Vitamin

Q. What is the pro-Vitamin for Vitamin A?

A. Beta carotene present in plants.

Q. What is the major function of Vitamin A?

A. In vision, as the Wald's visual cycle.

Q. What causes the nerve impulse in retina?

A. Photo-isomerisation of 11-cis retinal to all trans- retinal.

Q. How is all trans-retinal regenerated?  

A. Trans-retinal is taken to liver, where it is made to trans-retinol, then isomerised to cis-retinol and then to cis retinal.

Q. What are the enzymes required for this regeneration?                                                           

A. Alcohol dehydrogenase and retinol isomerase.

Q. What are other minor biological roles of Vitamin A?                                                               

A. Normal  maintenance of epithelium  and skin, glycoprotein   synthesis, reproduction,   cell differentiation, and anti-oxidant role.

Q. What are the sources of Vitamin A?    

A. Carrot, mangoes, papaya, green leafy vegetables, and fish liver oil.

Q. What  are the  deficiency  manifestations of Vitamin A?                                               

A. Night  blindness, xerophthalmia,  keratomalacia, and keratinisation of epithelium.

Q. Nyctalopia is due to the deficiency  of  which Vitamin?                                               

A. Vitamin A.

Q. What is the daily requirement of Vitamin A?

A. 750 to 1000 microgram.

Q. How cholecalciferol is synthesised?   

A. From 7-dehydro cholesterol in the malpighian layer of epidermis, by the action of ultra violet rays.

Q. How Vitamin D is activated?                

A. Cholecalciferol from skin reaches liver. There it is hydroxylated to form 25-hydroxy cholecalciferol (25-HCC). It then reaches kidney, where further hydroxylation takes place to form 1,25-dihydroxy cholecalciferol (DHCC).

Q. What is calcitriol?                                      

A. 1,25-dihydroxy cholecalciferol, or active Vitamin  D, contains three hydroxyl  groups at 1, 3 and 25 positions. So it is called calcitriol.

Q. Which Vitamin acts as a pro-hormone?

A. Vitamin D.

Q. What is the function of Vitamin D?    

A. It increases absorption of calcium from intestine, and it also increases mineralisation of bone.

Q. How Vitamin D increases absorption of calcium?

A. Calcitriol  binds to a cytoplasmic receptor. The hormone-receptor complex interacts with  DNA  and causes  transcription  of specific genes that code  for  calbindin.  Due  to  the  increased availability of  calcium  binding  protein,  the absorption of calcium is increased.

Q. How is Vitamin D deficiency manifested?

A. Rickets in children and osteomalacia in adults.

Q. In  renal disease, oral doses of Vitamin D may not be effective, why?                            

A. Hydroxylation and activation of Vitamin is taking place in kidney.

Q. What are the causes of rickets?             

A. Chronic renal failure, liver  diseases, and under- exposure to sunlight.

Q. What is the daily requirement of Vitamin D?

A. 5-10 microgram.

Q. What is the chemical nature of Vitamin E?

A. Alpha tocopherol.

Q. What is the function of Vitamin E ?     

A. Anti-oxidant.

Q. What is the relationship of selenium with Vita- min E?                                                        

A. They act synergistically as anti-oxidants.

Q. What is the source of Vitamin E?        

A. Vegetable oils are rich sources of Vitamin  E, e.g. wheat germ oil, sunflower oil, safflower oil, cotton seed oil, and palm oil.

Q. What is the normal daily requirement of Vitamin E?                                                               

A. 15 mg or 33 international units.

Q. What is the chemical nature of Vitamin K?

A. Naphthoquinone  derivative.

Q. What is menadione?                               

A. It is synthetic water soluble Vitamin  K, widely  used in clinical practice.

Q. What is the function of Vitamin K?    

A. Gamma carboxylation of clotting factors such as prothrombin.

Q. Deficiency of  Vitamin K  can occur in  which conditions?                                               

A. Obstructive  jaundice, antibiotic  therapy,  and administration of dicoumarol.

Q. What is the mechanism of action of dicoumarol?

A. It  competitively  inhibits  Vitamin  K  epoxide reductase.

Q. So, dicoumarol  is used for what purpose?

A. To prevent intravascular thrombosis.

Q. Excess dicoumarol will produce what?

A. Bleeding tendency.

Q. Which substance will inhibit Vitamin K?

A. Dicoumarol.

Q. Excess dose of Vitamin K in neonates may lead to which condition?

A. Hemolysis and jaundice.                        

Q. Bleeding tendency is common in the deficiency  of which?                                                  

A. Vitamin  K, Vitamin  C, and platelets.

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Difference between Primary Cell Wall and Secondary Cell Wall

 Difference between Primary Cell Wall and Secondary Cell Wall

Primary and Secondary Cell wall

Cell wall was first discovered by Robert Hooke in 1665.  In cells of bacteria, cyanobacteria, protists, fungi and plants, a thick, rigid, protective cell wall is present outside the plasma membrane. One of the most important differences between plant and animal cells is the presence of cell wall in plant cells.

Composition of cell wall: The composition of cell wall varies in different groups.

Bacteria: Peptidoglycan cell wall: Cell wall is composed of protein-lipid-polysaccharides having two important chemical components: N -acetyl glucosamine (NAG) and N-acetyl muramic acid (NAM). 

 Bacterial Cell wall & Gram Staining

Bacteria can be divided into two large groups on the basis of differential staining technique called Gram Stain developed by Christian Gram.

      Gram Staining

        Procedure

How Gram Stain Works? Gram Staining Principle: Step by step procedure with explanation

Gram Positive Bacteria

• Split into a low and high GC group which reflects the GC content of their DNA.
•   ex: Staphylococcus, Actinobacteria, Streptococcus, Lactobacillus, Cornyebacterium, Listeria ,Bacillus, Nocardia, Clostridium

Gram negative bacteria

•   Proteobacteria is the largest and diverse group of bacteria including many of the more commonly known gram-negative bacteria
•  Ex: Escherichia coli (E.coli),Helicobcater, Salmonella,Cyanobacteria, Klebsiella,Hemophilus, Neisseria, Pseudomona

A comparison of cell walls of gram positive and gram negative bacteria

Algae: Cell wall contains cellulose and a variety of glycoproteins. 

 Cell wall

In plant cells, apart from the plasma membrane, there is a hard protective covering which is called the cell wall.

Plant cell wall

Why do plants have cell wall? What would happen if we possess cell wall?

The advantage of having a cell wall is, it is primarily made up of cellulose, a tough substance that provides great protection from external injuries. Like two sides of a coin, this protection is delivered at the expense of free movement. We can move in search of food as we don’t have cell wall. Plants cannot move and have to find their food from the rooted site itself. Definitely, their capability of photosynthesis clearly overcomes this disadvantage.

In plant cells, cell wall is made up of cellulose, hemicelluloses, pectin and proteins.

Remember fungal cell wall is made up of chitin (polymer of N-acetyl glucosamine) and bacterium has a peptidoglyacan cell wall.

Cell wall is a stiff protective layer outside plasma membrane and provides mechanical support and also determines the shape of the cell.

Plant cell wall, what is it made up of?

Cell wall is composed of variety of polysaccharides. Cellulose microfibril is the basic building unit of primary wall surrounded by a matrix of hemicelluloses and pectins.  Secondary wall consists of cellulose microfibrils surrounded by lignin. The orientation of microfibrils is governed by cytoskeleton.

Cellulose is a linear, unbranched polymer, consisting of straight polysaccharide chains made of glucose units linked by 1-4 β glycosidic bond.

Structure of cell wall

What are microfibrils?

Each microfibril is a ribbon like flat fibre of 25-30 nm in diameter. Each microfibril cosists of three elementary fibrils (micelles). Each elementary fibril is made up of 100 cellulose units. Each cellulose molecule consists of 40-70 glucan chains.

Thus 1 microfibril=3x100x70=~21000 glucan chains. Microfibrils have great tensile strength similar to that of rubber.

In short 

Primary wall=cellulose microfibrils+ hemicelluloses+pectin

Middle lamella=pectin

Secondary wall=cellulose microfibrils+ lignin

What are Hemicelluloses?

Are short branched heteropolymers made up of various kinds of monosaccharides like xylose, mannose, glucose etc.

Eg: glucomannans, xyloglucans, xylans.

What are Pectins?

A water soluble branched polysaccharide made up of negatively charged D-galacturonic acid and D-glucoronic acid residues.

Other depositions include cutin (made of fatty acids), Suberin (water resistant coat comprising of fatty acids found in many plant cell walls and cork. Mineral deposits are present in the cell wall of families Cruciferae and Cucurbitaceae (Calcium deposits), whereas silicate deposits are common in the cell wall of Graminae, grass family.

Cell wall synthesis?

Primary cell wall is the first formed boundary of the cell that is formed when cell is increasing in size. It is thin and permeable but cutin deposition makes it impermeable at some sites. Cellulose is synthesised from uridine diphosphoglucose (UDP-glucose) by an enzyme complex, cellulose synthase. Matrix materials, hemicelluloses and pectin are synthesised in the golgi apparatus and transported to the plasma membranes as vesicles where it is discharged and deposited.

Why wood is so hard?

Secondary wall is thick and permeable and consists of cellulose microfibrils and lignin. Even the master digesters, fungus cannot degrade lignin that offers mechanical protection from fungal pathogens. Lignin deposition is very common in wood.

 The formation of the secondary  wall is not uniform in all the cells. The differentiation of various types off cells like parenchyma, collenchyma, sclerenchyma, fibres and tracheids are due to the secondary wall formation

Tertiary wall: In some tissues a tertiary cell wall is formed on the inner surface of the seconary cell wall. This layer is very thin and is found in the xylem tracheids of gymnosperms. It is composed of mainly of xylan, instead of cellulose.

Structure of lignocellulose:The main component of lignocellulose is cellulose, a beta(1–4)-linked chain of glucose molecules. Hydrogen bonds between different layers of the polysaccharides contribute to the resistance of crystalline cellulose to degradation. Hemicellulose, is composed of various 5- and 6-carbon sugars such as arabinose, galactose, glucose, mannose and xylose. Lignin is composed of three major phenolic components, namely p-coumaryl alcohol (H), coniferyl alcohol (G) and sinapyl alcohol (S). Lignin is synthesized by polymerization of these components and their ratio within the polymer varies between different plants, wood tissues and cell wall layers. Cellulose, hemicellulose and lignin form structures called microfibrils, which are organized into macrofibrils that mediate structural stability in the plant cell wall.

How adjacent cells are joined?

Middle lamella: It is thin amorphous layer of cemeting material present between the adjacent cells. It is chiefly made up of calcium pectate. It is formed at the time of cytokinesis by the joint activity of the cells between it lies. 

How plant cells communicate with each other?

Plasmodesmata

Plamodesmata and desmotubule

The cell wall is perforated by narrow pores or pits at many places.Through these narrow pores, fine strands of cytoplasm are extending. These strands are called plasmodesma connections or plasmodesmata. These connect the cytoplasm of adjacent cells and facilitate the movement of substance between cells.

 Plasmodesmata are narrow channels through the wall bound by plasmalemmma containing cytoplasm and often a desmotubule. The desmotubule is the cental core and is composed of protein subunits consists of modified membraneous structure  continuous with the endoplasmic reticulam of the adjoining cells.These membraneous structures probably originate from the ER. Desmotubule acts as valve to control the direction of flow of materials.The functions of plasmodesmata may vary. It is concerned with the transport of materials from one cell to another and also conduction of stimuli. In the case of certain parasites like Viscum, Loranthus, Cuscuta etc plasmodesmata connections exist between the haustoria and the cells of their host. Through these channels food and virus are transported.

Functions of the cell wall

• Cell walls primary function is mechanical support. It acts like a skeletal framework of the plants.
• Cell wall is tough and has high tensile strength. Still plant cell is fully permeable to water and solutes. Plant cell wall has minute water filled channels through which water,hormones and gases passes to and fro.
• Cell wall shows plasticity and elasticity during cell growth.
• It helps to maintain the balance of intracellular  osmotic pressure with that of its surrondings
• Lignification of secondary walls greatly enhances compressive strength permitting woody structures to reach the sky. Cell wall upon lignification becomes dead as it becomes impermeable and thus protoplasm has no access to take up solutes that is why lignified tissue is always dead.
• Lignin provided extra mechanical strength and also provides a water resistant channel for transport of solutes.

Additional Points

•   Algal cell wall:

 Algae typically possess walls constructed of glycoproteins and polysaccharides, however certain algal species may have a cell wall composed of silicic acid. Alginic acid is common polysaccharide in the cell walls of brown algae. Manosyl form microfibrils in the cell walls of a number of marine green algae, such as Acetabularia as well as as in the walls of the some red algae, like Bangia and Porphyra.Sulphonated polysaccharides occur in the cell walls of most algae, those common in red algae include agarose, carrageenan, porphyran, furcelleran and funoran.The group of  algae known as the diatoms synthesise their cell walls from silicic acid ( specially orthosilicic acid, H₂SiO₄).

Fungi: Cell wall is made up of chitin (Polymer of N-acetyl glucosamine)

Plants: Cell wall is made up of cellulose (β-D-glucose units), hemicellulose (Arabinose, mannose, xylose, galactose, etc.) and pectin (Galactose, galacturonic acid and arabinose).

A typical cell is made up of four layers: middle lamella, primary, secondary and tertiary wall. Middle lamella is the cementing layer between the cells. It is made up of Ca and Mg pectates. 

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biotechnology

1.   Highest capacity vector is

a) Cosmid

b) YAC

c) Yeast integrative vector

d) Bacteriophage vector

Ans: b

• Cosmid: A plasmid with a cos site of lambda phage.Insert size: 30-45 kb
• YAC: yeast artificial chromosome. Insert size: 1 Mb
• Bacteriophage vectors: refers to lambda and M13 phage vectors
• Lambda phage vectors: Insert size: 8-23 kb
• M13 vectors used for obtaining single stranded copies of cloned DNA that are suited for DNA sequencing
• BAC: Bacterial Artificial Chromosome. Insert size: 300 kb

2.   The C value denotes the total number of DNA in a

        a) Diploid

        b) Haploid

        c) Aneuploid

        d) Polyploid

      Ans:b

• C value: Haploid DNA content of an organism or the amount of DNA in haploid nuclei like a gamete.
• C value paradox: refers to the phenomenon that huge genomic content has nothing to do with the complexity of an organism.For eg: Protists has genome size much larger than humans.

3. Cdc mutants are useful for the study of 

a) Chromosome breakpoint

b) Apoptopsis

c) Various stages of Cell cycle

d) Homeodomain 

 Ans: c  cdc refers to cell division cycle

4.  RNA is very much susceptible to hydrolysis in alkali because

a) It contains Uracil residues in its structure

b) Its 2’ OH groove participate in intramolecular cleavage of phosphodiester backbone

c) Cleavage occurs in the glycosilic bonds of purine bases

d)Cleavage occurs in the glycosilic bonds of pyrimidine bases

Ans:b

• Because of the presence of 2’ OH group in the ribose sugar
• DNA is more stable than RNA because of the presence of H at 2’ position in DNA rather than OH in RNA.

5. Which one of the following is not a requirement of a PCR reaction?

a) DNA template

b) NTPs

c) MgCl2

d) Taq polymerase

Ans:d

Here the best option is MgCl2 even though it is required for DNA polymerase activity. (It can be avoided)

6. The heterozygosity of any locus can be ascertained by

 a) SNPs

b) RFLPs

c) FISH analysis

d) Either RFLP analysis or SNP

Ans: d

• SNP: single nucleotide polymorphism refers to the variation in the lengths of some DNA btw individuals due to single base changes

          Application: used in DNA finger printing

• RFLP: restriction fragment length polymorphism: refers to the variation in the restriction fragment length between individuals of a species.
•  FISH: florescence insitu hybridization: hybridizing fluorescently labelled DNA probes on to human chromosome preparations allows genes to be mapped directly to their chromosomal locations.

7.  Hormone pairs requires for a callus to differentiate are

 a) Auxin and Cytokinin

b) Auxin and Gibberellin

c) Cytokinin and Gibberellin

d) Ethylene and Gibberllin 

Ans: a

Auxin induces rooting whereas cytokinins promote shooting if supplied in optimum ratio.

8. Embryo rescue is a useful technique to 

 a) Grow /generate hybrids between different plant species

 b) Complete the growth of embryos susceptible to defects in seed development

 c) Break the dormancy of seeds

 d) All of the above

Ans:d 

9. Antibody diversity is generated by           

a) protein splicing

b) somatic mutations

c) allelic exclusion

d) interchromosomal recombination          

   Ans: b

• Somatic mutation: a mechanism by which point mutations are introduced into rearranged immunoglobulin variable region genes during activation and proliferation of B cells. It contributes significantly to antibody diversity.
• Allelic exclusion: a process that permits the expression of only one of the allelic forms of a gene. It contributes to specificity of IgGs

10. The precursor for penicillin-G biosynthesis during fermentation process is            

a)Phenylacetic acid

b)Acetic acid

c)Phenoxy acetic acid

d)None of the above

Ans:c 

11.Plastome is                

a)Plasmalemma protein

b)A type of plasmid

c)An organellar genome

d)None of the above

Ans: plastome: Genetic material of plastid like Chloroplast

12. Which of the following process require energy

a) ligation

b) restriction digestion

c) hybridization

d) transformation

Ans:a

Ligation: sealing of single stranded nicks or breaks by ligase enzyme. The process requires ATP and NADP⁺.

13) Enhanced axillary branching for multiple shoot production is promoted by               

a)2,4-D

b)Abscisic acid

c)BA (Benzyl adenine)

d)Gibberellic acid

Ans:a

2,4-D is an auxin and auxin is responsible for sustaining apical dominance there by promoting axillary branching

14.  Viral replication within cells is inhibited by             

a)IL-4

b)IL-1

c)IFN alpha

d)TNF alpha

Ans:c 

Interferons: are antiviral agents (proteins) secreted by virus infected cells and induces a virus resistant state to the surrounding cells by inhibiting its replication.

15.  In the course of cell cycle, the level of the protein cyclin abruptly falls during             

a) G1phase

b) G2 phase

c) S phase

d) M phase

 Ans:d                                    

16. Protein binding regions of DNA are identified by one of the following techniques

a) Finger printing

b) Foot printing

c) Southern blotting

d) Western blotting

  Ans: (b) Foot Printing

• Foot Printing: The identification of a protein-binding site on a DNA molecule by determining which phosphodiester bonds are protected from cleavage by DNase I.
• Southern Blotting: The technique used for identifying specific DNA fragment using labelled probe.

17.Si RNA (s) interfere at

a) Transcriptional level

b) Post -transcriptional level

c) DNA replication level

d)Translational level

Ans: (b) Post -transcriptional level

18. Plant secondary metabolites

a) Help to increase the growth rate of plant

b) Help in plant reproduction processes

c) Provide defence mechanisms against microbial attack

d) Make the plant susceptible to unfavourable conditions

Ans: (c) Provide defence mechanisms against microbial attack

19. Mobile genetic elements present in human genome are

P) LINES

Q) SINES

R) P elements

S) IS elements

(a) Q, R (b) P, Q  (c) P, R (d) Q, S

Ans: (b) P, Q

20. Meristems escape virus invasion because

a)Vascular system is absent in the meristem

b) of low metabolic activity in the meristem

c) of low endogenous auxin level

d) the virus inactivating system has low activity in the meristem

Ans: b) of low metabolic activity in the meristem

21. The enzyme that can be used in 5’end labeling of DNA are

P) Alkaline phosphatase

Q) DNA ligase

R) Terminal transferase

S) Polynucleotide kinase

(a) P, S (b) R, Q (c) P, R (d) R, S

Ans: (a) P, S

22. Identify the natural plant growth regulators from the following list

 P) Zeatin

Q) Benzylamino purine (BAP)

R) Indole Acetic Acid (IAA)

S)  2,4-Dichlorophenoxyacetic acid (2,4-D)

(a)P, Q (b) Q, S (c) P, R(d) R,S

Ans: c) P, R

23. Cells of meristemoid are best described as

a) differentiated and non dividing

b) Dedifferentiated and dividing

c) differentiated and dividing

d) dedifferentiated and non dividing

Ans: c) differentiated and dividing

24. Ultra filtration process cannot be used for

a)Fractionation of Proteins

b)Desalting

c) Harvesting of cells

d)Selective removal of solvents

Ans: c) Harvesting of cells

25. The number of replicons in a typical mammalian cell is

a) 40-200

b) 400

c) 1000-2000

d) 50000-100000

Ans: a) 40-200

26. What product will result from complete hydrolysis of soluble dextran

a) Sucrose only

b) Fructose only

c) Glucose &Fructose only

d) Glucose only

Ans: d) Glucose only

27. The mobility of DNA in agarose gel electrophoresis is solely based on its

a) Charge

b) Conformation

c) Size

d) None of these

Ans: c) Size

28. Which of the following fluorescent probes is used to monitor the progress of amplification of Real Time PCR?

a) SYBR Green

b) FITC

c) Cyan Blue

d) Rhodamine

Ans: b) FITC

29. Expression of which of the following reporter genes does not require addition of specific substrate for detection

a) Luciferase

b) β- Glucuronidase

c) β-Glucosidase

d) Green Fluorescent Protein (GFP)

        Ans: a) Luciferase

                                                                                 

30. Zinc fingers are characteristic of

(a) Blood clotting Proteins

(b) RNA Chaperones

(c) DNA binding proteins

(d) Lysosomal hydrolyses

Ans: DNA binding proteins

• Chaperons: proteins that assist in proper folding of proteins.

31. Multiplication of genetically identical copies of a cultivar by asexual reproduction is known as

(a) aclonal propagation

(b) clonal propagaation

(c) polyclonal propagation

(d) vegetative propagation

Ans: clonal propagaation

32. Parthenogenetic embryos in plant are those are formed by

(a) unfertilized eggs

(b) fertilized eggs

(c) male gametophyte

(d) sporophytic cells

Ans: unfertilized eggs

• Parthenogenesis: The development of an individual from an egg without fertilization.

33. Which one of the following is the growth factor used for growth of tissues and organs in plant tissue culture?

(a) Cytokinin

(b) Cysteine

(c) Cytidylate

(d) Cyclic AMP

Ans: Cytokinin

• Cysteine :Sulphur containing amino acid
• Cytidylate: an enzyme with kinase activity
• Cyclic AMP: second messenger in intracellular signal transduction

34. Which of the following techniques is best suited for immobilizing an affinity ligand?

(a) Physical adsorption

(b) Gel entrapment

(c) Cross-linking with a polymer

(d) Covalent linkage to a spacer arm

Ans: Covalent linkage to a spacer arm                                                   

 35.  To be a cloning vector, a plasmid does NOT require

a) an origin of replication

b) a restriction site

c) an antibiotic resistance marker

d) to have a high copy number

Ans: d) to have a high copy number

Properties of a Cloning Vector

• Should be able to replicate autonomously
• Easy to transform
• Should have suitable marker
• Unique restriction sites
• For expression vector, control elements like promoter, operator etc should be there.

36. Enzyme papain is used with success to

a) increase in meat production

b) leaven bread

c) ripen papaya fruit

d) tenderize meat

Ans: d) tenderize meat

• Source: papaya fruit

37. For protoplast fusion to be successful in plant cells

a) fusion agents other than polyethylene glycol should be used

b) cellwall of the two strains of cells should be compatible

c) DNA between the two cells should be compatible

d) osmolarity of the medium is not important

Ans: DNA between the two cells should be compatible

38. In animal cell culture, the addition of serum to media is essential for providing

a) amino acids for protein synthesis

b) nucleotides for DNA synthesis

c) growth factors

d) All of the above

Ans: All of the above

39. Which one of the following reactions is used for the purpose of recycling enzymes in bioprocesses?

a) isomerisation

b) immobilisation

c) phosphorylation

d) polymerisation

Ans: Immobilisation

• Technique used for the fixation of enzymes or cells on to solid supports
• Advantage: reuse of enzyme for many reactions
• Methods of enzyme immobilization:-adsorption, covalent bonding, entrapment, membrane confinement

------------------------------------------ Best Wishes: Dr.Ehab Aboueladab, Tel:01007834123 Email:ehab10f@gmail.com,ehababoueladab@yahoo.com ------------------------------------------