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

The correct answer is There are at least two proteins that are linked through non-covalent interactions.

Concept:

• Gel filtration chromatography separates proteins based on their size. A 150 kDa species isolated from a gel filtration column is typically a protein complex or a single protein of the stated molecular weight.
• 2-dimensional SDS-PAGE separates proteins based on two properties: their isoelectric point (1st dimension) and their molecular weight (2nd dimension). It is a powerful tool to analyze protein composition, including complexes.
• Non-covalent interactions are weak interactions such as hydrogen bonds, ionic interactions, and hydrophobic forces that help in forming protein complexes without covalent bonding.

Explanation:

• In the 2-dimensional SDS-PAGE analysis, multiple distinct spots are observed for the 150 kDa species. This indicates that the species is composed of at least two proteins of different isoelectric points and molecular weights.
• The proteins are likely linked by non-covalent interactions, as SDS disrupts non-covalent bonds but not covalent bonds like disulfide bridges. If the proteins were covalently linked, they would run as a single band or spot corresponding to the 150 kDa species.
• Therefore, the correct conclusion is that there are at least two proteins in the complex that are linked through non-covalent interactions.

The correct answer is EcoRV (V) - Smal (I); Aval (V) - Sall (I)

Concept:

• Restriction Enzymes are molecular scissors used in molecular biology to cut DNA at specific recognition sites. They generate either sticky ends (overhanging sequences) or blunt ends (no overhangs).
• For ligation, DNA fragments need compatible ends. Sticky ends with complementary sequences or blunt ends (which do not require sequence complementarity) can be ligated together.
• The recognition sites of the enzymes provided in the question are:
  • EcoRV: GAT^ATC (blunt ends)
  • Aval: C^YCGRG (sticky ends)
  • HindIII: A^AGCTT (sticky ends)
  • Sall: G^TCGAC (sticky ends)
  • Smal: CCC^GGG (blunt ends)
  • Xbal: T^CTAGA (sticky ends)

Explanation:

EcoRV (V) - Smal (I); Aval (V) - Sall (I): This is the correct combination.​

• EcoRV: Generates blunt ends.
• Smal: Also generates blunt ends.
• Blunt ends can be ligated together without sequence complementarity, making EcoRV (V) and Smal (I) compatible.
• Aval: Generates sticky ends with a 5' overhang (C^YCGRG).
• Sall: Generates sticky ends with a compatible 5' overhang (G^TCGAC).
• Aval (V) and Sall (I) have complementary sticky ends and can be ligated without further enzymatic treatment.

The correct answer is 33.3

Explanation:

• The extinction coefficient (ε) is a measure of how strongly a chemical species absorbs light at a given wavelength. It is expressed in units of M^-1cm^-1.
• The relationship between absorbance (A), extinction coefficient (ε), concentration (c), and path length (l) is given by the Beer-Lambert Law:
  • A = ε × c × l
• Here:
  • A: Absorbance (unitless, directly measured by a spectrophotometer).
  • ε: Extinction coefficient (M^-1cm^-1).
  • c: Concentration of the compound (M).
  • l: Path length of the cuvette (cm).
• By rearranging the equation, we can calculate the extinction coefficient:
  • ε = A / (c × l)
• In this question, the path length is given as 1 cm, and the absorbance values are plotted against the corresponding concentrations. The slope of the plot (ΔA/Δc) represents the extinction coefficient (ε) since the path length is constant.
• From the data provided, the slope of the plot of absorbance (A) against concentration (c) is calculated. If the absorbance increases by 0.1 units for every 0.003 M increase in concentration, the slope is:
  • Slope = ΔA / Δc = 0.1 / 0.003 = 33.3 M^-1cm^-1
• Since the path length (l) is 1 cm, the slope directly gives the extinction coefficient (ε).
• Therefore, the extinction coefficient is 33.3 M^-1cm^-1.

The correct answer is A and D only

Explanation:

DNA double-strand breaks (DSBs) are critical lesions that can lead to genetic instability if not properly repaired. Cells employ multiple repair pathways, including:

• Gene Conversion (GC) is a homologous recombination-based repair mechanism that uses a homologous sequence as a template to restore the correct DNA sequence.
• Single-Strand Annealing (SSA) is a homology-based pathway that anneals complementary single-strand DNA sequences, leading to loss of the intervening DNA sequence.
• Non-Homologous End Joining (NHEJ) is a repair mechanism that directly joins the broken DNA ends without requiring a homologous template, often resulting in insertions or deletions.

The experimental construct described has a neomycin resistance gene flanked by two inactive GFP genes:

• The first GFP gene is inactivated by the insertion of an I-SceI recognition sequence, which creates a DSB upon induction of the I-SceI endonuclease.
• The second GFP gene has a 99 bp deletion at its 5' end, rendering it inactive but still capable of serving as a homologous template during repair.
• The repair pathway utilized determines whether GFP expression and neomycin resistance are restored in the cell.
  • Option A (Correct): If the DSB is repaired by the Gene Conversion (GC) pathway, the first GFP gene can be repaired using the second GFP gene as a template. This restores GFP expression, making the cells GFP-positive. Additionally, the neomycin resistance gene remains intact, making the cells neomycin-resistant. Hence, this statement is correct.
  • Option B (Incorrect): This option incorrectly states that GC repair leads to GFP-positive but neomycin-sensitive cells. GC repair does not disrupt the neomycin resistance gene, so this statement is not correct. Neomycin resistance is conferred by the neomycin gene, which is part of the construct and should not be affected by the repair of the GFP gene unless there are large deletions or rearrangements that extend to it, which is not the primary mechanism of GC. So, cells should remain neomycin-resistant.
  • Option C (Incorrect): Single-Strand Annealing (SSA) pathway occurs when a DSB occurs between two direct repeat sequences. The DNA is resected to expose single-stranded regions, and if homologous repeats are present, they can anneal. This process typically leads to the deletion of the sequence between the repeats and one copy of the repeat.
    • The disrupted GFP has the I-SceI site and is flanked by a 350 bp region and a 730 bp region of original GFP sequence.
      The truncated GFP is further downstream.
    • SSA would require significant homologous regions flanking the break. While there are GFP sequences, SSA would likely involve deletion of the intervening sequence between homologous repeats.
    • More importantly, for GFP to become "positive" through SSA, the precise deletion of the I-SceI site without introducing other frame-shifts or deletions that maintain the GFP reading frame would be required. It is unlikely that SSA would precisely restore a functional GFP from the disrupted and truncated versions. SSA typically leads to deletions.
    • If SSA occurred by annealing the disrupted GFP with the truncated GFP (which has a 99 bp deletion), the resulting GFP would still be truncated and non-functional (GFP-negative)
  • Option D (Correct): If the DSB is repaired by Non-Homologous End Joining (NHEJ), the repair is imprecise and does not restore GFP functionality, so the cells remain GFP-negative. However, the neomycin resistance gene is not affected, and the cells remain neomycin-resistant. Hence, this statement is correct.

The correct answer is Near the BamHI site

Explanation:

• Circular DNA molecules, such as plasmids or certain viral genomes, replicate using a specific region known as the origin of replication (Ori). This region is where the replication machinery assembles and begins DNA synthesis.
• During replication, the DNA molecule is unwound at the origin, forming a "bubble," and the two strands are copied by DNA polymerases.
• In Southern blot analysis, different shapes formed by DNA fragments can provide important insights into the replication process and the structure of the DNA at the origin of replication.
• The patterns observed — "double Y," "bubble to Y," and "bubble arc" — reflect the status of DNA replication and the activity around the replication origin. 
  • Double Y Shape: The "double Y" shape is formed when there are two replication forks moving out from the origin of replication. It suggests that the region of the DNA is actively undergoing replication, and both forks are moving outward. This pattern typically appears at or near the origin of replication because this is where the replication process is initiated.
  • Bubble to Y Shape: The "bubble to Y" pattern indicates that replication is ongoing in the region. It suggests a unidirectional replication process. The "bubble" is the unwound portion of the DNA, and the "Y" shape represents the replication fork. The replication bubble itself is indicative of the DNA region where the strands are being separated for replication. It suggests that the restriction site lies close to or within the replication bubble.
  • Bubble Arc: The "bubble arc" indicates the presence of a replication bubble, showing that the DNA is being unwound at a specific origin of replication. The arc is usually associated with the region where the DNA begins to replicate.

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 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.

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.

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