Detailed Solution:

Phenylketonuria (PKU) is a genetic disorder caused by a specific enzyme deficiency.

1. Enzyme Deficiency:
   • PKU is due to deficiency of the enzyme phenylalanine hydroxylase (PAH).
   • PAH converts phenylalanine into tyrosine.
2. Consequences:
   • When PAH is deficient, phenylalanine accumulates to toxic levels.
   • This accumulation causes brain damage, intellectual disabilities, and developmental issues.
3. Other options: The other enzymes listed are involved in other biochemical pathways and are not the cause of PKU.

Therefore, the deficiency of **phenylalanine hydroxylase (PAH)** causes PKU.

Final Answer: Phenylalanine hydroxylase (PAH)

Detailed Solution:

The meninges are protective layers surrounding the brain and spinal cord.

1. Dura Mater: This is the outermost and toughest layer.
2. Arachnoid Mater: This is the middle, web-like layer, which provides cushioning
3. Pia Mater: The innermost and thin layer which is directly attached to the brain and spinal cord

Therefore, the correct order from outer to inner is Dura mater, Arachnoid, Pia mater.

Final Answer: Dura mater, Arachnoid, Pia mater

Detailed Solution:

Pancreatic juice and bicarbonate production is controlled by specific hormones.

1. Cholecystokinin (CCK):
   • CCK is released in response to fats and proteins in the small intestine.
   • CCK stimulates the pancreas to release an enzyme-rich juice which aids in digestion.
2. Secretin:
   • Secretin is released when acidic chyme enters the duodenum.
   • Secretin causes pancreas to release bicarbonate-rich juice to neutralize the acidity.

Therefore, the hormones that stimulate the production of pancreatic juice and bicarbonate are Cholecystokinin (CCK) and Secretin.

Final Answer: Cholecystokinin (CCK) and Secretin

Detailed Solution:

The menstrual cycle has specific phases which are arranged in a sequence.

1. Menstruation: This phase is the beginning of the cycle and marks the shedding of the uterine lining.
2. Follicular Phase: In this phase, follicles grow in the ovary in preparation for ovulation.
3. Ovulation: This is the release of a mature egg from the ovary.
4. Luteal Phase: In this phase, the corpus luteum develops and secretes progesterone.

Therefore, the correct sequence is Menstruation, Follicular phase, Ovulation, Luteal phase.

Final Answer: Menstruation, Follicular phase, Ovulation, Luteal phase

Detailed Solution:

The electron transport system utilizes specific electron acceptors.

1. Electron Transport System: Electrons pass through a chain of proteins in the inner mitochondrial membrane.
2. Oxygen as the Final Acceptor: The final electron acceptor at the end of the chain is oxygen, which combines with electrons and protons to form water.
3. Other Molecules: NADH and FADH are electron carriers. CO2 is a by product, not an acceptor.

Therefore, the last electron acceptor in ETS is Oxygen (O2).

Final Answer: Oxygen (O2)

Detailed Solution:

Apical dominance is a phenomenon where the apical bud inhibits lateral bud growth and specific hormones counter this effect.

1. Apical Dominance: Apical dominance is controlled by auxin which inhibits lateral bud growth.
2. Cytokinin:
   • Cytokinin is a hormone that promotes cell division in lateral buds.
   • It overcomes the inhibitory effect of auxin, and promotes lateral bud growth.
3. Other Options: Gibberellin and abscisic acid have other roles and do not reverse apical dominance

Therefore, cytokinin reverses apical dominance.

Final Answer: Cytokinin

Detailed Solution:

The female gametophyte originates from the megaspore lineage.

1. Megaspore Lineage: The megaspore mother cell gives rise to the megaspore which develops into the female gametophyte (embryo sac). The nucellus provides nutrition to the female gametophyte.
2. Microspore lineage: Microspore mother cell produces microspores, which form the male gametophyte (pollen grain).

Therefore, the female gametophyte does not develop from the microspore mother cell.

Final Answer: Microspore mother cell

Detailed Solution:

Inorganic phosphate plays specific roles in cellular respiration.

1. Glycolysis:
   • Inorganic phosphate (Pi) has a direct role in glycolysis.
   • It is used during the conversion of glyceraldehyde-3-phosphate (G3P) to 1,3-bisphosphoglycerate.
2. Other Options:
   • Inorganic phosphate may have indirect roles in Krebs and ETS.
   • There is no direct involvement of Pi in fermentation

Therefore, inorganic phosphate is directly involved in Glycolysis.

Final Answer: Glycolysis

Detailed Solution:

This question involves the principle of fluid flow through openings based on Torricelli’s theorem.

1. Efflux Velocity:
   • According to Torricelli's theorem the efflux velocity (v) = √(2gh).
2. Velocity from Hole A: The efflux velocity for hole A is va = √(2gh)
3. Velocity from Hole B: The efflux velocity for hole B is vb = √(2g(3h)) = √(6gh)
4. Equating Flow rates: Flow rate = Area x Velocity
   • Flow rate through hole A (area is L²): Qa = L²√(2gh)
   • Flow rate through hole B (area is πr²): Qb = πr²√(6gh)
   • Given Qa=Qb : L²√(2gh) = πr²√(6gh)
5. Solving for L:
   • L²√(2) = πr²√(6) or L² = πr²√(6)/√(2) = πr²√(3) or L = r√(π√(3)) = 3^(1/4) π^(1/2) r

Therefore, L = 3^(1/4)π^(1/2)r.

Final Answer: L = 3^(1/4)π^(1/2)r

Detailed Solution:

Heat required is related to heat capacity and temperature.

1. Heat at Constant Volume: The heat required (Q) to raise the temperature of a gas at constant volume is Q= nCvΔT
   • Here, n is the number of moles, Cv is the molar heat capacity at constant volume, and ΔT is change in temperature.
2. Number of Moles: At STP, a gas occupies 22.4L. Hence:

   n = 67.2 L/ 22.4 L/mol = 3 mol

3. Heat Capacity of Helium: For helium which is a monoatomic gas, Cv = (3/2)R. Hence:

   Cv = (3/2)*8.314 = 12.471 J/(mol.K)

4. Substitution: Putting all values together

   Q = nCvΔT = 3 mol * 12.471 J/mol.K * 20K = 748.26 J

Therefore, the heat required is approximately 747.9 J.

Final Answer: 747.9 J

Detailed Solution:

Hemoglobin's affinity for different gases varies depending on the chemical interactions with the heme group.

1. Carbon Monoxide (CO):
   • Hemoglobin has the highest affinity for carbon monoxide (200-300 times greater than O2).
   • This strong interaction prevents oxygen transport and is toxic.
2. Carbon Dioxide (CO2):
   • The affinity for carbon dioxide is less than that for CO.
   • CO2 binds to form carbaminohemoglobin, but is less strong than CO
3. Oxygen (O2): Hemoglobin's affinity for O2 is essential for respiration but is lower than CO and CO2.

Therefore, the decreasing order of affinity is CO > CO2 > O2.

Final Answer: CO > CO2 > O2

Detailed Solution:

The abundance of isotopes is calculated using the average atomic mass.

1. Define Abundances: Let the abundance of the lighter isotope (¹⁰B) be 'x%', then the abundance of heavier isotope (¹¹B) is (100-x)%.
2. Average Atomic Mass Formula: Average atomic mass can be expressed as

   average mass = (x * mass of isotope1 +(100-x) * mass of isotope 2)/100

3. Substitute Values: Substitute mass as 10 and 11 and equate with average atomic mass of 10.81

   10.81 = (10x + (100-x)11)/100

   1081 = 10x + 1100 - 11x

4. Solve for x:

   x = 1100-1081 = 19 %

Therefore, the abundance of the lighter isotope is 19%.

Final Answer: 19%