List of top Botany Questions asked in Telangana State Engineering Agriculture & Medical CET

Match the following % Table \begin{table}[h!] \renewcommand{\arraystretch}{0.9} % For better row height \centering \begin{tabular}{|l|l|} \hline List - I & List - II
\hline A) ABO blood groupings & I) Frame-shift mutations
\hline B) Starch synthesis in pea seeds & II) Sickle cell anemia
\hline C) Point mutation & III) Pleiotropy
\hline D) Deletion of DNA base pairs & IV) Co-dominance
\hline \end{tabular} \end{table}
The correct answer is

In Morgan’s experiments on Drosophila linkage, the percentage of white-eyed, yellow body recombinants and the percentage of white-eyed, miniature winged recombinants in F\textsubscript{2 generation respectively:}

Match the following: \renewcommand{\arraystretch}{1.3} \begin{table}[h!] \centering \setlength{\tabcolsep}{10pt} \begin{tabular}{|c|c|c|c|} \hline S. No & Crop & Variety & Resistance to Disease
\hline I & Wheat (గోధుమ) & Pusa sadabahar (A) & Leaf and stripe rust (W)
II & Brassica (క్రూసిఫెరే) & Pusa komal (B) & White rust (X)
III & Cow Pea (చిక్ పీస్) & Pusa swarnim (C) & Bacterial blight (Y)
IV & Chilli (మిర్చి) & Himagiri (D) & Leaf curl (Z)
\hline \end{tabular} \end{table}

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.

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.

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

Dual functions of codon AUG

Find out A, B, C, and D products of anaerobic respiration pathway cited below \includegraphics[width=0.4\linewidth]{5image.png}

Observe the diagrammatic representation of ATP synthesis in oxysome of mitochondria. Identify A, B and C respectively. \includegraphics[width=0.3\linewidth]{3image.png}

Assertion (A): Abscisic acid is called stress hormone.
Reason (R): Abscisic acid increases the tolerance of plants to various kinds of stresses.

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

Assertion (A): In Arachis, fruits are geocarpic.
Reason (R): Seeds are non-endospermic in Arachis.

Identify the enzyme of the following reaction: \[ \text{Phosphoglyceric acid + ATP} \xrightarrow{\text{Enzyme}} \text{Bisphosphoglyceric acid + ADP} \]

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:

Match the following: \[ \begin{array}{|c|c|} \hline \text{List-I (Element)} & \text{List-II (Significance)}
\hline A. \text{Copper} & I. \text{Activator of carboxylase}
B. \text{Zinc} & II. \text{Component of nitrogenase}
C. \text{Molybdenum} & III. \text{Chlorophyll structure}
D. \text{Magnesium} & IV. \text{Component of cytochrome C}
\hline \end{array} \]

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}

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}