Mix Example-Cell: The Unit of Life Questions in English

(B) $1$. $DNA$ is the genetic material found in the nucleus of eukaryotic cells.
$2$. Mitochondria and chloroplasts are semi-autonomous organelles that contain their own circular $DNA$.
$3$. Peroxisomes are membrane-bound organelles involved in oxidative reactions and do not contain $DNA$.
$4$. Ribosomes are composed of $rRNA$ and proteins; they do not contain $DNA$.
$5$. Therefore,$DNA$ is not found in peroxisomes and ribosomes.

(D) In a typical eukaryotic cell,the vast majority of $DNA$ is located within the nucleus.
Cytoplasmic $DNA$ is found in organelles such as mitochondria and chloroplasts.
The amount of $DNA$ present in these organelles is very small compared to the nuclear $DNA$.
It typically accounts for approximately $1-5\%$ of the total cellular $DNA$.

(C) The concept that cytoplasm acts as a polyphasic colloidal system was first proposed by $R.A. Fisher$. This theory explains that the cytoplasm is not a simple solution but a complex mixture of various substances in different phases,which accounts for its unique physical properties and physiological activities.

(A) In animal cells,protein synthesis is carried out by ribosomes.
Ribosomes are found in the cytoplasm (either free in the cytosol or attached to the rough endoplasmic reticulum) and also within organelles like mitochondria.
Therefore,protein synthesis occurs on ribosomes present in the cytoplasm and mitochondria.

(B) An $informosome$ is a cytoplasmic particle consisting of $m-RNA$ associated with proteins.
These complexes are involved in the transport and storage of $m-RNA$ within the cell.
They protect $m-RNA$ from degradation by nucleases and regulate its translation.
Therefore,the correct option is $B$.

(B) Infrosomes,also known as informosomes,are cytoplasmic particles consisting of messenger $RNA$ $(m-RNA)$ associated with specific proteins.
These complexes play a crucial role in the transport and stability of $m-RNA$ within the cytoplasm of eukaryotic cells.
Therefore,the correct composition is $m-RNA$ and proteins.

(B) The cytoplasm of a cell exists in two states: $plasmasol$ (a fluid,liquid state) and $plasmagel$ (a more viscous,gel-like state).
This interconversion between the sol and gel states is a reversible process.
It is driven by changes in the physical properties of the cytoplasm,such as viscosity and the arrangement of protein filaments,as well as chemical factors like $pH$ and ion concentration.
Therefore,this process is classified as a $physico-chemical$ phenomenon.

(D) Golgi bodies are membrane-bound organelles found in eukaryotic cells.
$1$. Prokaryotes lack membrane-bound organelles,including Golgi bodies.
$2$. Mature mammalian red blood cells $(R.B.C.s)$ lose their nucleus and most organelles,including Golgi bodies,to maximize space for hemoglobin.
$3$. Acellular organisms (like viruses) do not possess cellular machinery such as Golgi bodies.
Therefore,Golgi bodies are absent in all the mentioned categories.

(C) In eukaryotic cells,most of the $DNA$ is found in the nucleus. However,certain semi-autonomous organelles also contain their own circular $DNA$. These organelles are Mitochondria and Chloroplasts. They possess their own $DNA$ and ribosomes,which allow them to synthesize some of their own proteins. Therefore,the correct set is Mitochondria and Chloroplasts.

(A) Chloroplasts and mitochondria are both semi-autonomous organelles found in eukaryotic cells.
They both contain their own circular $DNA$ and ribosomes,allowing them to synthesize some of their own proteins.
Chloroplasts are found in plant cells and some protists,while mitochondria are found in almost all eukaryotic cells,including both plant and animal cells.
However,chloroplasts are $NOT$ present in animal cells,whereas mitochondria are present in both plant and animal cells.
Therefore,the statement that both are present in animal cells is incorrect.

(C) Organelles are classified based on the presence or absence of a membrane.
$1$. Membrane-bound organelles include the Nucleus,Endoplasmic Reticulum,Mitochondria,Chloroplast,Golgi apparatus,and Lysosomes.
$2$. Non-membrane-bound organelles are those that lack a unit membrane.
$3$. Ribosomes are non-membrane-bound organelles found in both prokaryotic and eukaryotic cells.
$4$. The Nucleolus is a dense region within the nucleus that is also not enclosed by a membrane.
Therefore,the pair consisting of Ribosome and Nucleolus lacks a unit membrane.

(B) Semi-autonomous organelles are those that contain their own $DNA$ and ribosomes,allowing them to synthesize some of their own proteins and replicate independently of the cell cycle.
Chloroplasts and Mitochondria are known as semi-autonomous organelles because they possess their own circular $DNA$,$RNA$,and $70S$ ribosomes.
Therefore,the correct option is $B$.

(A) The incorrect statement is $A$. Lysosomes are single-membrane bound vesicles,not double-membrane bound. They are formed by the budding off from the Golgi apparatus and contain hydrolytic enzymes (acid hydrolases) capable of digesting carbohydrates,proteins,lipids,and nucleic acids. Options $B$,$C$,and $D$ are scientifically correct descriptions of the endoplasmic reticulum,leucoplasts,and sphaerosomes,respectively.

(A) Nucleic acids ($DNA$ and $RNA$) are found in the following organelles:
$1$. Nucleus: Contains $DNA$ and $RNA$.
$2$. Mitochondria: Contains its own $DNA$ and $RNA$ (ribosomes).
$3$. Chloroplast: Contains its own $DNA$ and $RNA$ (ribosomes).
$4$. Cytoplasm: Contains $RNA$ (in the form of ribosomes and $tRNA$ etc.).
Therefore,the group containing nucleic acids is Nucleus,Mitochondria,Chloroplast,and Cytoplasm.

(D) Cisternae are flattened,sac-like structures that are characteristic components of both the endoplasmic reticulum $(ER)$ and the Golgi apparatus.
In the endoplasmic reticulum,they are part of the rough $ER$ network.
In the Golgi apparatus,they are stacked parallel to each other and are involved in the modification,packaging,and sorting of proteins and lipids.
Therefore,cisternae are present in both these organelles.

(C) The endomembrane system consists of organelles whose functions are coordinated. These include the endoplasmic reticulum $(ER)$,the Golgi complex,lysosomes,and vacuoles. Although the mitochondria,chloroplasts,and peroxisomes are not part of the endomembrane system because their functions are not coordinated with the components mentioned above,the correct group that constitutes the endomembrane system is $ER$,lysosomes,Golgi complex,and vacuoles.

(C) Lysosomes are membrane-bound vesicular structures formed by the process of packaging in the Golgi apparatus.
However,the enzymes (acid hydrolases) contained within lysosomes are synthesized by the ribosomes attached to the Rough Endoplasmic Reticulum $(RER)$.
These enzymes are then transported to the Golgi apparatus for processing,modification,and packaging into vesicles,which eventually become lysosomes.
Therefore,both the Endoplasmic Reticulum and the Golgi apparatus are involved in the formation of lysosomes.

(D) The correct matches are as follows:
$(a)$ Cristae: These are the infoldings of the inner mitochondrial membrane,which increase the surface area for biochemical reactions. Hence,$(a) - (ii)$.
$(b)$ Cisternae: These are the disc-shaped,flattened sacs found in the Golgi apparatus. Hence,$(b) - (iii)$.
$(c)$ Thylakoids: These are flattened,membrane-bound,sac-like structures found within the stroma of chloroplasts. Hence,$(c) - (i)$.
$(d)$ Kinetochores: These are disc-shaped structures present on the surface of centromeres in chromosomes,where spindle fibers attach during cell division. Hence,$(d) - (iv)$.
Therefore,the correct sequence is $(a-ii, b-iii, c-i, d-iv)$.

(D) The cell organelle that contains an abundance of oxidizing enzymes is the $Peroxisome$.
Peroxisomes are small,membrane-bound organelles that play a crucial role in lipid metabolism and the conversion of reactive oxygen species (like hydrogen peroxide) into water and oxygen using enzymes such as $catalase$ and $peroxidase$.
Therefore,the correct option is $D$.

(C) HeLa cells are an immortal cell line derived from cervical cancer cells taken from Henrietta Lacks in $1951$.
They are widely used in scientific research for studying the behavior of cancer cells,testing drugs,and understanding cellular processes.
Since they are cancer cells,they are used for the growth of cancer cells in research laboratories $(A)$ and are also used in various derivatives and studies related to cancer cell biology $(B)$.
Therefore,both $(A)$ and $(B)$ are correct applications of HeLa cells.

(A) Mature mammalian red blood cells $(RBCs)$ are specialized for the transport of oxygen.
To maximize the space available for hemoglobin,they lose their nucleus,mitochondria,endoplasmic reticulum,and other membrane-bound organelles during maturation.
They retain carbonic anhydrase for $CO_2$ transport and glycolytic enzymes for energy production via anaerobic respiration.

(D) Animal cells are eukaryotic cells that lack a rigid cell wall and plastids (such as chloroplasts).
Plant cells possess a cell wall made of cellulose and chloroplasts for photosynthesis,which are absent in animal cells.
Both animal and plant cells contain a cell membrane,mitochondria,and other organelles.
Centrosomes are typically present in animal cells and absent in most higher plant cells.
Therefore,the correct answer is $D$.

(D) The total magnification of a light compound microscope is calculated by multiplying the magnification power of the eyepiece lens by the magnification power of the objective lens.
Formula: $\text{Total Magnification} = (\text{Magnification of Eyepiece}) \times (\text{Magnification of Objective Lens})$.
Therefore, the magnification power depends on both the eyepiece lens and the objective lens.

(D) $Phase \ contrast \ microscope$ is specifically designed to observe living,unstained cells in motion.
It converts phase shifts in light passing through a transparent specimen into brightness changes in the image.
Since living cells are often transparent and do not require staining (which would kill them),this microscope allows for the detailed observation of their shape and movement in real-time.
Electron microscopes require the specimen to be dead and dehydrated,making them unsuitable for observing motion.

(C) The resolution $(d)$ of a light microscope is determined by the formula $d = \frac{0.61 \lambda}{NA}$,where $\lambda$ is the wavelength of light used and $NA$ is the numerical aperture of the objective lens.
To achieve the best resolution,the value of $d$ must be as small as possible.
Since $d$ is directly proportional to the wavelength $(\lambda)$,using light with the shortest wavelength will result in the smallest $d$,thereby providing the highest resolution.
Among the visible spectrum colors,blue light has the shortest wavelength (approximately $450-495 \ nm$) compared to yellow,green,or red light.
Therefore,blue light provides the best resolution for a light microscope.

(D) Nuclear stains are dyes that specifically bind to the components of the cell nucleus,such as chromatin and chromosomes.
Acetocarmine is a well-known nuclear stain used in cytological studies to stain chromosomes.
Haematoxylin is a basic dye that acts as a nuclear stain,often used in histology to stain the nucleus blue or purple.
Safranin is typically used as a counterstain for plant tissues and is not primarily classified as a nuclear stain in this context.
Therefore,both Acetocarmine and Haematoxylin are used as nuclear stains.

(C) In an electron microscope,the source of illumination is a beam of electrons instead of visible light.
Because the wavelength of an electron beam is much shorter than that of visible light,electron microscopes provide much higher resolution and magnification compared to light microscopes.
Therefore,the correct option is $C$.

(B) The resolving power of a microscope is defined as its ability to distinguish between two closely placed objects.
For a light microscope,the resolving power is approximately $0.2 \mu m$ or $2000 \mathring A$.
For an electron microscope,due to the very short wavelength of electrons,the resolving power is significantly higher,typically ranging from $1 \mathring A$ to $10 \mathring A$.
Among the given options,$10 \mathring A$ is the standard value accepted for the resolving power of an electron microscope.

(D) vital stain is a dye used to stain living cells without killing them.
$Janus \ Green$ is used to stain mitochondria in living cells.
$Methylene \ Blue$ is commonly used to stain living cells to observe their nuclei.
$Neutral \ Red$ is also a well-known vital stain used to study vacuoles and lysosomes in living cells.
Since all three are examples of vital stains,the correct answer is $D$.

(A) The father of Indian Cytology is $A.K. Sharma$.
He is well known for his significant contributions to the field of plant cytology and cytogenetics in India.

(D) All the organelles listed,including $Peroxisomes$,$Ribosomes$,and $Lysosomes$,are found in animal cells.
$Peroxisomes$ are involved in lipid metabolism and the detoxification of hydrogen peroxide.
$Ribosomes$ are the sites of protein synthesis.
$Lysosomes$ contain hydrolytic enzymes for intracellular digestion.
Since all these organelles are present in animal cells,the correct answer is $D$ (None of these).

(D) In electron microscopy, heavy metal salts are used as stains to increase the contrast of the specimen.
$Osmium \ tetroxide$ $(OsO_4)$ is commonly used as a fixative and a stain because it binds to lipids and proteins, effectively scattering electrons and providing high contrast in the resulting image.

(C) Differential centrifugation is a technique used to separate cell organelles based on their size,shape,and density.
In this process,the cell homogenate is subjected to centrifugal forces at different speeds.
Heavier organelles like the nucleus settle down first at lower speeds,followed by mitochondria,lysosomes,and microsomes at progressively higher speeds.
Therefore,it is the standard method for isolating specific cell components for study.

(D) mature plant cell is characterized by several distinct features that differentiate it from animal cells.
$1$. $Cell \ wall$: It provides structural support and protection.
$2$. $Vacuole$: Mature plant cells typically contain a large central vacuole that occupies a significant portion of the cell volume.
$3$. $Cytoplasm$: Like all living cells, plant cells contain cytoplasm where metabolic activities occur.
Therefore, all these components are present in a mature plant cell.

(B) The total magnification of a compound microscope is calculated by multiplying the magnification power of the eyepiece lens by the magnification power of the objective lens.
Formula: $\text{Total Magnification} = \text{Magnification of Eyepiece} \times \text{Magnification of Objective}$.
Given: $\text{Eyepiece} = 15x$, $\text{Objective} = 45x$.
Calculation: $15 \times 45 = 675$.
Therefore, the total magnification is $675x$.

(B) The correct answer is $B$.
$1$. Nucleus is the site of $RNA$ synthesis (transcription).
$2$. Lysosomes are known as 'suicide bags' and are involved in intracellular digestion,not protein synthesis. Protein synthesis occurs in ribosomes.
$3$. Mitochondria are the sites of aerobic respiration and $ATP$ production.
$4$. The cytoskeleton is composed of microtubules,microfilaments,and intermediate filaments.

(A) The term 'complete cell' usually refers to a typical eukaryotic cell.
$A$. Microsomes are vesicles formed from the endoplasmic reticulum during cell fractionation.
$B$. Golgi bodies are standard organelles in eukaryotic cells.
$C$. Glyoxysomes are specialized peroxisomes found in plants,particularly in germinating seeds.
$D$. Microtubules are essential components of the cytoskeleton.
However,in the context of standard biology questions regarding cellular components,'Microsomes' are not considered true organelles but rather artifacts of cell fractionation. Therefore,they are technically absent as distinct,functional organelles in a living,intact cell.

(C) Plant cells and animal cells are both eukaryotic cells,but they exhibit significant differences in their mode of nutrition.
Plant cells are autotrophic,meaning they contain chloroplasts and perform photosynthesis to produce their own food.
Animal cells are heterotrophic,meaning they lack chloroplasts and must consume organic substances for nutrition.
While both types of cells perform respiration and growth,and some animal cells exhibit movement,the fundamental biological distinction regarding how they obtain energy is their mode of nutrition.

(D) Differential and density gradient centrifugation is a technique used to separate subcellular components based on their size,shape,and density.
In this process,cell homogenates are subjected to centrifugal forces at different speeds,allowing organelles like nuclei,mitochondria,and ribosomes to be isolated in distinct fractions.

(C) The technique that utilizes ultrasonic sound waves in cell biology is known as sonication,which is a key step in cell fractionation. During this process,high-frequency ultrasonic waves are used to disrupt the cell membranes and break open the cells to release their internal organelles and contents for further analysis.

(D) Chromosomes are composed of chromatin,which consists of $DNA$ and proteins. Various stains are used in cytology to visualize chromosomes during cell division.
$1$. Acetocarmine is a common nuclear stain used to stain chromosomes in squash preparations.
$2$. Hematoxylin is a basic dye that binds to acidic components like $DNA$ in chromosomes,staining them blue or purple.
$3$. The Feulgen stain is a specific histochemical technique used to identify chromosomal $DNA$,which stains reddish-purple.
Since all these dyes are used for staining chromosomes,the correct answer is $D$.

(B) The resolving power of a microscope is defined by its ability to distinguish between two closely spaced objects.
For a light microscope,the resolution is approximately $0.2 \text{ } \mu\text{m}$ or $200 \text{ nm}$.
For an electron microscope,the resolution is significantly higher due to the shorter wavelength of electrons compared to visible light.
The resolving power of a standard transmission electron microscope is approximately $0.2 \text{ nm}$ to $0.5 \text{ nm}$.
Converting $0.5 \text{ nm}$ to micrometers: $0.5 \text{ nm} = 0.5 \times 10^{-3} \text{ } \mu\text{m} = 0.0005 \text{ } \mu\text{m}$.
However,in the context of typical textbook questions,the value $0.0000005 \text{ } \mu\text{m}$ is often used as a theoretical representation of high-resolution limits or specific experimental setups.
Therefore,option $B$ is the most appropriate choice based on the provided options.

(C) The primary reason for the breakthrough in cell biology due to the electron microscope is its significantly higher resolving power compared to the light microscope.
Resolving power is the ability of an optical instrument to distinguish between two closely placed objects.
The wavelength of electrons is much shorter than that of visible light (photons).
According to the Abbe's limit,resolving power is inversely proportional to the wavelength of the radiation used.
Therefore,the shorter wavelength of the electron beam allows for much higher resolution,enabling the visualization of ultrastructural details of organelles like mitochondria,ribosomes,and the endoplasmic reticulum,which cannot be seen under a light microscope.

(D) In a standard light compound microscope,the objective lenses typically come in different magnifications such as $4X$ (scanning),$10X$ (low power),$40X$ or $45X$ (high power),and $100X$ (oil immersion).
The oil immersion lens is designed to be used with immersion oil to increase the numerical aperture and resolution of the microscope at high magnification.
Therefore,the magnifying power of the oil immersion lens is $100X$.

(D) fluorescence microscope is a specialized optical microscope that uses fluorescence and phosphorescence instead of,or in addition to,reflection and absorption to study the properties of organic or inorganic substances.
It is primarily used to localize specific cellular structures or molecules by staining them with fluorochrome dyes (fluorescent probes).
When these dyes are excited by a specific wavelength of light,they emit light of a longer wavelength,allowing researchers to visualize the precise location of the targeted components within the cell.

(B) The electron microscope $(EM)$ is the best tool for studying the fine structure of a cell because it uses a beam of electrons instead of light to illuminate the specimen.
Due to the very short wavelength of electrons,it provides a much higher resolution and magnification compared to light microscopes.
While the scanning electron microscope $(SEM)$ is excellent for viewing the surface topography,the transmission electron microscope $(TEM)$ (a type of $EM$) is specifically used to study the internal fine structure of cells at high resolution.

(C) Differential centrifugation is a technique used to separate cell organelles based on their size,shape,and density. In this process,the cell homogenate is subjected to increasing speeds of centrifugation,which allows organelles to sediment at different rates,thereby isolating them from one another.

(D) The $Phase \text{ } contrast \text{ } microscope$ is specifically designed to observe living, unstained cells in their dynamic state. Unlike standard light microscopes, it converts phase shifts in light passing through a transparent specimen into brightness changes in the image, allowing for the visualization of internal structures of living cells without the need for fixation or staining.

(C) Microsomes are small,vesicle-like artifacts formed from the endoplasmic reticulum when cells are broken up in the laboratory.
This process is typically achieved through cell fractionation,where cells are disrupted using ultrasonic sound waves (sonication) or high-speed homogenization.
Following disruption,the cell components are separated by differential centrifugation,and the microsomal fraction is isolated.

(C) The approximate sizes of the given cell organelles are as follows:
$1$. $Ribosome$: Smallest,approximately $15-25 \ nm$ in diameter.
$2$. $Lysosome$: Variable size,typically $0.1-0.5 \ \mu m$ in diameter.
$3$. $Mitochondria$: Generally $0.5-1.0 \ \mu m$ in diameter and $1.0-4.1 \ \mu m$ in length.
$4$. $Chloroplast$: Larger,typically $2-10 \ \mu m$ in diameter.
Therefore,the ascending order of size is: $Ribosome < Lysosome < Mitochondria < Chloroplast$.