List of practice Questions

Match the following \begin{table}[h!] \centering \begin{tabular}{|l|l|} \hline List - I & List - II
\hline A. Operator site & I. Binding site for RNA polymerase
B. Promoter site & II. Binding site for repressor
C. Regulator gene & III. Codes for Enzyme
D. Structural gene & IV. Codes for repressor
\hline \end{tabular} \caption{Matching the following} \end{table} The correct answer is
The separated DNA fragments can be visualised after staining with 'X' and exposure to 'Y'. The 'X' and 'Y' are
Assertion (A): \( \sigma \) (Sigma) and \( \rho \) (Rho) factors are essential for transcription in bacteria.
Reason (R): \( \sigma \) (sigma) factor helps in initiation and \( \rho \) (Rho) factor helps in termination of transcription.
Dual functions of codon AUG
Identify wrong matching I. \= Roundup ready = Soyabean \hspace{1cm} = Herbicide tolerant
II. \= Transgenic Plant = Papaya \hspace{1cm} = Pseudomonas resistant
III. \= Taipei variety = Rice \hspace{1cm} = rich in Vitamin B
IV. \= Flavr savr = Tomato \hspace{1cm} = Bruise resistant
Assertion (A): Protoxin of Bacillus thuringensis become active in gut of insects.
Reason (R): Alkaline pH of intestine gut solubilizes the crystals of toxin to make it active.
Observe the diagrammatic representation of ATP synthesis in oxysome of mitochondria. Identify A, B and C respectively. \includegraphics[width=0.3\linewidth]{3image.png}
Find out A, B, C, and D products of anaerobic respiration pathway cited below \includegraphics[width=0.4\linewidth]{5image.png}
Assertion (A): Abscisic acid is called stress hormone.
Reason (R): Abscisic acid increases the tolerance of plants to various kinds of stresses.
Assertion (A): In Arachis, fruits are geocarpic.
Reason (R): Seeds are non-endospermic in Arachis.
Identify the wrong match.
I. \hspace{2cm} \= Acidy amino acid \hspace{1cm} \= = \hspace{1cm} \= Valine
II. \hspace{2cm} \= Neutral amino acid \hspace{1cm} \= = \hspace{1cm} \= Tryptophan
III. \hspace{2cm} \= Aromatic amino acid \hspace{1cm} \= = \hspace{1cm} \= Tyrosine
IV. \hspace{2cm} \= Basic amino acid \hspace{1cm} \= = \hspace{1cm} \= Lysine
Identify correct statements
Match the following \begin{table}[h!] \centering \begin{tabular}{|l|l|} \hline List - I & List - II
\hline A. Toxin & I. Cellulose
\hline B. Drug & II. Concanavalin A
\hline C. Lectin & III. Curcumerin
\hline D. Polymer substance & IV. Ricin
\hline \end{tabular} \caption{Match the following} \end{table} The correct answer is:
Identify the enzyme of the following reaction: \[ \text{Phosphoglyceric acid + ATP} \xrightarrow{\text{Enzyme}} \text{Bisphosphoglyceric acid + ADP} \]
Match the following: \includegraphics[width=0.5\linewidth]{1image.png} The correct answer is
Primary endosperm nucleus is formed by the fusion of
Identify the correct matching: \renewcommand{\arraystretch}{1.3} \begin{table}[h!] \centering \setlength{\tabcolsep}{15pt} \begin{tabular}{|c|l|c|l|} \hline Column I & Plant Name & Column II & Feature
\hline I & Nerium & = Verticillaster & = Sunken stomata
II & Vallisnaria & = Free floating hydrophyte & = Epihydrophily
III & Tribulus & = Emphemeral & = Xerophyte
IV & Rhizophora & = Halophyte& = Vivipary
\hline \end{tabular} \end{table}
Based on the figure given below, which of the following statements are not correct? \includegraphics[width=0.4\linewidth]{2image.png}
Match the following:

\begin{tabular}{|l|l|} \hline List - I & List - II
\hline A. Glut - 4 & I. Hormone
B. Antibody & II. Enzyme
C. Trypsin & III. Fights infectious agents
D. Insulin & IV. Enables glucose transport into cells
\hline \end{tabular}
The correct answer is
Choose the correct combination \begin{table}[h!] \centering \begin{tabular}{|c|c|c|} \hline Substrates & End products & Enzyme class
\hline I. Argenosuccinic acid & Arginine + Fumaric acid & 3
\hline II. Fructose 1,6 bisphosphate & Fructose 6 phosphate + iP & 5
\hline III. Glutamic acid + NH$_3$ + ATP & Glutamine + ADP + iP & 6
\hline IV. Malate + NAD$^+$ & Oxaloacetate + NADH + H$^+$ & 1
\hline \end{tabular} \end{table}
Match the following: \[ \begin{array}{|c|c|} \hline \text{List-I (Element)} & \text{List-II (Significance)}
\hline A. \text{Copper} & I. \text{Activator of carboxylase}
B. \text{Zinc} & II. \text{Component of nitrogenase}
C. \text{Molybdenum} & III. \text{Chlorophyll structure}
D. \text{Magnesium} & IV. \text{Component of cytochrome C}
\hline \end{array} \]
Characters of sunflower ovule. I. Micropyle lie close to funicle
II. Inverted ovule
III. Ovule curvature is 180°
IV. Curved embryo sac
Which of the following is not a function of cytoskeleton?
‘A’ diploid chromosome number is 4. ‘B’ haploid chromosome number is twice to that of A. ‘C’ diploid chromosome number is thrice to that of A. A, B and C respectively are
Identify true sentences regarding leaf modifications in Nepenthes: \begin{itemize} \item[A.] Petiole upper part is modified into tendril. \item[B.] Petiole lower part is modified into phyllode. \item[C.] Lamina is modified into pitcher. \item[D.] Petiole lower part is modified into lid of pitcher. \end{itemize}