Explore topic-wise InterviewSolutions in Current Affairs.

Alcoholic fermentation and Lactic acid fermentation Ethyl alcohol and lactic acid.

Yes. As a result of respiration, the energy is released in the form of ATP. This energy is used for the active absorption of salt.

ATP is utilized. There is no production of ATP during this process. It does not produce any beneficial product.

1. Peter Mitchel 

2. Fg-F, particle 

3. Proton gradient leads to ATP production

• 6C – Citric acid, 5C- Alpha keto glutaric acid, 
• 4C- Succinic acid/succinyl CoAl malic acid/ OAA, 
• 2C- Acetyl CoA.

1. RuBP carboxylase, PEPcase, ATPase, 

2. Pyruvate dehydrogenase, ATPase, cytochrome oxidase, Hexokinase, Lactate dehydrogenase 

3. ATPase

1. ATP synthesis 

2. Flow of proton 

3. In the presence C₂ , oxidation takes place and ADP combines with inorganic phosphate to form ATP 

4. Crystal

1. Amount of CO₂ released during respiration / Amount of O₂ absorbed during respiration, 

2. Amount of CO₂ released = Amount O₂ absorbed

1. Athletes felt muscular pain is due to the inadequacy of O₂. In the absence of O₂ partial oxidation takes place. It involves the formation of pyruvic acid followed by lactic acid. 

2. Yeast & Lactobacillus. 

3. After the formation of pyruvic acid, 2 NADPH molecules are utilised for the formation of Lactic acid. So the net gain of ATP in Anaerobic respiration is 2 ATPs.

1. Amount of CO₂ released/Amount of O₂ absorbed 

2. In glucose amount of CO₂ released = amount of O₂ absorbed 

3. Organic acids

1. Yes 

2. The reason are:

• Synthesis of ATP takes place in mitochondria, 
• Inner membrance acts as electron transport chain. 
• Presence of ATP – synthetase enzyme. 
• Enzymes for aerobic respiration in mitochondrial matrix

1. Hans Krebs, citric acid(Tricarboxylic acid) 

2. oaa+acetyl CoA citric acid since it has three -COOH group 

3. Cytoplasm 

4. 5^th step (Succinic acid to malic acid)

The energy-releasing process by oxidation of organic food materials in the living cell is respiration. During this process, the energy contained in the food is released and is trapped in the ATP molecules. NADH⁺H⁺ and FADH₂ formed during various steps of respiration are oxidised, and protons (H⁺) and electrons (e ) are released.

These electrons are transported to the oxygen through a series of electron carries in the electron transport system (ETS), and their energy is stored in ATP molecules. So, respiration is process of oxidative production of ATP. 45% of energy released during the oxidation of 1 glucose molecule is stored in 38 ATP molecules.

1. Melvin Calvin 

• Carboxylation 
• Reduction 
• Regeneration of RUBP 

2. Stroma 

3. 12 NADPH₂ and 18 ATP molecules.

Certain insectivorous plants, like bladderwort and venus fly trap, are partially heterotrophic.

1. Cytoplasm 

2. 3 Carbon (Pyruvic acid) 

3. Fructose 1, 6 biphosphate

1. The regeneration of NAD fails to produce ATP as the electrons are not shifted to oxygen 

2. The end product of anaerobic respiration can be further oxidized to release energy.

The name of reaction is Glycolysis.

If energy will be released at one go then most of it will be most in the form of heat. Cells should be in a position to utilize all the energy to synthesize something. To facilitate proper usage of energy, it is released in a stepwise manner during respiration.

Anaerobic respiration also called fermentation involves the production of energy from food nutrients in the absence of oxygen.

Lactic acid and Ethyl alcohol.

(b) EMP/glycolysis

It is possible to make calculations of the net gain of ATP for every glucose molecule oxidised, but in reality this can remain only a theoretical exercise. 

These calculations can be made only on certain assumptions that: 

1. There is a acquential, orderly pathway functioning, with one substrate forming the next and with glycolysis, TCA cycle and ETS pathway following one after another. 

2. The NADH synthesised in glycolysis is transferred into the mitochondria and undergoes oxidative phosphorylation. 

3. None of the intermediates in the pathway are utilised to synthesise any other compound.

4. Only glucose is being respired – no other alternative substrates are entering in the pathway at any of the intermediary stages.

1. There is incomplete oxidation of glucose during anaerobic respiration, while there is complete oxidation during aerobic respiration. 

2. The pyruvic acid formed during glycolysis is first converted to Acetyl coenzyme A, which undergoes citric acid cycle to produce critic acid. At the end of citric acid cycle NADH⁺H⁺ is released. 

3. In both (b) and (c) glycolysis is the first step cellular respiration. The product of glycolysis is further utilized by either fermentation or critic acid cycle.

(a) 2 PGAL are formed in Glycolysis and none in kerbs cycle

Glycolysis is significant for energy production in RBC

(a) death of cells

(b) fucoxanthin

Mannitol is the reserve food material of Brown algae.

(A) Both A and R are correct and R is the correct explanation of A 

ΔG = ΔH - TΔS

∵ Entropy of liquid is more than solid

∴ on melting the entropy increases and ΔG becomes more negative and hence it becomes easier to reduce metal

Correct option is (A) Both Statement I and Statement II are correct.

O₂, Cu²⁺ and Fe³⁺ are paramagnetic,

∴ Weakly attracted by magnetic field.

NaCl and H2O are diamagnetic, 

∴ Weakly repelled by magnetic field.

Mole of toluene = \(\frac 5{92}\) mole

Mole of benzaldehyde = \(\frac 5{92} \times\frac{92}{100}\)

Weight of benzaldehyde = \(\frac 5{92} \times\frac{92}{100} \times 106 = 530 \times 10^{-2} \, gram\)

x =  530 

(4) MnO₂

It is fact

Fehling solution is a complex of Cu⁺⁺ 

Cu⁺⁺ = 3d⁹

No. of unpaired e^- = 1

M.M \(= \sqrt{1 (1 + 2)} = \sqrt 3 = 1.73\) BM

Answer: a) IgA

Immunoglobulin A (IgA) is an antibody that plays a crucial role in the immune function of mucous membranes.
Colostrum is a milky fluid that comes from the breasts of humans, cows, and other mammals the first few days after giving birth before true milk appears.
Immunoglobulin abundant in colostrum is IgA in the form of secretory IgA or sIgA. Much of these are produced by plasma cells in the mammary tissue.

L.H.S.  = X + X’Y

= X.1 + X’Y    (X . 1 = X property of 0 and 1) 

= X(1 + Y) + X’Y   (1 + Y = 1 property of 0 and 1) 

= X + XY + X’Y

= X + Y(X + X’)

= X + Y.1   (X + X’ =1 complementarity law)

= X + Y   (Y . 1 = Y property of 0 and 1)

= R.H.S. Hence proved

The desired Canonical Sum-of-Product form is as following;

F = ∑(2, 4, 7) = x’yz’ + xy’z’ + xyz

(X + Y)’= X’.Y’

Now to prove DeMorgan’s first theorem, we will use complementarity laws. Let us assume that P = x + Y where, P, X, Y are logical variables. Then, according to complementation law

P + P’ =1 and P . P’= 0

That means, if P, X, Y are Boolean variables hen this complementarity law must hold for variables P. In other words, if P i.e., if (X + Y)’= X’.Y’then

(X + Y) + (XY)’must be equal to 1. (as X + X’= 1)

(X + Y) . (XY)’must be equal to 0. (as X . X’= 0)

Let us prove the first part, i.e.,

(X + Y) + (XY)’ = 1

(X + Y) + (XY)’= ((X + Y) +X’).((X + Y) +Y’) (ref. X + YZ = (X + Y)(X + Z))

= (X + X’+ Y).(X + Y +Y’)

= (1 + Y).(X + 1) (ref. X + X’=1)

= 1.1 (ref. 1 + X =1)

= 1

So first part is proved.

Now let us prove the second part i.e.,

(X + Y) . (XY)’= 0

(X + Y) . (XY)’ = (XY)’ . (X + Y) (ref. X(YZ) = (XY)Z) = (XY)’X + (XY)’Y (ref. X(Y + Z) = XY + XZ)

= X(XY)’ + X’YY’

= 0 .Y + X’ . 0 (ref. X . X’=0)

= 0 + 0 = 0

So, second part is also proved, Thus: X + Y = X’ . Y’

The desired Canonical Product-of-Sum form is as following;

H = π(2, 4, 6, 7) = (U + V’ + W)(U’ + V + W)(U’ + V’ + W)(U’ + V’ + W’)

LHS = XY + YZ + Y’Z

= XY + Z(Y + Y’) ( Y + Y’ = 1) 

= XY + Z

Given:

Alloy weight = 1500 gram

Calculation:

Let the copper and brass be 15x and 8x respectively

Tin  = 25 % of 8x

= 2x

Gold = 2.5 (25 % of 8x)

= 2.5 × (25/100) × 8x

= 5x

According to question:

⇒ gold + copper + brass + tin = 1500

⇒ 5x + 15x + 8x + 2x = 1500

⇒ 30x = 1500

⇒ x = 50

⇒ gold = 2.5  × 2(50) = 250 gram

⇒ Copper = 15 × 50 = 750 gram

⇒ brass = 8 × 50 = 400 gram

⇒ Tin = 2(50) = 100 gram

The major social groups of Sri Lanka are: the Sinhala speakers (74 per cent) and the Tamil speakers (18 per cent).

Among Tamils there are two sub-groups: Tamil natives of the country are called ‘Sri Lankan Tamils’ and the rest, whose forefathers came from India as plantation workers during colonial period are called ‘Indian Tamils’. 

Sri Lankan Tamils are concentrated in the north and east of the country. 

Most of the Sinhala-speaking people are Buddhists, while most of the Tamils are Hindus or Muslims. There are about 7 per cent Christians who are both Tamils and Sinhalas.

DeMorgan’s theorems state that

(i) (X + Y)’= X’.Y’

(ii) (X.Y)’= X’ + Y’

(i) (X + Y)’= X’.Y’ 

Now to prove DeMorgan’s first theorem, we will use complementarity laws. Let us assume that P = x + Y where, P, X, Y are logical variables. Then, according to complementation law 

P + P’ =1 and P . P’= 0 

That means, if P, X, Y are Boolean variables then this complementarity law must hold for variables P. In other words, if P i.e., if (X + Y)’= X’.Y’then

(X + Y) + (XY)’must be equal to 1. (as X + X’= 1)

(X + Y) . (XY)’must be equal to 0.

(as X . X’= 0) Let us prove the first part, i.e.,

(X + Y) + (XY)’ = 1

(X + Y) + (XY)’= ((X + Y) +X’).((X + Y) +Y’) (ref. X + YZ = (X + Y)(X + Z))

= (X + X’+ Y).(X + Y +Y’) 

= (1 + Y).(X + 1) (ref. X + X’=1)

= 1.1 (ref. 1 + X =1)

= 1

So first part is proved.

Now let us prove the second part i.e.,

(X + Y) . (XY)’= 0

(X + Y) . (XY)’ = (XY)’ . (X + Y)   (ref. X(YZ) = (XY)Z)

= (XY)’X + (XY)’Y    (ref. X(Y + Z) = XY + XZ)

= X(XY)’ + X’YY’ 

= 0 .Y + X’ . 0     (ref. X . X’=0) 

= 0 + 0 = 0

So, second part is also proved, Thus: X + Y = X’ . Y’

(ii)  (X.Y)’= X’ + Y’ 

Again to prove this theorem, we will make use of complementary law i.e., 

X + X’= 1 and X . X’= 0

If XY’s complement is X + Y then it must be true that

(a) XY + (X’+ Y’) = 1 and (b) XY(X’+ Y’) = 0 

To prove the first part

L.H.S = XY + (X’+Y’)

= (X’+Y’) + XY      (ref. X + Y = Y + X) 

= (X’+Y’ + X).(X’+Y’ + Y)     (ref. (X + Y)(X + Z) = X + YZ)

= (X + X’+Y’).(X’ + Y +Y’) 

= (1 +Y’).(X’ + 1)      (ref. X + X’=1)

= 1.1 (ref. 1 + X =1)

= 1 = R.H.S

Now the second part i.e.,

 XY.(X + Y) = 0

L.H.S = (XY)’.(X’+Y’)

= XYX’ + XYY’ (ref. X(Y + Z) = XY + XZ) 

= XX’Y + XYY’ 

= 0.Y + X.0   (ref. X . X’=0) 

= 0 + 0 = 0 = R.H.S. 

XY.(X’ + Y’)= 0 and XY + (Xʹ +Y’) = 1 

(XY)’= X’ + Y’. Hence proved.

There are 2 Pair and 1 Quad that reduce as given below:

Pair-1(m₃ + m₇) reduces to A’CD

Pair-2(m₄ + m₇) reduces to A’BD’

Quad(m₁+ m₅ + m₉ + m₁₃ ) reduces to BD

Simplified Boolean expression for given K-map is

F(A, B, C, D) = A’CD +’ A’BD’ + BD

The desired Canonical Sum-of-Product form is as following;

G = ∑(2, 3, 4, 6, 7) = P’QR’ + P’QR + PQ’R’ + PQR’ + PQR

The equivalent Boolean Expression for the given Logic Circuit is: F = (U’ + V).(V’ + W) 

(a)  X + X’ = 1

To prove this law, we will make a following truth table :

0 + 1 = 1 and 1 + 0 = 1

From truth table it is prove that X + X’ = 1

(b) X . X’= 0

To prove this law, we will make a following truth table :

0 . 1 = 0 and 1 . 0 = 0

The get() and put() functions perform I/O byte by byte. On the other hand, read() and write() functions let you read and write structures and objects in one go without creating need for I/O for individual constituent fields.

Example:

file.get(ch);

file.put(ch);

file.read((char *)&obj, sizeof(obj));

file.write((char *)&obj, sizeof(obj)); 

(a) X + X = X 

To prove this law, we will make a following truth table :

 0 + 0 = 0 and 1 + 1 = 1 0 . 0 = 0 and 1 . 1 = 1

The error-status flags store the information on the status of a file that is being currently used. The current state of the I/O system is held in an integer, in which the following flags are encoded: