Numerical Problems for ISC Class 11 Biology

Chapter: Cardiac Cycle

Q1: A person has a heart rate of 75 beats per minute. Calculate the duration of their cardiac cycle.

Solution: The duration of one cardiac cycle is the reciprocal of the heart rate. Duration = 60 seconds / Heart Rate Duration = 60 / 75 = 0.8 seconds.

Q2: If a cardiac cycle is 0.8 seconds long, and the ventricular systole lasts for 0.3 seconds, what is the duration of ventricular diastole?

Solution: Total cycle duration = Ventricular Systole + Ventricular Diastole 0.8 s = 0.3 s + Ventricular Diastole Ventricular Diastole = 0.8 - 0.3 = 0.5 seconds.

Q3: In a normal cardiac cycle of 0.8s, atrial systole is 0.1s, and ventricular systole is 0.3s. Calculate the duration of joint diastole.

Solution: Joint diastole is the period when both atria and ventricles are relaxed. Total cycle = Atrial Systole + Ventricular Systole + Joint Diastole is incorrect. The duration of joint diastole is the total cycle duration minus the time when either atria or ventricles are in systole. Joint Diastole = Total Cycle - (Atrial Systole + Ventricular Systole - Overlap) However, a simpler way is: Joint Diastole = Total Cycle Duration - Duration of Ventricular Systole - Duration of Atrial Diastole that is not part of ventricular systole. A simpler calculation is to consider the total time spent in systole by either chamber. Atrial systole is 0.1s and ventricular systole is 0.3s. The total duration of the cardiac cycle is 0.8s. Joint diastole = Total cycle duration - (time of atrial systole + time of ventricular systole) is incorrect. The correct way: Joint Diastole = Total Cycle duration - (duration of atrial systole + duration of ventricular systole - duration of overlap) The period of joint diastole is when both chambers are in diastole. Ventricular diastole = 0.5s. Atrial systole occurs for 0.1s at the end of ventricular diastole. So, Joint Diastole = Ventricular Diastole - Atrial Systole = 0.5 - 0.1 = 0.4s. Alternatively, Total Cycle = 0.8s. Ventricular Systole = 0.3s. Atrial Systole = 0.1s. The heart is in some form of systole for 0.1s (atria) + 0.3s (ventricle) = 0.4s. The rest of the time it is in joint diastole. Joint Diastole = 0.8s - 0.4s = 0.4s.

Chapter: Enzyme Kinetics

Q1: An enzyme has a Vmax of 100 µM/s and a Km of 2mM. What is the velocity of the reaction when the substrate concentration is 2mM?

Solution: According to the Michaelis-Menten equation: v = (Vmax * [S]) / (Km + [S]) Given: Vmax = 100 µM/s, Km = 2mM, [S] = 2mM. v = (100 * 2) / (2 + 2) = 200 / 4 = 50 µM/s. This is consistent with the definition of Km as the substrate concentration at which the reaction velocity is half of Vmax.

Q2: The Vmax for an enzyme-catalyzed reaction is 60 µM/min. The Km is 1 mM. What is the substrate concentration when the velocity is 20 µM/min?

Solution: Using the Michaelis-Menten equation: v = (Vmax * [S]) / (Km + [S]) 20 = (60 * [S]) / (1 + [S]) 20 * (1 + [S]) = 60 * [S] 20 + 20[S] = 60[S] 20 = 40[S] [S] = 20 / 40 = 0.5 mM.

Q3: A Lineweaver-Burk plot gives a y-intercept of 0.05 (µM/min)⁻¹ and an x-intercept of -0.2 (mM)⁻¹. Calculate Vmax and Km.

Solution: The y-intercept = 1/Vmax. 0.05 = 1/Vmax Vmax = 1 / 0.05 = 20 µM/min. The x-intercept = -1/Km. -0.2 = -1/Km Km = 1 / 0.2 = 5 mM.

Q4: An enzyme has a Km of 5x10⁻⁵ M. A competitive inhibitor is present at a concentration of 2x10⁻⁵ M, and the inhibitor constant Ki is 4x10⁻⁵ M. Calculate the apparent Km (Km_app).

Solution: For competitive inhibition, the apparent Km is calculated as: Km_app = Km * (1 + [I]/Ki) Km_app = 5x10⁻⁵ * (1 + (2x10⁻⁵ / 4x10⁻⁵)) Km_app = 5x10⁻⁵ * (1 + 0.5) Km_app = 5x10⁻⁵ * 1.5 = 7.5x10⁻⁵ M.