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This section includes 436 Mcqs, each offering curated multiple-choice questions to sharpen your Bioinformatics knowledge and support exam preparation. Choose a topic below to get started.
201. |
Pitt Rivers was a--------- |
A. | ethnologist |
B. | sociologist |
C. | philologist |
D. | none |
Answer» B. sociologist | |
202. |
‘Ithaca’ was a work of------------ |
A. | schliemann |
B. | woolley |
C. | wheeler |
D. | petrie |
Answer» B. woolley | |
203. |
Who excavated the city of Troy -------- |
A. | schliemann |
B. | woolley |
C. | wheeler |
D. | petrie |
Answer» B. woolley | |
204. |
‘Trojan Antiquities is a work of--------- |
A. | schliemann |
B. | woolley |
C. | wheeler |
D. | petrie |
Answer» B. woolley | |
205. |
Schliemann was belongs to---------- |
A. | germany |
B. | england |
C. | france |
D. | canada |
Answer» B. england | |
206. |
Early India and Pakistan is a work of---------- |
A. | marshall |
B. | wheeler |
C. | mackay |
D. | woolley |
Answer» C. mackay | |
207. |
‘Civilization of the Indus valley and Beyond’ was written by ---------- |
A. | marshall |
B. | wheeler |
C. | mackay |
D. | woolley |
Answer» C. mackay | |
208. |
Mohenjo-Daro and the Indus civilization is a work of-------- |
A. | marshall |
B. | wheeler |
C. | mackay |
D. | woolley |
Answer» B. wheeler | |
209. |
Marshall was the Director General of-------- |
A. | asi |
B. | isi |
C. | csi |
D. | irdp |
Answer» B. isi | |
210. |
‘Digging up the past’ is the work of----------- |
A. | woolley |
B. | wheeler |
C. | marshall |
D. | pitt rivers |
Answer» B. wheeler | |
211. |
‘The monuments of Sanchi’ was written by--------- |
A. | marshall |
B. | wheeler |
C. | mackay |
D. | woolley |
Answer» B. wheeler | |
212. |
‘A guide to Taxila’ is a work of---------- |
A. | woolley |
B. | wheeler |
C. | marshall |
D. | pitt rivers |
Answer» D. pitt rivers | |
213. |
Leonard Woolley conducted excavation at--------- |
A. | ur |
B. | mohenjo-daro |
C. | harappa |
D. | memphis |
Answer» B. mohenjo-daro | |
214. |
‘Royal Cemetery’ was related to---------- |
A. | ur |
B. | kish |
C. | memphis |
D. | nippur |
Answer» B. kish | |
215. |
Which of the following is untrue about dot plot method and its applications? |
A. | This method gives a direct visual statement of the relationship between two sequences |
B. | One of its advantages is the identification of sequence repeat regions based on the presence of parallel diagonals of the same size vertically or horizontally in the matrix |
C. | It is not useful in identifying chromosomal repeats |
D. | The method can be used in identifying nucleic acid secondary structures through detecting self-complementarity of a sequence |
Answer» D. The method can be used in identifying nucleic acid secondary structures through detecting self-complementarity of a sequence | |
216. |
Self complementarity of DNA sequences cannot be identified using a dot plot. |
A. | True |
B. | False |
Answer» C. | |
217. |
A sequence can be aligned with itself to identify internal repeat elements. |
A. | True |
B. | False |
Answer» B. False | |
218. |
A root x4 – 3x + 1 = 0 needs to be found using the Newton-Raphson method. If the initial guess x0 and 0, then the new estimate x1 after the first iteration is |
A. | -3 |
B. | \(\frac 12\) |
C. | 3 |
D. | \(\frac 13\) |
Answer» E. | |
219. |
f(z) = (z − 1)−1 − 1 + (z − 1) − (z − 1)2 + ⋯ is the series expansion of |
A. | \(\frac{-1}{z(z-1)} ~for ~|z - 1| < 0\) |
B. | \(\frac{1}{z(z-1)}~ for~ |z - 1| < 0\) |
C. | \(\frac{1}{(z-1)^2} ~for ~|z - 1| < 0\) |
D. | \(\frac{-1}{(z-1)} ~for~ |z - 1| < 0\) |
Answer» C. \(\frac{1}{(z-1)^2} ~for ~|z - 1| < 0\) | |
220. |
For the equation f(x) = x2 – x – 1 = 0, a root lies between 1 and 2. The root of equation at second interval by bisection method is |
A. | 1.51 |
B. | 2 |
C. | 1.66 |
D. | 1.75 |
Answer» E. | |
221. |
Match the application to appropriate numerical method. Application Numerical MethodP1:Numerical integrationM1:Newton-Raphson MethodP2:Solution to a transcendental equationM2:Runge-Kutta MethodP3:Solution to a system of linear equationsM3:Simpson’s 1/3-ruleP4:Solution to a differential equationM4:Gauss Elimination Method 1) P1—M3, P2—M2, P3—M4, P4—M12) P1—M3, P2—M1, P3—M4, P4—M23) P1—M4, P2—M1, P3—M3, P4—M24) P1—M2, P2—M1, P3—M3, P4—M4 |
A. | a |
B. | b |
C. | c |
D. | d |
Answer» C. c | |
222. |
In order to evaluate the integral \(\displaystyle\int_{{0}}^{{1}} e^xdx \) with Simpson’s 1/3rd rule, values of the function ex are used at x = 0.0, 0.5 and 1.0. The absolute value of the error of numerical integration is |
A. | 0.000171 |
B. | 0.00044 |
C. | 0.000579 |
D. | 0.002718 |
Answer» D. 0.002718 | |
223. |
A curve is drawn to pass through the followingx11.522.533.54y22.42.72.832.62.1 The area bounded by the curve, x-axis and lines x = 1, x = 4. The volume of solid generated by revolving this area using Simpson's 3/8 rule is |
A. | 68.54 |
B. | 65.38 |
C. | 63.58 |
D. | 64.38 |
Answer» C. 63.58 | |
224. |
A river is 80 metre wide. The depth ‘d’ in metres at a distance ‘x’ metres from one bank is given, by the following table:X:01020304050607080D:047912151483Hence the area of c / s of the river using Simpson’s rule is: |
A. | 713 sq. met. |
B. | 710 sq. met. |
C. | 715 sq. met. |
D. | 716 sq. met. |
Answer» C. 715 sq. met. | |
225. |
f (x) = x2 + 1If xi is very close to the root then according to Newton Raphson iterative procedure, xi+1 is |
A. | \(\dfrac{x^2_i - 1}{2x_i}\) |
B. | \(\dfrac{2x_i}{x^2_i - 1}\) |
C. | \(\dfrac{2x_i}{x^2_i + 1}\) |
D. | \(\dfrac{x^2_i + 1}{2x_i}\) |
Answer» B. \(\dfrac{2x_i}{x^2_i - 1}\) | |
226. |
Let \(I = \mathop \smallint \limits_{{x_0}}^{{x_1}} f\left( x \right)dx.\) Then which of the following is false? |
A. | \(I\sim\frac{h}{2}\left[ {{y_0} + {y_n} + \frac{1}{2}\left( {{y_1} + {y_2} + \ldots + {y_{n - 1}}} \right)} \right]\) |
B. | \(I\sim\frac{h}{3}\left[ {{y_0} + {y_n} + 4\left( {{y_1} + {y_3} + \ldots + {y_{n - 1}}} \right) + 2\left( {{y_2} + {y_4} + \ldots + {y_{n - 2}}} \right)} \right],\) n is even |
C. | \(I\sim h\left( {{y_0} + {y_1} + {y_2} + \ldots + {y_{n - 1}}} \right)\) |
D. | \(I\sim\frac{h}{{140}}\left( {41{y_0} + 216{y_1} + 27{y_2} + 272{y_3} + 27{y_4} + 216{y_5} + 41{y_6}} \right)\;\;\) |
Answer» B. \(I\sim\frac{h}{3}\left[ {{y_0} + {y_n} + 4\left( {{y_1} + {y_3} + \ldots + {y_{n - 1}}} \right) + 2\left( {{y_2} + {y_4} + \ldots + {y_{n - 2}}} \right)} \right],\) n is even | |
227. |
Consider the following statements regarding the convergence of the Newton-Raphson procedure: 1. It does not converge to a root when the second differential coefficient changes sign2. It is preferred when the graph of (X) is nearly horizontal where it crosses the X-axis3. It is used to solve algebraic and transcendental equationsWhich of these statements are correct? |
A. | 1, 2, and 3 |
B. | 1 and 2 only |
C. | 2 and 3 only |
D. | 1 and 3 only |
Answer» E. | |
228. |
If one root of the equation f(x) = 0 is near to x0, then the first approximation of this root as calculated by Newton Raphson method is the abscissa of the point, where the following straight line intersects the x-axis” |
A. | The straight line through the point (x0, y = f(x0)) having the gradient \(\frac{1}{{f'\left( {{x_0}} \right)}}\) |
B. | Tangent to the curve y = f(x) at the point (x0, y = f(x0)) |
C. | Passing through the point (x0, y = f(x0)) |
D. | Normal to the curve y = f(x) at the point (x0, y = f(x0)) |
Answer» C. Passing through the point (x0, y = f(x0)) | |
229. |
Back substitution is required in the following method (s) in the solution of linear simultaneous equation: |
A. | Gauss-Elimination method |
B. | Gauss-Jordan method |
C. | Iterative method |
D. | All of the above |
Answer» B. Gauss-Jordan method | |
230. |
If 2.5 is the initial root of the equation x3 - x - 10 = 0 then by method of Newton - Raphson, the next approx root will be equal to- |
A. | 2.3089 |
B. | 2.5395 |
C. | 2.676 |
D. | 2.6657 |
Answer» B. 2.5395 | |
231. |
As soon as a new variable is found by iteration, it is used immediately in the linear equations. This method is called: |
A. | Relaxation Method |
B. | Gauss Seidal Method |
C. | Gauss Jordan Method |
D. | Jacobi Method |
Answer» C. Gauss Jordan Method | |
232. |
Given thatx :44.24.44.64.85.05.2log x :1.38631.43511.48161.52611.56861.60941.6484 Evaluate \(\mathop \smallint \nolimits_4^{5.2} \log x\;dx\) by Trapezoidal Rule. |
A. | 1.827887 |
B. | 1.827655 |
C. | 1.827867 |
D. | 1.8278 |
Answer» C. 1.827867 | |
233. |
Consider an ordinary differential equation. \(\frac{{{\rm{dx}}}}{{{\rm{dt}}}} = 4{\rm{t}} + 4.\) If x = x0 at t = 0, the increment in x calculated using Runge-Kutta fourth order multi-step method with a step size of Δt = 0.2 is |
A. | 0.22 |
B. | 0.44 |
C. | 0.66 |
D. | 0.88 |
Answer» E. | |
234. |
During the determination of roots of equations x2 + 2xy = 6 and x2 - y2 = 3 using the Newton-Raphson method, the values of Jacobin matrix ‘D’ is found to be ______. If initial approximation is (1.414, 0.517). |
A. | - 4 |
B. | - 8 |
C. | - 12 |
D. | + 4 |
Answer» D. + 4 | |
235. |
Function f is known at the following points:x00.30.60.91.21.51.82.42.73.0f(x)00.090.360.811.442.253.245.767.299.0 The value \(\mathop \smallint \limits_0^3 f\left( x \right)dx\) computed using the trapezoidal rule is |
A. | 8.983 |
B. | 9.003 |
C. | 9.017 |
D. | 9.045 |
Answer» E. | |
236. |
In regula falsi method the point of intersection of curve AB and x axis is replaced by: |
A. | Point of intersection of y axis and curve AB |
B. | Point of intersection of y axis and chord AB |
C. | Point of intersection of x axis and chord AB |
D. | Point of intersection of x axis and y axis |
Answer» D. Point of intersection of x axis and y axis | |
237. |
Match the CORRECT pairs:NumericalIntegration Scheme Order of Fitting PolynomialP. Simpson’s 3/8 Rule 1. FirstQ. Trapezoidal Rule 2. SecondR. Simpson’s 1/3 Rule3. Third |
A. | P-2; Q-1; R-3 |
B. | P-3; Q-2; R-1 |
C. | P-1; Q-2; R-3 |
D. | P-3; Q-1; R-2 |
Answer» E. | |
238. |
Numerical integration using trapezoidal rule gives the best result for a single variable function, which is |
A. | linear |
B. | parabolic |
C. | logarithmic |
D. | hyperbolic |
Answer» B. parabolic | |
239. |
In Newton-cotes formula, if f(x) is interpolated at equally spaced nodes by a polynomial of degree four then it represents |
A. | Trapezoid rule |
B. | Simpson rule |
C. | Three-eigth rule |
D. | Booles rule |
Answer» E. | |
240. |
If f(x) is a polynomial of degree n in x, then nth difference of this polynomial is |
A. | Constant |
B. | Variable |
C. | Zero |
D. | Ones |
Answer» B. Variable | |
241. |
Gauss forward interpolation formula involves |
A. | Given difference above the central line and odd differences on the central line |
B. | Even difference below the central line and odd differences on the central line |
C. | Odd differences below the central line and even differences on the central line |
D. | Odd differences above the central line and even differences on the central line |
Answer» D. Odd differences above the central line and even differences on the central line | |
242. |
Given thatx\(\frac{1}{{1 + {x^2}}}\)0110.520.230.140.058850.038560.027 Evaluate \(\mathop \smallint \nolimits_0^6 \frac{{dx}}{{1 + {x^2}}}\) using Simpson’s 3 / 8 rule. |
A. | 1.3574 |
B. | 1.3569 |
C. | 1.3576 |
D. | 1.3571 |
Answer» E. | |
243. |
Considering four subintervals, the value of \(\mathop \smallint \limits_0^1 \frac{1}{{1 + x}}dx\) by Trapezoidal rule is: |
A. | 0.695 |
B. | 0.687 |
C. | 0.6677 |
D. | 0.3597 |
Answer» B. 0.687 | |
244. |
From the given data, the maximum value of y is given asx-1123y-2115123 |
A. | At x = 1.1743, ymax = 15.171612 |
B. | At x = 0.1743, ymax = 15.171612 |
C. | At x = -1.1743, ymax = 15.171612 |
D. | At x = 2.1743, ymax = 15.171612 |
Answer» B. At x = 0.1743, ymax = 15.171612 | |
245. |
A gradually varied flow profile can be governed by equation \(\dfrac{dy}{dx} = f(x,y)\) where x is distance and y is the depth of water above the bed level. Which of the following methods can be used for solution? |
A. | Linear regression method |
B. | Simplex method |
C. | Gauss elimination method |
D. | Runge-Kutta Method |
Answer» E. | |
246. |
\(f(x)=f(0)+x\nabla f(0)+\frac{x(x+1)}{2!}\nabla^2f(0)+....+\frac{x(x+1)...(x+n-1)}{n!} \nabla^n f(0)\) represents |
A. | Newton backward difference formula |
B. | Newton forward difference formula |
C. | Gauss's forward formula |
D. | Newton divided difference formula |
Answer» B. Newton forward difference formula | |
247. |
For k = 0, 1, 2,……, the steps of Newton-Raphson method for solving a non-linear equation is given as\({x_{k + 1}} = \frac{2}{3}{x_k} + \frac{5}{3}x_k^{ - 2}\)Starting from a suitable initial choice as k tends to ∞, the iterate xk tends to |
A. | 1.7099 |
B. | 2.2361 |
C. | 3.1251 |
D. | 5 |
Answer» B. 2.2361 | |
248. |
In solving ordinary differential equation y’ = 2x, y(0) = 0 using Euler’s method, the iterate yn, n ϵ N satisfy |
A. | yn = 2Xn |
B. | \({y_n} = X_n^2\) |
C. | yn = XnXn-1 |
D. | yn = Xn + Xn-1 |
Answer» D. yn = Xn + Xn-1 | |
249. |
P (0, 3), Q (0.5, 4), and R (1, 5) are three points on the curve defined by. Numerical integration is carried out using both Trapezoidal rule and Simpson’s rule within limits x = 0 and x = 1 for the curve. The difference between the two results will be |
A. | 0 |
B. | 0.25 |
C. | 0.5 |
D. | 1 |
Answer» B. 0.25 | |
250. |
By Simpson’s \({\frac{1}{3}^{rd}}\) rule, the value of \(\mathop \smallint \limits_1^7 \frac{{dx}}{x}\) is |
A. | 1.958 |
B. | 1.458 |
C. | 1.658 |
D. | 1.358 |
Answer» B. 1.458 | |