Category: Cell Biology Objective Questions

Cell Biology Multiple Choice Questions & Answers (MCQs)on “Genes, Chromosomes and Genomes – 1”.

1. The segregation of alleles on one trait did not have any effect on the segregation of alleles on a different trait. This is based on ____________
A. Mendel’s law of Heredity
B. Mendel’s law of Dominance
C. Mendel’s law of Independent Assortment
D. Mendel’s law of Segregation
Answer: C
Clarification: There are three Mendel’s laws n Heredity. 1. The Mendel’s law of Dominance states that when a dominant individual is crossed with a recessive individual, the F1 generation offspring will exhibit the dominant trait. 2. According to Mendel’s law of Segregation, a pair of allele governing a single trait segregates from each other during the formation of gametes. 3. According to Mendel’s law of Independent Assortment, the segregation of alleles on one trait did not have any effect on the segregation of alleles on a different trait.

2. Genes that show tendency to be inherited together is known as ____________
A. Linkage group
B. Homologous group
C. Co-dependent genes
D. None of the mentioned
Answer: A
Clarification: Some genes are packed together on a chromosome and tend to pass on from parents to offsprings as a group. These genes form a unit during the formation of gametes and are not separated during gamete formation. These genes on the same chromosome act like they are linked to one another and hence, are known as linkage groups.

3. Giant polytene chromosomes are found in ___________
A. Egg of fruit fly
B. Salivary gland of larvae of fruit fly
C. Salivary gland of adult fruit fly
D. All of the mentioned
Answer: B
Clarification: Polytene chromosomes are a type of giant chromosomes which are many times longer in length than normal chromosomes. Polytene chromosomes are found in the salivary gland of a larval fruit fly and show several distinct darkly stained bands.

4. An example of co-dominance is ____________
A. Mouse coat colour
B. Human ABO blood group system
C. Human hand size
D. Human eye colour
Answer: B
Clarification: When two different alleles are crosses and neither of them is dominant and recessive and both get expressed in the phenotype offspring, then such a genetic scenario is known as codominance. In the ABO Blood group system of humans, a person with A protein has blood type A and a person with B protein has blood type B. If a person with blood type A mates with person with blood type B, then the dominant A and B genes will be co-expressed in the offspring and he will have blood type ‘AB’.

5. Loops in lampbrush chromosomes represent site of _____________
A. Replication
B. Transcription
C. Cell division
D. Crossing over
Answer: b
Clarification: Lampbrush chromosomes are a type of giant chromosomes found in the growing oocytes of amphibians. Twin loops arise on either side of the chromosome in meiotic prophase. This is due to the active transcription of many genes.

6. How can it be determined whether the parent progeny is homozygous or heterozygous?
A. Test cross
B. Back cross
C. Monohybrid cross
D. Reciprocal cross
Answer: A
Clarification: Test cross is the cross of a F1 generation offspring with the homozygous recessive parent. If the offspring phenotype is dominant: recessive = 1:1, the F1 offspring is heterozygous dominant. If all the plant exhibits dominant phenotype, the F1 offspring is homozygous dominant. In this way, the genotype of an offspring can be determined.

7. The expression of Holandric genes causes the following genetic trait ______________
A. Haemophilia
B. Colour blindness
C. Down’s syndrome
D. Hypertrichosis
Answer: D
Clarification: Y-linked genes are called Holandric genes. Hypertrichosis is the genetic trait by which hair appears on the earlobe of an individual. The gene for Hypertrochosis is present in the Y chromosome and thus this trait is only expressed in males. This trait is governed by holandric gene.

8. The method of DNA replication is _____________
A. conservative
B. semi-conservative
C. non-conservative
D. disruptive
Answer: B
Clarification: In DNA replication , the two daughter strands unwind and are used as template for the formation of a new DNA strand. As a result, one strand in conserved and a new strand is produced by taking the conserved strand as a template. Thus, DNA replication is a semi-conservative process.

9. The DNA binding proteins bind at the _______________
A. Minor groove
B. Major groove
C. Phosphate molecules
D. Pentose sugars
Answer: B
Clarification: The DNA double strands are present as a double-helix as they wound around each another. Due to base stacking, alternative minor and major grooves are formed in the helix. Major groove provide the site for DNA protein binding.

10. During DNA denaturation, which of the following occurs?
A. Unwinding of DNA double strand
B. Absorbance of UV rays
C. Decrease in hydrophobic interactions of base stacking
D. All of the mentioned
Answer: D
Clarification: DNA denaturation takes place either due to thermal treatment or due to treatment in saline solution. During denaturation, DNA double strands get separated and there is an increase in the absorbance of UV rays by the dissolved DNA. Once the two DNA strands have separated, the hydrophobic interactions that result from base stacking are greatly decreased.

11. Human Genome Project was focused on discovering the details of ___________
A. sNRPs
B. 3D DNA structure
C. Junk DNA
D. VNTRs
Answer: C
Clarification: The Human Genome Project was an international scientific research project with the goal of determining the sequence of nucleotide base pairs that make up human DNA from a period of 1990 to 2003. The main purpose of these projects was to find out the base sequence of human junk DNA which was different for every individual.

Cell Biology Problemson “Mitochondrial Structure and Function – 2”.

1. Glycolysis takes place in _____
A. Outer membrane of mitochondria
B. Inner membrane of mitochondria
C. Mitochondrial matrix
D. Cytosol
Answer: D
Clarification: The process of cellular respiration involves the following steps- Glycolysis, Kreb’s cycle and electron oxidation. The Glycolysis stage occurs in cytoplasm and the Kreb’s cycle occurs in the matrix of mitochondria.

2. The first stable compound of Kreb’s cycle is _____________
A. Citrate
B. Cis- Aconitate
C. Oxaloacetate
D. Malate
Answer: A
Clarification: Kreb’s cycle has Citrate as its first stable compound. That is why it is also called Citric Acid Cycle. Citrate is a compound containing three carboxylic acid groups and it is thus also called Tricarboxylic Acid cycle or TCA cycle.

3. How many ATP molecules are produced per glucose molecules in eukaryotic Glycolysis?
A. 2 ATP
B. 3 ATP
C. 4 ATP
D. 6 ATP
Answer: A
Clarification: The process of cellular respiration involves the following steps- Glycolysis, Kreb’s cycle and electron oxidation, which are involved in energy production by utilizing carbohydrate. Energy is produced in the form of Adenosine triphosphate (ATP) molecules with 2 ATP molecules produced per glucose molecule in Glycolysis.

4. How many ATP molecules are produced per Krebs’ Cycle in eukaryotes?
A. 2 ATP
B. 36 ATP
C. 38 ATP
D. 24 ATP
Answer: D
Clarification: A total of 6 NADH, 2 ATP, and 2 FADH2 molecules are being produced. 3 ATP molecules are produced per NADH molecules. Thus, 18 (6X3) ATP molecules are produced as Krebs’ cycle produces 6 NADH molecules. 1 FADH2 molecule produces 2 ATP, so for 2 FADH2 molecules, 4 ATP molecules are produced. 2 GTP molecules equivalents to 2 ATP molecules produced in Krebs’ cycle. Thus, in one Krebs’ cycle, (18+4+2) = 24 molecules of ATP are produced.

5. After glycolysis, which of the following is transported across the inner mitochondrial membrane into the matrix?
A. Pyruvate
B. Acetyl CoA
C. ATP molecules
D. Coenzyme A
Answer: A
Clarification: Each pyruvate molecule produced in cytosol during Glycolysis is transported across the inner mitochondrial membrane into the mitochondrial matrix. Pyruvate is then decarboxylated to produce Acetyl group (CH₃COO^–) The Acetyl group is transferred to Coenzyme A to produce acetyl CoA.

6. The TCA cycle produces ____________
A. 2 NADH
B. 6 NADH
C. 8 NADH
D. 4 NADH
Answer: B
Clarification: The primary products of the TCA cycle pathway are the reduced coenzymes, NADH and FADH₂, which contain high energy electrons removed from various substrates as they are oxidized. A total of 6 NADH are formed during one Krebs’ cycle process.

7. The TCA cycle produces _____________
A. 2 FADH₂
B. 3 FADH₂
C. 4 FADH₂
D. 1 FADH₂
Answer: A
Clarification: The primary products of the TCA cycle pathway are the reduced coenzymes, NADH and FADH₂, which contain high energy electrons removed from various substrates as they are oxidized. A total of 2 FADH₂ are formed during one Krebs’ cycle process.

8. The TCA cycle produces ___________
A. 2 GTPs
B. 1 GTP
C. 3 GTPs
D. 4 GTPs
Answer: A
Clarification: The primary products of the TCA cycle pathway are the reduced coenzymes, NADH and FADH₂, which contain high energy electrons removed from various substrates as they are oxidized. The Krebs’ cycle produces two GTP molecules during one cycle. 2 GTP molecules equivalents to 2 ATP molecules.

9. In the electron transport chain, each pair of electron transferred from NADH to oxygen releases sufficient energy to produce __________
A. 3 ATPs
B. 1 ATP
C. 2 ATPs
D. 4 ATPs
Answer: A
Clarification: Electron is transported from the reduced cofactor NADH to oxygen by electron transport chain. During the transfer, it releases sufficient energy to drive the formation of approximately 3 ATP molecules.

10. In the electron transport chain, each pair of electron donated by FADH2 releases sufficient energy to produce __________
A. 3 ATPs
B. 1 ATP
C. 2 ATPs
D. 4 ATPs
Answer: C
Clarification: Electron is donated by the reduced cofactor FADH₂ to oxygen by electron transport chain. During the transfer, it releases sufficient energy to drive the formation of approximately 2 ATP molecules.

11. Total number of ATPs formed by oxidation of one glucose molecule is ___________
A. 36 ATPs
B. 30 ATP
C. 32 ATPs
D. 34 ATPs
Answer: A
Clarification: For oxidation of one glucose molecules, 2 ATP molecules are produced by glycolysis. 24 ATP molecules are produced by Kreb’s cycle and 10 ATP molecules is produced by electron oxidative chain. This adds up to 36 molecules ATP released by oxidation of one molecule of glucose.

12. Which of the following is not involved in electron chain transport system?
A. Complexes I, II, II, IV
B. Ubiquinone
C. Cytochrome C
D. All of them are involved
Answer: D
Clarification: Electron carriers can be isolated as part of four distinct, asymmetric, membrane-spanning complexes known as complexes I, II, II and IV. Two of the electron transport chain complexes, Ubiquinone and Cytochrome C are not part of any complexes. Ubiquinone is present as a pool of molecules dissolved in lipid bilayer and Cyt C is soluble protein in intermembrane space.

13. NADH and FADH₂ is associated with respectively ____________
A. Complexes I and complex II
B. Complexes II and complex III
C. Complexes I and complex III
D. Complexes III and complex IV
Answer: A
Clarification: When NADH is the electron donor, electrons enter the respiratory chain via complex I, which transfers electrons to Ubiquinone. FADH₂ remains covalently attached to succinate dehydrogenase, a component of complex II. When FADH₂ is donor, electron is directly passed to Ubiquinone.

To practice all areas of Cell Biology Problems,

Cell Biology Multiple Choice Questions on “Cytoplasmic Membrane Systems – Endomembrane System”.

1. Which of the following together represent an endomembrane system?
A. macromolecules of the cell
B. cell receptors
C. cytoplasmic structures
D. nuclear structures
Answer: C
Clarification: The cytoplasmic organelles of a cell namely endoplasmic reticulum, Golgi complex, lysosomes, vacuoles; together represent the endomembrane system where individual components function as a part of a coordinated unit.

2. Organelles of the endomembrane system are stable and static.
A. True
B. False
Answer: B
Clarification: Organelles of the endomembrane system are dynamic in nature and experience a continuous flux while they perform the regular activities inside the cell.

3. Proteins are synthesized in which of the following organelle of the endomembrane system?
A. Endoplasmic reticulum
B. Golgi complex
C. Lysosomes
D. Vacuoles
Answer: A
Clarification: Proteins are synthesized in the endoplasmic reticulum and then modified in the Golgi complex. From the Golgi complex, the proteins are further transported to different destinations depending upon their function and fate.

4. There are ______ types of secretory activities of a cell.
A. one
B. two
C. three
D. four
Answer: B
Clarification: Secretory activities lead to the discharge of proteins synthesized in the endoplasmic reticulum. There two types of secretory activities of a cell namely constitutive secretion and regulated secretion.

5. In regulated secretion, materials are __________
A. secreted
B. stored
C. degraded
D. aggregated
Answer: B
Clarification: In regulated secretion activities of the cell, materials are stored in membrane bound packages and released only when there is an invoked response.

6. Which type of endomembrane secretion occurs in nerve cells?
A. constitutive
B. regulatory
C. non-continuous
D. intermittent
Answer: B
Clarification: The nerve cells secrete neurotransmitters in response to an extracellular stimulus, the secretion is therefore termed as regulatory secretion.

7. What of the following molecules is not transported through the secretory pathway of endomembrane system?
A. nucleic acids
B. complex polysaccharides
C. proteins
D. lipids
Answer: A
Clarification: In the secretory pathway or the biosynthetic pathway biomolecules like proteins, lipids and complex polysaccharides are transported across the cell.

8. In the endocytic pathway, materials are discharged from the cell.
A. True
B. False
Answer: B
Clarification: Endocytic pathway is the opposite of secretory pathway. In the endocytic pathway materials move from the outer surface of the cell to the cytoplasmic compartments like endosomes and lysosomes.

9. Which of the following biomolecules are contained in the lysosomes?
A. nucleic acids
B. ribonucleic acids
C. proteins
D. polysaccharides
Answer: C
Clarification: The lysosomes play a vital role in digestion and waste removal from the cell. These organelles are composed of soluble proteins that carry out this function.

10. Which type of signals direct the proteins to their appropriate cellular destinations?
A. sorting signals
B. apoptotic signals
C. ubiquitylation
D. degradation signals
Answer: A
Clarification: Sorting signals encoded in amino acid sequence of proteins or in the attached sugar molecules are responsible for directing the proteins to their appropriate cellular destinations.

Cell Biology Interview Questions and Answerson “Two Fundamentally Different Classes of Cells – 2”.

1. Escherichia coli and Vibrio cholerae are ________________
A. lophotrichous; monotrichous
B. amphitrichous; lophotrichous
C. peritrichous; monotrichous
D. peritrichous; amphitrichous
Answer: C
Clarification: Monotrichous bacteria have single flagellum (eg. Vibrio cholerae). Lophotrichous bacteria have multiple flagella in same spot which helps in unidirectional movement. Peritrichous bacteria (eg. E.coli) has flagella projecting in all directions. Amphitrichous bacteria have single flagellum on opposite ends.

2. Nitrogen fixation can be done by ______________
A. All prokaryotes
B. Certain cyanobacteria
C. Certain archaebacteria
D. None of the mentioned
Answer: B
Clarification: Certain cyanobacteria have the unique property to fix nitrogen i.e., conversion of nitrogen gas into reduced ammonia. Certain cyanobacteria remain in a symbiotic relationship in the roots of legume plants such as peas and help in nitrogen fixation. Eg. Rhizobium, Azotobacter, etc.

3. F-plasmid contains ______________
A. Only OriV
B. Only OriT
C. Both OriV and OriT
D. None of the mentioned
Answer: C
Clarification: The F-plasmid is an episome formed during bacterial conjugation. It contains its own origin of replication called as ‘OriV’ and Origin of transfer called as ‘OriT’.

4. Which is the main building block of cilia?
A. Tubulin
B. Nexin
C. Dyenin
D. Actin
Answer: A
Clarification: Tubulin proteins are the main building blocks of cilia in prokaryotic organisms. Dyenin forms bridges between neighbouring microtubule doublets in structure of cilia. Nexin is present between microtubule doublets and prevent them to slide over one another.

5. Which of the following is not a difference between cilia and flagella?
A. Cilia is short hairlike; flagella is long thread-like
B. Nexin present in cilia; Nexin absent in flagella
C. Axoneme present in cilia; Axoneme absent in flagella
D. Rapid rotational motion of cilia; Undulating, sinusoidal slow movement of flagella
Answer: C
Clarification: Inside both cilia and flagella, is a microtubule-based cytoskeleton called axoneme. Axoneme is the central strand of either a cilium or flagellum which is composed of an array of microtubules. Microtubules are arranged in nine pairs around two central microtubule (9+2 combination).

6. Which of the following is absent in Gram- negative bacteria and present in Gram- positive bacteria?
A. Teichoic acids
B. Periplasmic space
C. Outer membrane
D. Lipopolysaccharide
Answer: A
Clarification: Teichoic acids are present in Gram positive and absent in Gram negative bacteria and they stain purple by Gram stain. Periplasmic space and outer membrane are present in Gram negative while they are absent in Gram positive bacteria. Lipolysaccharide content is high in Gram negative and almost none in Gram positive.

7. Exotoxins are produced by _______________
A. All prokaryotes
B. Gram positive bacteria
C. Gram negative bacteria
D. Archaea
Answer: B
Clarification: Exotoxins are produced by Gram positive bacteria. Endotoxins are produced by Gram negative bacteria.

8. Which of the following statements is false?
A. Gram negative bacteria is less resistant to physical disruption than Gram positive.
B. Gram positive bacteria is inhibited by basic dyes
C. Gram negative bacteria cell wall is thin and single layered.
D. Gram negative bacteria is more resistant to antibiotics than Gram positive.
Answer: C
Clarification: The cell wall of Gram negative bacteria is thick and two layered. However, the cell wall of Gram positive bacteria is thick and single layered.

9. Which of the following are a Gram negative bacteria?
A. Neisseria gonorrheae
B. Mycoplasma pneumonia
C. Cornybacterium diptheriae
D. Streptococcus pyogenes
Answer: A
Clarification: Neisseria gonorrheae is an example of Gram negative bacteria. However, Mycoplasma pneumonia, Cornybacterium diptheriae, Streptococcus pyogene are some examples of Gram positive bacteria.

To practice all areas of Cell Biology for Interviews,

Cell Biology Questions and Answers for Fresherson “Genes, Chromosomes and Genomes – 2”.

1. In FISH technology, which fluorescence confirms a BCR/ABL translocation?
A. Red signal
B. Green signal
C. Yellow signal
D. Orange signal
Answer: C
Clarification: In FISH, the green signal indicates presence of the BCR gene and the red signal indicate the presence of the ABL gene. The red-green fusion (yellow) signal confirms a BCR/ABL translocation.

2. The nucleosome is composed of how many histone proteins?
A. 7
B. 9
C. 8
D. 10
Answer: B
Clarification: The nucleosome is composed of 8 histone proteins known as Histone Octamer and a separate H1 protein. The histone proteins present are two H2A, two H2B, two H3 and two H4 proteins along with a single H1 protein. The DNA strand wounds around the DNA Octamer to produce a nucleosome unit.

3. An example of polyploidy is _________
A. Xenopus laevis
B. Xenopus tropicalis
C. Xenopus amieti
D. Xenopus andrei
Answer: A
Clarification: Polyploidy is the phenomenon by which the entire genome is duplicated in an organism. The organism becomes triploid (3n), tetraploid (4n) etc in place of the normal diploid (2n). Xenopus laevis has double the set of chromosome as Xenopus tropicalis is an example of polyploidy.

4. Hybrid vigour is referred to __________
A. superior phenotype of a hybrid
B. increased fertility of a hybrid
C. increased mortality of a hybrid
D. decreased function of a hybrid
Answer: A
Clarification: Hybrid vigour is the phenomenon of increased vigour in the hybrids as compared to both of its parents. This phenomenon came to light in the 20th century in corn F1 hybrids. The resulting hybrid plant had a higher growth rate, was phenotypically superior and had increased yield as compared to both the parents.

5. Hybrid vigour results from ____________
A. dominance
B. co-dominance
C. over dominance
D. incomplete dominance
Answer: c
Clarification: In case of dominance, the expression of dominant allele masks the expression of recessive allele. In co-dominance, both the alleles are equally expressed. In over dominance, the combination of genotypes from two different parents leads to supplementing the effect of each other and thus the effects lead to increased vigour.

6. Application of non-ionizing ultraviolet radiation causes _________
A. formation of pyrimidine dimers
B. deletion of pyrimidine bases
C. formation of methyl guanine
D. formation of methyl thymine
Answer: A
Clarification: During application of UV rays, as photons are absorbed by DNA molecules, an excited state is produced which allows for the rearrangement of electrons resulting in the formation of photoproducts.These photoproducts are cyclobutane pyrimidine dimer (CPD. and 6-4 pyrimidine –pyrimidone, both of which are pyrimidine dimers.

7. During translation of proteins, the aminoacyl tRNA arrives at ____________
A. Ribosomal A site
B. Ribosomal P site
C. Ribosomal T site
D. Ribosomal S site
Answer: A
Clarification: During translation, the aminoacyl tRNA carrying the desired amino acid approaches the A site on the ribosome. It then further moves to the P site and becomes peptidyl tRNA, which initiates the formation of a peptide bond between the amino acid and the growing peptide chain. The peptidyl tRNA leaves through the E site of ribosome.

8. Which of the following is a consensus sequence?
A. Promoter sequence
B. Enhancer sequence
C. Terminator sequence
D. All of the mentioned
Answer: C
Clarification: Consensus sequence is the most common version of conserved gene sequence which has minute variations from gen to gene. A promoter sequence, enhancer sequence and terminator sequence of a particular gene is conserved and are consensus sequences.

9. Which of the following is the start codon in protein translation?
A. AUG
B. UAA
C. UAG
D. UGA
Answer: A
Clarification: The start codon indicates the start of protein translation after the 5’-untranslated region (UTR) and the ribosome binds to the start codon in the mRNA. AUG is the sole start codon and codes for the amino acid Methionine. UAA, UAG and UGA are the three stop codons that indicate the termination of protein translation.

10. What is the approximate fraction of genetic variation in the nuclear genome that is expected to have a harmful effect on gene function?
A. 50%
B. 25%
C. 10%
D. 1%
Answer: D
Clarification: Genetic variation due to mutation causes harmful effects on gene function. Ad low as 1% genetic variation can cause noticeable harmful changes the expression and function of genes. It may lead to congenital diseases or may be fatal.

11. Which of the following statements is correct according to Chargaff’s rules?
A. All DNA molecules contain the same proportions of A, C, G and T
B. Single-stranded RNA molecules contain the same amount of A and U
C. In double-stranded DNA, the amount of T equals the amount of C
D. In double-stranded DNA, the amount of G equals the amount of C
Answer: D
Clarification: Chargaff rule states that the amount of A is equal to the amount of T whereas the amount of C is equal to the amount of G. This is because during base pairing, A binds with T with two hydrogen bonds and C binds with G with three hydrogen bonds.

To practice all areas of Cell Biology for Freshers,

Cell Biology Puzzleson “Mitochondrial Structure and Function – Peroxisomes”.

1. Who first discovered peroxisomes?
A. Christian de Duve
B. Boveri
C. J Rhodin
D. Van Beneden
Answer: A
Clarification: Peroxisomes are small cell organelles found in the cytoplasm of most eukaryotic organisms. They we first described by J Rhodin in the year 1954. But peroxisomes were discovered to be a cell organelle by Christian de Duve in the year 1967.

2. What is the diameter of peroxisomes?
A. 0.2 – 0.5 µm
B. 0.1 – 1.0 µm
C. 1 – 5 µm
D. 1 – 10 µm
Answer: B
Clarification: Peroxisomes are usually 0.1 – 1.0 µm in diameter. They do not have a definite size, but they vary in the range with respect to specific organisms.

3. What is the major role of peroxisomes in our body?
A. Breakdown of Formaldehyde
B. Breakdown of proteins
C. Breakdown of Hydrogen Peroxide
D. Breakdown of Phthalates
Answer: C
Clarification: The peroxisomes have an enzyme called catalase. This enzyme is used to breakdown Hydrogen Peroxide into water.

4. Which cells in our body contains abundant peroxisomes?
A. Liver cells
B. Reproductive cells
C. Cardiac cells
D. Brain cells
Answer: A
Clarification: The liver cells have a high amount of peroxisomes so that they can detoxify the chemicals in our body. The peroxisomes also aid in the production of bile acids from cholesterol.

5. What enzyme is used to detoxify alcohol in our body?
A. Catalase
B. Peroxidase
C. Urea – catalase
D. Amylase
Answer: B
Clarification: Peroxidase is one of the enzymes produced by peroxisomes. They breakdown alcohol by transferring a hydrogen atoms to oxygen forming hydrogen peroxide which is later converted to water.

6. Which of the following enzyme produced by peroxisomes are present in plant cell, but absent in human cell?
A. Catalase
B. Peroxidase
C. Uric acid oxidase
D. D – amino acid oxidase
Answer: C
Clarification: The peroxisomes in human cells lack Uric acid catalase which is present in plant cells. Due to this lacking, the humans are exposed to a disease called gout.

7. Among which of the following Peroxisomes are absent?
A. Bacillus subtilis
B. Zea mays
C. Homo habilis
D. Euphlyctis hexadactylus
Answer: A
Clarification: Peroxisomes are only present in eukaryotic cells. Prokaryotic cells are devoid of peroxisomes thus Bacillus subtilis is a bacteria and it lacks peroxisomes.

8. What cell organelle assists in the oxidation of fatty acids along with peroxisomes?
A. Ribosome
B. Nucleus
C. Endoplasmic Reticulum
D. Mitochondria
Answer: D
Clarification: Peroxisomes and mitochondria helps in breaking long fatty acid chains to simpler substances by β-oxidation. This was found by Lazarow and De Duve in 1976.

9. Which genetic disorder is associated with dysfunction of peroxisomes?
A. Parkinson’s disease
B. Down’s syndrome
C. Zellweger Syndrome
D. Bubble Boy Syndrome
Answer: C
Clarification: Zellweger is an autosomal recessive systemic disorder caused by the impairment of peroxisomes biogenesis. They affect the brain, nerve, face and liver.

10. Which organelle is used in the production of white matter in the nervous system?
A. Mitochondria
B. Peroxisomes
C. Endoplasmic Reticulum
D. Ribosomes
Answer: B
Clarification: Peroxisomes are found to be a vital cell organelle for the production of white matter of the cell. The process is known as myelination.

To practice all Puzzles on Cell Biology,