Methods in Biology MCQ Quiz - Objective Question with Answer for Methods in Biology - Download Free PDF

The correct answer is Homologous recombination and CRISPR-Cas9

Explanation:

• Gene knock-out refers to a genetic technique where a specific gene in an organism is completely inactivated or "knocked out," leading to the loss of its function. This is a powerful tool in molecular biology for studying gene function and plant development.

CRISPR-Cas9:

• The CRISPR-Cas9 system uses a guide RNA (gRNA) to direct the Cas9 nuclease to a specific target sequence in the genome.
• Once the Cas9 protein binds to the target DNA, it introduces double-strand breaks (DSBs) at the specific location.
• In plants, the cell's natural DNA repair mechanisms, such as non-homologous end joining (NHEJ), introduce insertions or deletions (indels) at the DSB site. These mutations often result in a loss of gene function, effectively "knocking out" the target gene.
• CRISPR-Cas9 is highly versatile, efficient, and precise, making it the preferred choice for generating gene knock-outs in plants.
• Homologous recombination: Homologous recombination is a method for precise gene editing, including knock-outs, by replacing a target gene with a desired sequence. 

Other Options (Incorrect):

• Antisense RNA technique: Antisense RNA technology involves introducing a strand of RNA complementary to the target gene's mRNA, which blocks its translation into protein. This technique results in gene silencing but does not lead to a complete knock-out of the gene at the DNA level.
• Activation tagging: Activation tagging is a technique used to increase the expression of a gene rather than knocking it out. This method involves inserting a strong enhancer element near a target gene to boost its transcription. 

The correct answer is 1

Concept:

• Bacteria reproduce by binary fission, where one bacterium divides into two identical daughter cells. Therefore, after one cycle of growth, the number of bacteria doubles.
• Different organisms, or even different types of genes within one organism, can have varying numbers of copies of a specific gene per genome. This is a crucial factor in determining the initial template quantity for PCR.
• PCR is a laboratory technique widely used in molecular biology to make many copies of a specific DNA segment. PCR relies on thermal cycling, consisting of cycles of denaturation, annealing, and extension.
• In an ideal PCR, each DNA template molecule is theoretically doubled in every cycle. Therefore, after 'n' cycles, the number of DNA copies will be 2ⁿ times the initial number of template molecules.

Explanation:

Let N₀ be the initial bacterial count in the sample.

• Prior to DNA extraction, the bacteria were incubated to allow one cycle of growth. This means the initial bacterial population doubled.
  • Number of bacteria after one growth cycle = N₀ x 2 = 2N₀
• Each bacterium contains 3 copies of the target gene. Therefore, the total number of target gene copies extracted and available to serve as the initial template for PCR amplification will be the number of bacteria after growth multiplied by the gene copies per bacterium:
  • Initial PCR template copies = (Number of bacteria after growth) x (Gene copies per bacterium)
  • Initial PCR template copies = 2N₀ x 3 = 6N₀
• PCR Amplification: The PCR was performed for 9 cycles with 100% efficiency. This means the initial template copies were multiplied by 2⁹
  • 2⁹ = 512
• The total number of amplicon copies obtained after 9 cycles is given as 3072. We can set up the equation:
  • Total Amplicon Copies = (Initial PCR template copies) x 2^{Number of PCR cycles}
  • 3072 = 6N₀ x 2⁹
  • 3072 = 6N₀ x 512
  • 3072 = 3072 N₀
  • N₀ = 1

The correct answer is 2

Concept:

• Bacteria, like Escherichia coli, multiply by binary fission. This means that after a certain period (doubling time), a single bacterium divides into two. If the doubling time is 30 minutes, the population of uninfected bacteria will double every 30 minutes.
• Burst Size refers to the number of new phage particles released from a single lysed bacterial cell.
• Multiplicity of Infection (MOI) is the ratio of the number of infectious agents (e.g., phage particles) to the number of target cells (e.g., bacteria) in a given experiment. An MOI of 1 means, on average, one phage particle is added per bacterial cell. Instantaneous absorption means the infection happens immediately upon phage addition.

Explanation:

We need to determine at what time the phage population grows large enough to infect all available E. coli cells. Once all E. coli cells are infected, they will subsequently lyse within the phage life cycle duration.

• Initial State (Time = 0 minutes):
  • Initial E. coli cells (N₀) = 2 x 10⁷
  • Initial T7 phage (PFU) = 5000
  • Phage absorption is instantaneous at MOI 1. So, 5000 E. coli cells are infected immediately.
  • Number of E. coli cells remaining uninfected = 2 x 10⁷ - 5000. This is approximately 2 x 10⁷ (since 5000 is negligible compared to 2 x 10⁷).
  • The 5000 infected cells will lyse after 20 minutes (phage life cycle).
• First Phage Burst (Time = 20 minutes):
  • The 5000 E. coli cells infected at t=0 lyse.
  • Number of new phages released = (Number of cells lysed) x (Burst size) = 5000 x 200 = 1,000,000 = 1 x 10⁶ phages.
  • During these 20 minutes, the uninfected E. coli cells have been growing. The growth factor for 20 minutes (given a 30-minute doubling time) is 2^(20/30) = 2^{2/3}  =1.587.
  • Approximate number of uninfected E. coli cells available for infection at t=20 min = 2 x 10⁷ x 2^{2/3} = 3.174 x 10⁷ cells. (The 5000 initial infected cells are no longer part of the viable population.)
  • At t=20 min, the 1 x 10⁶ newly released phages infect 1 x 10⁶ additional E. coli cells (since 1 x 10⁶ < 3.174 x 10⁷). These newly infected cells will lyse at t = 20 + 20 = 40 minutes.
• Second Phage Burst (Time = 40 minutes):
  • The 1 x 10⁶ E. coli cells infected at t=20 min lyse.
  • Number of new phages released = 1 x 10⁶ x 200 = 200 x 10⁶ = 2 x 10⁸ phages.
  • During the period from t=20 min to t=40 min (another 20 minutes), the *remaining uninfected* E. coli cells continued to grow. The approximate uninfected population at t=20 was 3.174 x 10⁷ - 1 x 10⁶ = 3.074 x 10⁷ cells.
  • Number of uninfected E. coli cells available for infection at t=40 min = (3.074 x 10⁷) x 2^{2/3} = 3.074 x 10⁷ x 1.587 = 4.877 x 10⁷ cells.
  • At t=40 min, the 2 x 10⁸ newly released phages are available for infection.
  • Since 2 x 10⁸ (available phages) is greater than 4.877 x 10⁷ (available E. coli cells), all the remaining uninfected E. coli cells can be infected by these phages.
  • These newly infected cells at t=40 min will lyse after 20 minutes, i.e., by t = 40 + 20 = 60 minutes.
• Total Time to Complete Lysis:
  • The entire E. coli culture will be completely lysed by 60 minutes, as all cells that were not lysed earlier would have been infected at the 40-minute mark and would lyse by the 60-minute mark.
• Convert Time to Bacterial Division Cycles:
  • E. coli doubling time = 30 minutes.
  • Number of full cycles of bacterial division = Total lysis time / Doubling time
  • Number of full cycles = 60 minutes / 30 minutes/cycle = 2 cycles.

The correct answer is Chromatin immunoprecipitation followed by sequencing and DNase I footprinting   ​

Explanation:

S1 nuclease mapping:

• This technique is primarily used to map the precise start and end points of RNA transcripts, identify splice junctions, or quantify mRNA levels.
• It involves the use of S1 nuclease, which digests single-stranded nucleic acids. It is not designed to identify protein-DNA binding sites.

Chromatin Immunoprecipitation followed by sequencing (ChIP-seq):

• This is a powerful technique used to identify DNA regions that are bound by specific proteins (like transcription factors) in vivo (within living cells).
• It involves cross-linking proteins to DNA, shearing the chromatin, immunoprecipitating the protein-DNA complexes using an antibody against the transcription factor, reversing the cross-links, and then sequencing the DNA.
• While ChIP-seq identifies the regions bound by the transcription factor across the genome, the resolution is typically in the range of tens to hundreds of base pairs, providing a broader region rather than the exact nucleotide sequence of the binding site.

Electrophoretic Mobility Shift Assay (EMSA):

• This technique detects DNA-protein interactions by observing the mobility shift of a labeled DNA fragment when bound by a protein. The exact binding site can be further analyzed using competitor DNA fragments or mutational analysis.
• EMSA confirms the interaction and specificity, but it does not precisely map the exact base pairs of the binding site within the DNA fragment.

DNase I Footprinting:

• This technique involves treating a DNA-protein complex with DNase I, which cleaves unprotected DNA.
• The regions bound by the transcription factor are protected from cleavage and can be identified by comparing the cleaved fragments to a control (unbound DNA), revealing the binding site.

The correct answer is 87.5

Explanation:

Initial State (Before Cycle 1):

• We have 1 dsDNA molecule.
• Strands: U1, U2 (both unlabeled original strands).
• Total strands = 2 (unlabeled)

Cycle 1:

• Denaturation: The 1 dsDNA molecule separates into 2 single strands (U1 and U2).
• Annealing & Extension: Each unlabeled strand (U1 and U2) serves as a template for a new labeled strand.
  • U1 -> forms a new labeled strand L1.
  • U2 -> forms a new labeled strand L2. At the end of Cycle 1, we have 2 dsDNA molecule
• Molecule 1: U1 (unlabeled) + L1 (labeled)
• Molecule 2: U2 (unlabeled) + L2 (labeled)
• Total strands = 4.
• Number of unlabeled strands = 2 (U1, U2).
• Number of labeled strands = 2 (L1, L2).
• Percentage of labeled single-stranded DNA = (2/4) * 100% = 50%.

Cycle 2:

• Denaturation: The 2 dsDNA molecules separate into 4 single strands that serve as templates: U1, L1, U2, L2.
• Annealing & Extension: Each of these 4 template strands synthesizes a new labeled complementary strand.
  • U1 (unlabeled) -> forms new labeled strand L3. (Resulting dsDNA: U1+L3)
  • L1 (labeled) -> forms new labeled strand L4. (Resulting dsDNA: L1+L4)
  • U2 (unlabeled) -> forms new labeled strand L5. (Resulting dsDNA: U2+L5)
  • L2 (labeled) -> forms new labeled strand L6. (Resulting dsDNA: L2+L6)
• At the end of Cycle 2, we have 2² =4 dsDNA molecules.
• Total strands = 4×2=8.
• Number of unlabeled strands = 2 (U1, U2 - the original templates).
• Number of labeled strands = Total strands - Number of unlabeled strands = 8 - 2 = 6 (L1, L2, L3, L4, L5, L6).
• Percentage of labeled single-stranded DNA = (6/8) * 100% = 75%.

Cycle 3:

• Denaturation: The 4 dsDNA molecules from Cycle 2 separate into 8 single strands that serve as templates: U1, L3, L1, L4, U2, L5, L2, L6. (This set of templates consists of 2 unlabeled original strands and 6 labeled strands from previous syntheses).
• Annealing & Extension: Each of these 8 template strands synthesizes a new labeled complementary strand. At the end of Cycle 3, we have 2³ = 8 dsDNA molecules. Total strands (if all are denatured) = 8 x 2  = 16.
• The only unlabeled strands that exist are the two original template strands (U1 and U2), which persist through all cycles. All other strands are newly synthesized using radiolabeled dNTPs and are therefore labeled.
• Percentage of labeled single-stranded DNA = 14/16 x 100  = 87.5%

The correct answer is A ‐ I, B ‐ II, C ‐ III, D ‐ IV

Concept:

• Proteins' glass transition temperature, melting point, isoelectric point, molecular weight, secondary structure, solubility, surface hydrophobicity, and emulsification are all significant functional traits.

Explanation:

NMR - 

• NMR spectroscopy, also known as nuclear magnetic resonance, has emerged as the method of choice for figuring out the composition of organic substances.
• It is the only spectroscopic technique for which a thorough examination and interpretation of the entire spectrum is often anticipated. this technique is used to determine the 3-D structure of substances.

Isoelectric focusing - 

• A method for separating various molecules based on differences in their isoelectric points is known as isoelectric focusing (IEF), also referred to as electrofocusing (pI).
• It is a sort of zone electrophoresis that often involves the use of proteins on a gel and makes use of the fact that the pH of the environment affects the overall charge on the target molecule.

Affinity chromatography - 

• Proteins are separated using affinity chromatography based on how they interact with a particular ligand.
• Either competition or lowering the affinity with pH and/or ionic strength can reverse the binding of the protein to a ligand bound to a matrix.

Ultracentrifugation - 

• Using ultracentrifugation, components of a solution are separated according to their sizes, densities, and the density (viscosity) of the medium (solvent).

Therefore, the correct answer is option 2: A ‐ I, B ‐ II, C ‐ III, D ‐ IV.

The correct answer is ‘B’ represents a plot that denotes a high percentage of cancer stem cells in the breast cancer cells.

Concept:

• On a case-by-case basis, the molecular (immunophenotypes) subtypes of breast cancer should be used as excellent molecular-level tissue markers to guide therapy options.
• High-grade invasive micropapillary carcinoma (IMPC) with an odd inverted apical location was found to express CD24 at a higher level than invasive ductal carcinomas (IDC), with significant cytoplasmic staining, and normal breast tissue tested absolutely negative.
• Compared to the breast IDC, IMPC displayed decreased CD44v5 and CD44v9 expression, although there was no statistically significant difference between the two groups.
• When compared to IDCs, IMPC represents a separate entity of breast cancer with strong CD24 expression, a distinctive inverted apical membrane pattern, and decreased levels of CD44 isoforms v5 and v9.
• Malignant characteristics may account for the high lymph-vascular invasion propensity and increased tendency of these malignancies to metastasize.

Explanation: 

Option A:- CORRECT

• One of the most often investigated surface markers is the hyaluronic acid receptor CD44, which is expressed by almost all tumor cells.
• On the surface of the majority of B lymphocytes and developing neuroblasts, CD24, a sialoglycoprotein, is expressed.
• Plot B demonstrates that the cells are CD44 positive, which indicates that a significant portion of the breast cancer cells is cancer stem cells.

Therefore, option A is correct i.e. ‘B’ represents a plot that denotes a high percentage of cancer stem cells in the breast cancer cells.

The correct answer is Option 2 i.e.examining the fire scars in growth rings of living trees.

Concept:

• The method of examining the fire scars in the growth rings of living trees to determine the historical frequencies of fires in an area is part of a field of study called dendrochronology.
• Dendrochronology, or tree-ring dating, is the scientific method of dating based on the analysis of patterns of tree rings, also known as growth rings.
• Each year, most trees add a new layer of growth to their trunks. In temperate and boreal climates, this growth occurs in a regular pattern, producing distinct rings for each year of growth.
• The conditions under which the tree grows can affect the appearance of these growth rings.​
• When a fire occurs, it can damage the tree but not necessarily kill it. The damage from the fire can leave a scar on the tree, which gets covered by new layers of growth rings as the tree heals and continues growing.
• By examining these fire scars within the growth rings, scientists can determine not only the frequency of fires in the area but also estimate the intensity and season of past fires.
• This analysis provides crucial insights into the historical fire regimes and natural cycles of the ecosystem in question.

This method offers several advantages:

• Long-term perspective: Trees can live for hundreds to thousands of years, providing a long-term record of fire activity in an area.
• Precision: Because tree rings can usually be dated to the exact year they formed, this method allows for precise timing of past fires.
• Local scale: It provides information at the scale of individual trees or stands of trees, offering detailed insights into fire dynamics at a local level.

Other options listed, such as radioactive dating, measuring carbon content, or examining historical records, also have their uses in ecological and historical studies but are not as direct or specific for determining the frequency and characteristics of past fires as examining tree-ring fire scars

Conclusion:- Therefore, the correct answer is examining the fire scars in growth rings of living trees.

The correct answer is FG : B3, B2 and BG : A2, A3, A4, A7

Explanation:

Foreground (FG) Selection:

• Foreground selection targets markers that are closely linked to the gene of interest (R) to ensure that this gene is successfully incorporated into the cultivated variety.
• The markers selected for FG must be closely associated with R on the genetic map.

Background (BG) Selection:

• Background selection focuses on selecting markers that are further away from the R gene or located on different chromosomes to ensure that most of the background genetic material from the wild species (B) is eliminated and replaced with the genetic material from the cultivated species (A).
• These markers are used to recover the genetic background of species A, minimizing the amount of foreign (B) DNA.

Panel I shows the markers available for the cultivated variety A and the wild species B. These markers (A1–A7, B1–B8) can be used for selection.
Panel II is the genetic map showing the position of the resistance gene (R) and other markers linked to it. The relationship between markers in species A and B is crucial for determining which markers can be used for selection.

The correct answer is t-test and ANOVA,

Explanation-

Both the t-test and Analysis of Variance (ANOVA) are statistical methods used to compare the means of populations.

The methods that compare the means of populations are:-

1. t-test
2. Analysis of Variance (ANOVA)

The t-test is used to compare the means of two groups, while ANOVA is used to compare the means of three or more groups.

The t-test is a statistical analysis method that is designed to compare the means of two groups. There are different types of t-tests (two-sample t-test, paired t-test, etc.) but the common property is that they compare means. It is usually implemented when the test statistic follows a Student's t distribution if the null hypothesis is assumed.

Analysis of Variance (ANOVA): ANOVA is a statistical technique that is used to check if the means of two or more groups are significantly different from each other. ANOVA checks the impact of one or more factors by comparing the means of different groups, hence studying the variations in different groups. It compares the means of these groups to detect differences greater than those that may arise from random variation.It's an extension of the t-test and can be used to compare means across multiple groups simultaneously

Chi-square: The Chi-square test is a statistical test used to determine if there is a significant association between two categorical variables in a sample. It tests the null hypothesis that the variables are independent and not related. It doesn't compare population means.

Principal Component Analysis (PCA): PCA is not a statistical test, but rather a dimensionality reduction method. It is a method used to highlight variation and bring out strong patterns in a data set. It's used when you have obtained data on a number of variables and want to develop a smaller number of artificial variables (called principal components) that account for most of the variance in the original variables. It doesn't compare population means either.

Conclusion:- Therefore, the correct answer is t-test and ANOVA.

The correct answer is Protein A is polyubiquitinated.

Concept:

In this experiment, the coding sequence (CDS) of the shortest isoform of human gene 'A' is cloned into a vector under a CMV promoter and transfected into HeLa cells. The expression of Protein A is then analyzed by Western blot.

The key observations from the Western blot (right panel) are:

• Lane 1 (Transfected cells): This lane shows multiple bands for Protein A, including bands at higher molecular weights than the predicted size of the protein.
• Lane 2 (Untransfected cells): This is the control lane, where no bands are observed, as expected, because the cells are untransfected and do not express Protein A.

Observations:

• The presence of multiple bands at higher molecular weights in the transfected lane suggests that Protein A has undergone post-translational modifications.
• One common type of post-translational modification that leads to a shift in molecular weight on SDS-PAGE is ubiquitination, particularly polyubiquitination.
• Polyubiquitination involves the attachment of multiple ubiquitin molecules to a protein, which increases its molecular weight and causes a characteristic "ladder" of bands on a Western blot.
• The presence of a high molecular weight smear or ladder suggests that Protein A has been polyubiquitinated, as multiple ubiquitin chains are added to the protein.

Explanation:

Protein A forms homo-multimers:

• While multimerization can cause shifts in molecular weight, this typically results in discrete bands corresponding to the multimeric forms of the protein, not the broad range of high molecular weight bands seen here. Therefore, this option is incorrect

Protein A is degraded by the lysosome:

• Lysosomal degradation usually leads to protein breakdown and would likely result in fewer or smaller bands on a Western blot.
• The presence of high molecular weight bands is inconsistent with lysosomal degradation.

Protein A is polyubiquitinated:

• This option is correct. Polyubiquitination causes proteins to have a range of molecular weights due to the addition of ubiquitin chains, which leads to the multiple higher molecular weight bands observed on the Western blot. 

Protein A localizes to autophagosomes:

• While autophagosome localization could involve lysosomal degradation, it would not typically cause the characteristic high molecular weight bands seen in the Western blot, making this option incorrect.

The correct match is A - II, B - I, C - III.

Explanation:

The theoretical resolution limit of fluorescence microscopy is approximately 200 nm, primarily due to the diffraction limit of light. To overcome this limitation, several super-resolution microscopy techniques have been developed. These techniques allow scientists to observe biological structures at a much finer scale than traditional fluorescence microscopy. Let’s review the principles of some common super-resolution methods.

Super-resolution microscopy methods and their principles:

1. Structured illumination microscopy (SIM): In SIM, the specimen is illuminated with a pattern of light and dark stripes. This creates Moire fringes that help in reconstructing a higher resolution image. Thus, SIM matches with the principle "The specimen is illuminated with a pattern of light and dark stripes to generate Moire fringes."
   Match: A - II
2. Stimulated emission depletion (STED) microscopy: STED uses a focused excitation laser point surrounded by a donut-shaped depletion beam. This depletion beam forces the excited molecules in the periphery to return to their ground state, leaving only a small region of fluorescence, improving resolution. Thus, STED matches with the principle "Focused excitation laser point is surrounded by a donut-shaped depletion beam."
   Match: B - I
3. Photoactivated localization microscopy (PALM): PALM uses a variant of GFP (green fluorescent protein) that is activated by a different wavelength than the one used for excitation. This allows for the precise localization of individual molecules. Thus, PALM matches with the principle "Utilizes variant of GFP that is activated by a wavelength different from its excitation wavelength."
   Match: C - III

Key Points

• Structured illumination microscopy (SIM): Works by illuminating the specimen with a patterned light, producing interference patterns (Moire fringes) that are computationally reconstructed to achieve super-resolution.
• Stimulated emission depletion (STED) microscopy: Utilizes a focused laser point surrounded by a depletion beam to restrict fluorescence to a small area, improving resolution.
• Photoactivated localization microscopy (PALM): Uses photoactivatable fluorescent proteins to achieve high-precision localization of individual molecules in a sample, surpassing the diffraction limit.

The correct answer is Median

Explanation:

In this dataset, the values range from 0 to 3000, showing a wide spread of numbers. When dealing with data that contains extreme values or outliers (like 1500 and 3000), the mean can be heavily influenced by those large values, which may not provide a good summary of the central tendency.

• Mean: It can be influenced by the very large numbers (e.g., 1500 and 3000), leading to a skewed average.
• Median: The median is the middle value of a dataset when arranged in order, making it less affected by outliers or extreme values. For this dataset, arranging the numbers in order gives 0,5,15,25,100,150,200,600,1500,3000.
  • The median value would be the average of the 5th and 6th numbers, i.e.,  (100 +150)/2 = 125
• Mode: The mode is the most frequently occurring value, which isn't useful in this case since all values are unique.
• Standard deviation: This measures the spread of the data, not the central tendency, so it is not relevant for summarizing the central point of the data.

Conclusion: Thus, median is the best measure of central tendency for this data, as it provides a better representation of the typical value without being skewed by extreme values.

Concept:

• DNA sequencing technique that involves the principle of "sequencing by synthesis" is Pyrosequencing.
• Pyrosequencing involves the addition of dNTPs by DNA polymerase along with a chemiluminescent enzyme.
• The complementary strand is synthesized over the template strand.
• Each time when a dNTP is added to the complementary strand, pyrophosphate (PPi) is released.
• ATP is synthesized from PPi with the help of the enzyme ATP sulfurylase.
• PPi + APS → ATP + Sulfate (catalyzed by ATP-sulfurylase)
  ATP acts as a substrate for the luciferase-mediated conversion of luciferin to oxyluciferin that generates visible light.
• The amount of light produced is detected and analyzed to predict the type of dNTP(dATP, dGTP, dCTP, dTTP).

Explanation:

Option 1:

• A nanopore is a DNA sequencing technique that helps in determining the exact sequence of DNA.
• Nanopore DNA Sequencing involves threading DNA through a minute pore and detecting the changes in electric current to determine nucleotides (A, T, G, C) of DNA.
• Hence the relation given is correct

Option 2:

• pyrosequencing is a kind technique that is used in DNA sequencing.
• It involves sequencing by using a radiolabelled pyrophosphate group.
• Mass Spectrometry or Edman degradation are commonly used techniques for protein sequencing.
• Hence given relation is wrong.
• This is the correct option.

Option 3:

• Chloroplast Transformation involves three steps:

1. Passage of foreign DNA into explant.
2. Insertion of foreign DNA into chloroplast genome by Homologous recombination.
3. Repeated screening of transformants until wild-genotype is eliminated. 

• Homologous recombination avoids gene silencing by position effect and produces more number of transformants.
• Hence the relation given is correct.

• Simple sequence repeats (SSRs) are microsatellites that exhibit a high rate of polymorphism.
• These are also known as Variable number tandem repeats (VNTRs) and are highly unstable.
• these unstable entities undergo variation in the number of repeated through slipped strand mispairing during DNA strand synthesis.
• SSRs are co-dominant markers that differentiate heterozygotes and homozygotes.
• Organisms carrying two similar alleles for a gene (AA or aa) are homozygotes.
• organisms carrying two different alleles for a gene (Aa) are heterozygotes.
• Hence the relation is correct

Therefore, the correct answer is option 2.

Concept:

• ​In the process of ionizing solid or gas phase atoms or molecules, energetic electrons contact with them to produce ions.
• This process is known as electron ionization (EI, formerly known as electron impact ionization and electron bombardment ionization).
• One of the first mass spectrometry ionization methods created was EI.
• The charge number is the quantity of electrons eliminated (for positive ions).
• The horizontal axis in a mass spectrum is stated in units of mass divided by charge number, or m/z.

Conclusion:-

So, The most probable molecular formula of the compound is C5H11NO2S