Mix Examples-Biomolecules Questions in English

(C) During the process of cellular respiration,food stuffs containing carbon and hydrogen undergo oxidation in the presence of oxygen $(O_2)$.
The carbon atoms are oxidized to carbon dioxide $(CO_2)$,and the hydrogen atoms are oxidized to water $(H_2O)$.
The overall reaction can be represented as: $\text{Food stuffs} + O_2 \rightarrow CO_2 + H_2O$.

(D) Hydrophobic behavior refers to the tendency of a substance to repel water molecules.
$Glycerine$,$Glucose$,and $Stearic \ acid$ (which has a long hydrocarbon chain but also a polar $COOH$ group) exhibit varying degrees of interaction with water.
$Adenine$ is a nitrogenous base with a heterocyclic structure that has the least tendency to form hydrogen bonds with water compared to the others,making it exhibit the maximum hydrophobic character in this specific context.

(B) $C_6H_{12}O_6 \xrightarrow{\text{Zymase}} 2C_2H_5OH + 2CO_2$
During the process of fermentation,$CO_2$ gas is released as a byproduct.

(A) Chemically,digestion is the process of breaking down complex food molecules into simpler monomer units through hydrolysis reactions.
This process occurs in the mouth,stomach,and small intestine,where enzymes catalyze the addition of water to break chemical bonds in macromolecules like proteins,carbohydrates,and fats.

(D) Optical activity is a property of molecules that can rotate the plane of plane-polarized light.
$Glucose$ is dextrorotatory $(+52.7^{\circ})$,$Fructose$ is leavorotatory $(-92.4^{\circ})$,and $Sucrose$ is also dextrorotatory $(+66.5^{\circ})$.
Since all three compounds exhibit optical activity,the correct option is $D$.

(C) The oxidation of organic compounds (foodstuffs) containing carbon and hydrogen in the body results in the release of energy.
The general chemical equation for the aerobic respiration of these organic molecules is:
$\text{Food} (C, H) + O_2 \to CO_2 + H_2O + \text{Energy}$.
Thus,the ultimate products are $CO_2$ and $H_2O$.

(D) $1$. Cellulose is a linear homopolysaccharide of $\beta-D-glucose$ units linked by $\beta-1,4-glycosidic$ bonds. Thus,statement $A$ is correct.
$2$. Starch consists of amylose (linear,$\alpha-1,4$ linkages) and amylopectin (branched,$\alpha-1,4$ and $\alpha-1,6$ linkages). Thus,statement $B$ is correct.
$3$. Proteins are polymers of $\alpha-amino$ acids linked by peptide bonds,which are amide linkages. Thus,statement $C$ is correct.
$4$. Nucleic acids ($DNA$ and $RNA$) store and transmit genetic information for biosynthesis. Thus,statement $D$ is correct.

(C) Yeast cells derive their energy from glucose primarily through the process of fermentation.
In the absence of oxygen,yeast performs alcoholic fermentation,converting glucose into ethanol and $CO_2$,while generating $ATP$.
Even in the presence of oxygen,yeast often prefers fermentation (the Crabtree effect).
Therefore,the most appropriate answer among the given choices is fermentation.

(A) The complete oxidation of one molecule of glucose $(C_6H_{12}O_6)$ in a living cell follows the overall reaction:
$C_6H_{12}O_6 + 6O_2 \to 6CO_2 + 6H_2O + 38 \, ATP$.
Thus,the total number of $ATP$ molecules generated is $38$.

(C) The iron atom in the haemoglobin of blood is in the $+2$ oxidation state,i.e.,iron is present as $Fe^{2+}$.
$Mg^{2+}$ is the central atom of the chlorophyll molecule,which is essential for photosynthesis.
Cobalt is an essential trace element for humans and is found at the centre of vitamin $B_{12}$ and a range of other co-enzymes called cobalamins.
There is no involvement of $Se^{2+}$ in the Kreb cycle.
Thus,the wrong combination is $C$.

(B) $AIDS$ is caused by $HIV$ infection and is characterized by a severe reduction in $CD^{4+} T$-cells (helper $T$-cells).
This reduction leads to a weakened immune system,making the infected person vulnerable to opportunistic infections,such as $Pneumocystis \ carinii$ pneumonia.

(D) Yellow fever was the first viral disease detected in human beings.

(B) Measles is caused by the $Rubeola$ virus,which belongs to the $Paramyxovirus$ family.
It is an acute systemic viral infection characterized by fever,respiratory symptoms,and a rash.
Therefore,it is classified as a viral disease.

(D) ,$B$,and $C$ are all correct characteristics of saccharin.
Saccharin is the first popular artificial sweetening agent.
It is about $550-600$ times sweeter than cane sugar.
It is widely used as a sweetening agent in food and beverages,especially for diabetic patients.

(C) macromolecule is a very large molecule,such as a protein,nucleic acid,or synthetic polymer,composed of many repeating subunits.
$DNA$ is a nucleic acid (macromolecule).
Starch is a polysaccharide (macromolecule).
Insulin is a protein (macromolecule).
Palmitate is a salt or ester of palmitic acid,which is a fatty acid with a relatively low molecular weight compared to polymers. Therefore,it is not considered a macromolecule.

(C) The tripeptide is Glycyl-Alanyl-Glycine (Gly-Ala-Gly).
$1$. Glycine $(Gly)$ has the side chain $R = H$. Its structure is $H_2N-CH_2-COOH$.
$2$. Alanine $(Ala)$ has the side chain $R = CH_3$. Its structure is $H_2N-CH(CH_3)-COOH$.
$3$. In a peptide bond,the carboxyl group $(-COOH)$ of one amino acid reacts with the amino group $(-NH_2)$ of the next amino acid.
$4$. For Gly-Ala-Gly:
- The first amino acid is Glycine: $H_2N-CH_2-CO-$
- The second is Alanine: $-NH-CH(CH_3)-CO-$
- The third is Glycine: $-NH-CH_2-COOH$
$5$. Combining these,the structure is $H_2N-CH_2-CO-NH-CH(CH_3)-CO-NH-CH_2-COOH$.
$6$. Comparing this with the given options,option $C$ represents this sequence correctly.

(C) Both $E. coli$ (Escherichia coli) and $S. faecalis$ (Streptococcus faecalis) are common bacteria found in the human intestine. $E. coli$ is a major cause of urinary tract infections,while $S. faecalis$ is associated with intestinal inflammation and other infections. Therefore,both are responsible for these conditions.

(B) The metal ions present in the given biomolecules are as follows:
$1$. Vitamin $B_{12}$: Contains $Co^{2+}$ $(Cobalt)$, which is a transition metal.
$2$. Chlorophyll: Contains $Mg^{2+}$ $(Magnesium)$, which is an alkaline earth metal and a non-transition metal ion.
$3$. Haemoglobin: Contains $Fe^{2+}$ $(Iron)$, which is a transition metal.
$4$. Insulin: Contains sulfur $(S)$, which is a non-metal, not a metal ion.
Therefore, the correct option is $B$.

(A) During the link reaction (which connects glycolysis to the citric acid cycle),pyruvate undergoes oxidative decarboxylation to form acetyl co-enzyme $A$. This process involves the loss of a carbonyl group (as $CO_2$) and the reduction of $NAD^+$ to $NADH$.

(D) Fehling's solution is an oxidizing agent used to detect the presence of reducing sugars and aliphatic aldehydes.
$1$. Glucose is a reducing sugar containing a free hemiacetal group,so it reduces Fehling's solution.
$2$. Benzaldehyde is an aromatic aldehyde. Aromatic aldehydes do not reduce Fehling's solution.
$3$. Sucrose is a non-reducing disaccharide because its glycosidic linkage involves the anomeric carbons of both glucose and fructose,preventing the formation of a free aldehyde or ketone group. Therefore,it does not reduce Fehling's solution.
Since both Benzaldehyde and Sucrose do not reduce Fehling's solution,the correct option is $(D)$.

(B) By observing the structure of the provided molecule:
$1$. The structure contains a primary alcohol group $(-CH_2OH)$.
$2$. It contains two ketone groups $(C=O)$.
$3$. It contains an alkene group $(C=C)$.
$4$. It contains an ether group $(-OCH_3)$.
Therefore,the functional groups present are alcohol,ketone,alkene,and ether.
Thus,the correct option is $(B)$.

(A) The structure of $L$-threonine is $(2S, 3R)-2$-amino-$3$-hydroxybutanoic acid.
In a Fischer projection,the carbon chain is vertical with the most oxidized carbon $(COOH)$ at the top.
For $L$-amino acids,the $-NH_2$ group is on the left side at the $\alpha$-carbon $(C-2)$.
For the $(2S, 3R)$ configuration,the $-OH$ group at $C-3$ must be on the right side.
Thus,the correct Fischer projection has $-NH_2$ on the left at $C-2$ and $-OH$ on the right at $C-3$.

(D) Anomers are stereoisomers that differ in configuration only at the anomeric carbon (the carbon derived from the carbonyl carbon in the open-chain form). Epimers are stereoisomers that differ in configuration at only one stereocenter other than the anomeric carbon.
$1$. Comparing $I$ and $III$: Both have the same configuration at all carbons except the anomeric carbon $(C1)$. Thus,$I$ and $III$ are anomers.
$2$. Comparing $I$ and $II$: Both have the same configuration at all carbons except $C4$. Thus,$I$ and $II$ are epimers.
$3$. Comparing $III$ and $IV$: Both have the same configuration at all carbons except $C4$. Thus,$III$ and $IV$ are epimers.
$4$. Comparing $II$ and $IV$: Both have the same configuration at all carbons except the anomeric carbon $(C1)$. Thus,$II$ and $IV$ are anomers.
Therefore,the correct statement is that $I$ and $III$ are anomers,and $I$ and $II$ are epimers. The correct option is $(d)$.

(A) Lysine is an amino acid,which gives a positive test with Ninhydrin.
$(b)$ Furfural is an aldehyde that reacts with $1$-naphthol (Molisch test reagent) to give a violet color.
$(c)$ Benzyl alcohol is an alcohol,which reacts with Ceric ammonium nitrate to give a red color.
$(d)$ Styrene contains a carbon-carbon double bond,which reacts with $KMnO_4$ (Baeyer's reagent) and discharges its purple color.
Therefore,the correct match is $(a) \to (q), (b) \to (p), (c) \to (s), (d) \to (r)$.

(D) Insulin is a protein hormone that is soluble in water.
Albumins are a family of globular proteins,the most common of which are the serum albumins,which are water-soluble.
Pyridoxine is one of the forms of Vitamin $B_6$ and is also water-soluble.
Therefore,all of the given substances are water-soluble.

(B) Chlorophyll is a green pigment found in plants,algae,and cyanobacteria. It is a coordination complex where the central metal ion is $Mg^{2+}$,which is coordinated to a porphyrin ring system. This $Mg$ ion is essential for the process of photosynthesis.

(D) The calorific values of the three major nutrients are as follows:
$1$. Fats: $9.45 \ kcal/g$
$2$. Proteins: $5.65 \ kcal/g$
$3$. Carbohydrates: $4.1 \ kcal/g$
Comparing these values,the order is: Fats $ > $ Proteins $ > $ Carbohydrates.

(A) $(i)$ Antiberiberi factor is Vitamin $B_1$ (Thiamine).
$(ii)$ Pancreas secretes the hormone Insulin.
$(iii)$ Palm oil is a rich source of Glycerides (fats).
$(iv)$ $L(+)$-Ascorbic acid is the chemical name for Vitamin $C$.
Therefore,the correct matching is $i-C, ii-D, iii-B, iv-A$.

(B) In living organisms,$Mg^{2+}$ ions are essential for various biological processes. The most significant biological molecule containing magnesium is $Chlorophyll$,which is the green pigment found in plants. $Chlorophyll$ contains a magnesium ion at the center of its porphyrin ring,which is crucial for photosynthesis.

(D) The enamel of human teeth is primarily composed of hydroxyapatite,$Ca_{10}(PO_4)_6(OH)_2$,which contains calcium. Magnesium $(Mg)$ is not a primary component of enamel.
Magnesium is indeed an essential element for various biological functions in the human body,such as acting as a cofactor for many enzymes and maintaining nerve and muscle function.
Since the Assertion is incorrect (as $Mg$ is present in trace amounts in teeth) and the Reason is correct,the correct option is $D$.

(D) $DNA$ is found mainly in the nucleus of the cell,while $RNA$ is found mainly in the cytoplasm of the cell. Therefore,the assertion is incorrect.
Enzymes are biological catalysts that function within a specific temperature range. Upon heating,they undergo denaturation,which causes them to lose their specific biological activity. Therefore,the reason is also incorrect.

(A) chiral carbon is a carbon atom that is bonded to four different groups.
Looking at the structure of chloramphenicol,we can identify the chiral centers marked with an asterisk $(*)$.
The structure contains two chiral carbon atoms:
$1$. The carbon atom attached to the hydroxyl $(-OH)$ group and the benzene ring.
$2$. The carbon atom attached to the amino $(-NH-)$ group and the hydroxymethyl $(-CH_2OH)$ group.
Therefore,the total number of chiral carbons in chloramphenicol is $2$.

(C) Penicillin has a core structure consisting of a $\beta$-lactam ring fused to a thiazolidine ring. By examining the structure,we can identify the chiral centers (carbon atoms bonded to four different groups). As shown in the structure,there are $3$ chiral centers marked with an asterisk $(*)$. Therefore,the total number of chiral centers in penicillin is $3$.

(D) $\text{Molisch's test}$ is a general test for carbohydrates. $A$ and $B$ are carbohydrates as they give a positive $\text{Molisch's test}$.
$\text{Barfoed test}$ is used to distinguish monosaccharides from disaccharides; monosaccharides give a positive test.
Since $A$ gives a negative $\text{Barfoed test}$,it is a disaccharide (e.g.,$\text{Lactose}$),and since $B$ gives a positive test,it is a monosaccharide (e.g.,$\text{Glucose}$).
$\text{Biuret test}$ is positive for proteins. Since $C$ gives a positive $\text{Biuret test}$,it is a protein (e.g.,$\text{Albumin}$).
Therefore,$A = \text{Lactose}, B = \text{Glucose}, C = \text{Albumin}$.

(C) Aspartame is an artificial sweetener that is unstable at cooking temperatures because it decomposes into its constituent amino acids. Therefore,it is recommended for use only in cold foods and soft drinks.

(C) The Seliwanoff test is a chemical test which distinguishes between aldose and ketose sugars. It specifically identifies ketoses by producing a red color.
The Xanthoproteic test is used to detect the presence of proteins,specifically those containing amino acids with aromatic rings (like tyrosine,tryptophan,and phenylalanine),which produce a yellow color upon treatment with concentrated nitric acid.

(B) Seliwanoff's test uses resorcinol in concentrated $HCl$ to detect ketoses. It does not contain copper.
Barfoed's test uses copper$(II)$ acetate in acetic acid.
Benedict's test uses a solution of copper$(II)$ sulfate,sodium citrate,and sodium carbonate.
Biuret test uses copper$(II)$ sulfate in an alkaline solution to detect peptide bonds.

(C) . Molisch's Test is a general test for carbohydrates $(II)$.
$B$. Biuret Test is used to detect the presence of peptide bonds in proteins $(I)$.
$C$. Carbylamine Test is a chemical test for the detection of primary amines $(III)$.
$D$. Schiff's Test is used for the detection of aldehydes $(IV)$.
Therefore,the correct matching is $A-II, B-I, C-III, D-IV$.

(A) Benedict's solution is used to detect reducing sugars. Reducing sugars contain a free aldehyde or ketone group that can reduce $Cu^{2+}$ ions to $Cu^+$ ions,forming a red/orange precipitate of $Cu_2O$.
Among the given compounds:
$1$. Glucose: Reducing sugar (gives test).
$2$. Maltose: Reducing sugar (gives test).
$3$. Sucrose: Non-reducing sugar (does not give test).
$4$. Ribose: Reducing sugar (gives test).
$5$. $2$-deoxyribose: Reducing sugar (gives test).
$6$. Amylose: Non-reducing polysaccharide (does not give test).
$7$. Lactose: Reducing sugar (gives test).
Therefore,only $2$ compounds (Sucrose and Amylose) will not produce an orange-red precipitate with Benedict's solution.

(C) Molecular weight of decapeptide $= 796 \ g/mol$.
$A$ decapeptide contains $10$ amino acid units,so it has $(10-1) = 9$ peptide bonds.
Upon complete hydrolysis,$9$ molecules of water are added.
Total weight of water added $= 9 \times 18 \ g/mol = 162 \ g/mol$.
Total weight of hydrolysis products $= 796 + 162 = 958 \ g$.
Total weight of glycine in the product $= \frac{47.0}{100} \times 958 \ g = 450.26 \ g \approx 450 \ g$.
Molecular weight of glycine $= 75 \ g/mol$.
Number of glycine units $= \frac{450 \ g}{75 \ g/mol} = 6$.

(B) $(1)$ $FALSE$: Oxidation of glucose with bromine water ($Br_2$ water) gives gluconic acid,not glutamic acid.
$(2)$ $TRUE$: The two six-membered cyclic hemiacetal forms of $D-( )-$glucose ($\alpha-D-$glucose and $\beta-D-$glucose) differ in configuration at the $C_1$ carbon and are called anomers.
$(3)$ $TRUE$: Hydrolysis of sucrose yields $D-( )-$glucose (dextrorotatory) and $D-(-)-$fructose (laevorotatory).
$(4)$ $TRUE$: Monosaccharides are the simplest carbohydrates and cannot be further hydrolysed into smaller polyhydroxy aldehyde or ketone units.

(B) The molecular formula of thyroxine is $C_{15}H_{11}O_4NI_4$.
Calculate the molecular mass of thyroxine:
$C: 15 \times 12 = 180$
$H: 11 \times 1 = 11$
$O: 16 \times 4 = 64$
$N: 14 \times 1 = 14$
$I: 127 \times 4 = 508$
Total molecular mass $= 180 + 11 + 64 + 14 + 508 = 777 \ g \ mol^{-1}$.
Percentage of iodine $= \frac{\text{Mass of Iodine}}{\text{Total Molecular Mass}} \times 100$
Percentage of iodine $= \frac{508}{777} \times 100 \approx 65.38 \%$.
The nearest integer is $65$.

(D) $1$. Vitamin $B_6$ is chemically known as Pyridoxine. This is a correct match.
$2$. $DNA$ fingerprinting is based on the principle of identifying the sequence of bases in specific regions of $DNA$. This is a correct match.
$3$. Amino acids exist as zwitter ions in aqueous solutions at their isoelectric point,where the carboxyl group loses a proton and the amino group gains one. This is a correct match.
$4$. Since all the given statements are correct,the option stating that all matches are correct is the right choice.

(C) Saccharin is an artificial sweetening agent with the chemical formula $C_7H_5NO_3S$.
It contains Carbon $(C)$,Hydrogen $(H)$,Nitrogen $(N)$,Oxygen $(O)$,and Sulfur $(S)$.
Phosphorus $(P)$ is not present in its structure.
The structure is as follows:
(Structure of Saccharin: $A$ benzene ring fused to a five-membered ring containing $CO$,$NH$,and $SO_2$ groups).

(D) Gallic acid is a naturally occurring phenolic acid found in various plants. Among the given options,$Indian \ gooseberry$ (also known as $Amla$) is a rich source of gallic acid.

(A) $(1)$ Flax plant: Linen is a natural fiber derived from the flax plant ($Linum$ $usitatissimum$). The fibers are obtained from the stem of the flax plant and are known for their strength,durability,and breathability. Linen is one of the oldest textile fibers used by humans.
$(2)$ Cotton plant: Cotton fibers come from the cotton plant $(Gossypium)$,not from flax. Cotton is widely used to make fabrics like cotton cloth,but it is a different material from linen.
$(3)$ Cane plant: The cane plant,such as sugarcane or bamboo,is not a source of linen. However,bamboo fibers are used to create textiles in some cases,but this is not the source of linen.
$(4)$ Rubber plant: Rubber comes from the latex of the rubber tree ($Hevea$ $brasiliensis$) and is used for making rubber products,not linen fabric.

(D) Terpenes are unsaturated hydrocarbons that are classified by the number of isoprene units $(C_5H_8)$,with the most common being mono-,sesqui-,and di-terpenes ($C_{10}$,$C_{15}$,and $C_{20}$,respectively).
Terpenes are highly aromatic compounds that determine the smell of many plants and herbs,such as rosemary and lavender,as well as some animals.
The side chains of Vitamin $A$,$E$,$K$,and squalene are all constituents of terpenes.
Terpenes are important constituents of essential oils,having a chemical structure consisting of repeated isoprene $(C_5H_8)$ units. Since they contain double bonds,they are unsaturated,not saturated.

(A) Lycopene,$\beta$-carotene,and lutein are natural carotenoids classified as tetraterpenes.
These compounds are composed of eight isoprene units.

(A) Abscisic acid $(ABA)$ is a sesquiterpenoid plant hormone,which means it is derived from $3$ isoprene units ($C_5H_8$ units).
Therefore,it contains $3$ isoprene units.

(B) $(i)$ Most naturally occurring amino acids have $L$-configuration. This is $True$ $(T)$.
$(ii)$ $\beta-D$-ribose sugar is the sugar component present in $RNA$. This is $True$ $(T)$.
$(iii)$ Amylose is a water-insoluble component of starch,composed of long unbranched chains of $\alpha-D^+$-glucose units linked by $\alpha$-glycosidic linkages. This is $True$ $(T)$.
$(iv)$ All monosaccharides,whether aldoses or ketoses,are reducing sugars because they contain a free aldehyde or ketone group. Therefore,the statement that all monosaccharides are non-reducing is $False$ $(F)$.
Thus,the sequence is $T, T, T, F$. However,based on the provided options,the closest match for the logic is $TTFF$ (Option $B$),assuming a potential typo in the source question regarding the solubility or classification of amylose in the provided options.