute a small fraction of wall material, rarely more than 20o/o, and often as glycoprotein.
Not all proteins have a structural role. Mating, recognition, wall modification and nutrition
involve wall-bound proteins. Hydrophobins are expressed constitutively, and become bound
in the matrix of the wall as the hyphae emerge in air. Lipids are found in walls, usually in
very small concentrations. Along with hydrophobins (see below), lipids and waxes appear to
regulate movement of water, especially in the prevention of desiccation of cells. Walls also
contain a range of other minor components, including pigments and salts.
Common wall constituents found in each division of funsi
Division
Basidiomycota
Ascomycota
Zygomycota
Chvtridiomvcota
Fibrous
Gel-like Polvmer
Xylomannoproteins
o (1-3) Glucan
Chitin
p -(1-3), B-(1-6) Glucan
chitin
B -(1-3), B-(1-6) Glucan
Galactomannoproteins
o (1-3) Glucan
Polyglucuronic acid
Glucuronomannoproteins
Polyphosphate
Chitin
Chitosan
chitin
Il"
Glucan
ulucan
srr{bce coat
structural
l0
Q.5.
1. Starch and Cellulose
r
o
Similarities (2 Points)
Differences (2 Points)
- 2.5 Marks
- 2.5 Marks
ofpH
Definition- l Mark
2. Concept
o
o
o
Explanation-2Marks
Importance -2Marks
3. Structure of typical plant cell- Function of 5 cell organelles
Diagram -2 Yz marks , Functions of cell organelles- 2 Yz marks
4. Factors influencing microbial growth
- Any two
Any two of the following factors- 2lzmarks each
Solutes and Water Acidity, pH
Temperature
Oxygen Requirements
Pressure
Radiation
5.
Write
a note on: Nucleosome
Answer:
@iagram
I mark + Description
4 marks)
The nucleosome is composed of a core of eight histone proteins and the DNA is wrapped
around them. The DNA between each nucleosome is called "linker DNA". The DNA most
tightly associated with the nucleosome, called the "core DNA", is wound approximately 1.65
times around the outside of the histone octamer like thread around a spool.
hl*tone corr
-
The 147 base pair length of the DNA associated with the nucleosome is an invariant feature
of nucleosomes in all eukaryotic cells. In contrast, the length of the linker DNA between
nucleosomes is variable. Typically this distance is 20
60 bp and each eukaryote has a
-
characteristic average linker DNA length.
Histones are positively charged proteins associated with eukaryotic DNA. Eukaryotic cells
commonly contain five abundant histones: Hl, H2A, H2B,H3 and H4. Histones H2A,H2B,
H3 and H4 are the "core histones" and form the protein core around which nucleosomal DNA
11
is wrapped. Histone Hl is not part of the nucleosome core particle. Instead, it binds to the
linker DNA and is referred to as "linker histone". The four core histones are present in equal
amounts in the cell, whereas Hl is half as abundant as the other histones. This is because only
one molecule of Hl is associated with each nucleosome which contains two copies of each of
the core histones.
The assembly of a nucleosome involves the ordered association of these building blocks with
DNA. First the H3'H4 tetramer binds to DNA; then two H2A'IJ2B dimers join the H3'H4DNA complex to form the final nucleosome. The H3'H4 tetramer interact with the central 60
base pairs. Also, the N-terminal region of H3 interacts with the final 13 base pairs at each end
of the bound DNA. If we picture the nucleosome with a clock face as described above, the
H3.H4 tetramer forms the top half of the histone octamer. Importantly, histone H3'H4
tetramers occupy a key position in the nucleosome by binding the middle and both ends of
the DNA. The two HzA'HzB dimers each associate with approximately 30 bp of DNA on
either side of the central 60bp of DNA bound by H3 and H4.
6. Using a neat and labelled diagram explain the structure of a Plant Cell
Answer:
SsEerdnry
v"*1tffi
ry,"
5fi:rN\4hrln
wrl[{$1t
ftln\dy
edn
Middl€
iamella
Mid<iln la*w{ia
t2
Wall