1- An eg. of primary active transport is the movt. of
a- glucose into muscle
b. K into a nerve cell
c- proteins across capillaries
d- Na into an endothelial cell
c- oxygen across the alveolar wall.
Ans:b???????
a- glucose into muscle or adipose cells - facilitated diffusion.
e- oxygen across the alveolar wall - simple diffusion.
c- proteins across capillaries-filtration.
There are three basic ways for things to cross the capillary wall:
filtration
diffusion
diapedesis
Na+-K+ pump is the primary active transport. The Na+-K+ pump is an antiport, it transports K+ into the cell and Na+ out of the cell at the same time, with the expenditure of ATP.These concentration gradients are established by the active transport of both ions called the Na+/K+ ATPase,It uses the energy from the hydrolysis of ATP to actively transport 3 Na+ ions out of the cellnand for each 2 K+ ions pumped into the cell.
In resting skeletal muscle, there is a much higher concentration of calcium ions (Ca2+) in the sarcoplasmic reticulum than in the cytosol. Activation of the muscle fiber allows some of this Ca2+ to pass by facilitated diffusion into the cytosol where it triggers contraction.
After contraction, this Ca2+ is pumped back into the sarcoplasmic reticulum. This is done by a Ca2+ ATPase that uses the energy from each molecule of ATP to pump 2 Ca2+ ions by active transport.
Na+-glucose secondary active transport
2- which of the following vessels contain the highest velosity blood in the body
a- aorta
b vena cava
c vasa recta
d pulmonary vein
e pulmonry artery
Ans:venacava - largest veins.(specifically inferior venacava is the largest compared to superior venacava).
Blood pressure is highest in aorta, drops sharply in arterioles and again in capillaries, correlating with an increase in total cross-sectional area of vessels.
In arteries, blood velocity is accounted for by blood pressure; as pressure decreases due to increased cross-sectional area, so does velocity.
In capillaries, blood velocity is slowest due to greatest total cross-sectional area.
Slow capillary velocity allows more time for exchange of nutrients, gases, and waste products with tissues.
In the veins, blood velocity increases due to:
i. Compression of the veins during skeletal muscle contraction aided by valves in veins.
ii. Reduction in cross-sectional area as small venules join to form veins.
iii.Decreased blood pressure during inspiration whenever chest expands.
3- At which stage in the process of gene cloning r restriction nucleases used
a isolation of mRNA
b synthesis of double stranded cDNA
c insertion of cDNA into vector DNA
d Expression of the cloned gene by the bacteria
e introduction of recombenant DNA into a bacterial cell
Ans:e e introduction of recombenant DNA into a bacterial cell ???????
Gene cloning is producing a clone of organisms carrying an altered copy of a specific gene.
Gene cloning is usually used either to produce genetically altered organisms or to study genes
TOOLS FOR GENE CLONING include:
SCISSORS: RESTRICTION ENZYMES.(RESTRICTION ENZYMES are sequence-specific nucleases) Eg.E.coli.
GLUE: DNA LIGASE
SIEVE: GEL ELECTROPHORESIS
VEHICLE: PLASMID OR VIRAL VECTORS
The essence of cell-based DNA cloning involves four steps:
1.Construction of recombinant DNA molecules by in vitro covalent attachment (ligation) of the desired DNA fragments (target DNA) to a replicon (any sequence capable of independent DNA replication). This step is facilitated by cutting the target DNA and replicon molecules with specific restriction endonucleases before joining the different DNA fragments using the enzyme DNA ligase.
2.Transformation. The recombinant DNA molecules are transferred into host cells (often bacterial or yeast cells) in which the chosen replicon can undergo DNA replication independently of the host cell chromosome(s).
3.Selective propagation of cell clones involves two stages. Initially the transformed cells are plated out by spreading on an agar surface in order to encourage the growth of well-separated cell colonies. These are cell clones (populations of identical cells all descended from a single cell). Subsequently, individual colonies can be picked from a plate and the cells can be further expanded in liquid culture.
4.Isolation of recombinant DNA clones by harvesting expanded cell cultures and selectively isolating the recombinant DNA.
5 reduced serum potassium might produce which of the following
a reduced memb. threshold potential
b elevated membrane thrshold potential
c hyperpolarised cell membrane pot.
d hypopolarised cell memb. pot.
e Hyperactive neuromuscular reflexex
Ans:c
hyperkalemia depolarizes the memebrane.
hypokalemia hyperpolarizes the membrane.(the resting membrane potential to more negative values).
6 Which of the FA has the greatest number of double bonds
a Oleic
b Stearic
c Palmitic
d arachidonic
Ans: (d) arachidonic acid.
Common Name Carbon atoms Double bonds
Butyric acid 4 0
Palmitic Acid 16 0
Palmitoleic Acid 16 1
Stearic Acid 18 0
Oleic Acid 18 1
Linoleic Acid 18 2
Linolenic acid 18 3
Arachidic Acid 20 0
Arachidonic Acid 20 4
7 Which of the foll represents the pressure change when the diaphragm and external intercostal muscles simultaneously contract
a Alveolar vol. decreases
b Alv pressure increses
c Intra pleural decr.
d intra pleural pre increases
Ans:c.Intrapleural pressure decreases.
Diaphragm contracts to expand the thorax for inspiration. The external intercostal muscles of the thorax are also often involved in inspiratory breathing.When these muscles contract, they act to raise the ribs and elevate the sternum.The contraction of the diaphragm increased the thoracic volume by lowering the bottom border of the thorax, while the external intercostal muscles increase thoracic volume by raising the top border of the thorax.
The major muscles involved are the internal intercostal muscles and abdominal wall muscles that constrict the abdomen like the internal and external obliques and the rectus abdominis muscles. The internal intercostal muscles lower and constrict the rib cage to decrease thoracic volume, while the abdominal wall muscles cause constriction of the abdomen, forcing the abdomen to push against the diaphragm, shoving it back upward into the thorax.
At rest:atmospheric pressure:760mmHg and Intra-alveolar pressure:760mmHg.
During Inspiration: atmospheric pressure is 760mmHg(no change) but Intra-alveolar pressure is 758mmHg(reduced).
Boyles law states when the temperature is constant and a chambers volume is increased, the pressure in the chamber decreases and vice versa.
Increased volume = decreased pressure
Decreased volume = increased pressure
9 Each of the followin conditions alters the rate of product formation by an enzyne except one
a ph
b Temperature
c ionic atrength
d Isoelectric point
e substrate conc.
Ans
???????????
Factors that influence enzyme reaction rates
1.pH
2.Temperature
3.Salt concentration
4.Enzyme inhibitions
Noncompetitive inhibition
Competitive inhibition
5.Enzyme Saturation
6.Cofactors
Coenzymes
Prosthetic groups
a.The pH can stop enzyme activity by denaturating (altering) the three dimentional shape of the enzyme by breaking weak bonds such as ionic, and hydrogen.
b.Increases in temperature generally lead to increases in reaction rates. There is a limit to the increase because higher temperatures lead to a sharp decrease in reaction rates. This is due to the denaturating (alteration) of protein structure resulting from the breakdown of the weak ionic and hydrogen bonding that stabilize the three dimentional structure of the enzyme.
e.Increasing the substrate concentration increases the rate of reaction (enzyme activity). However, enzyme saturation limits reaction rates. An enzyme is saturated when the active sites of all the molecules are occupied most of the time. At the saturation point, the reaction will not speed up, no matter how much additional substrate is added.
Please let me know if they r wrong.
