Biophysical Method MCQ Quiz - Objective Question with Answer for Biophysical Method - Download Free PDF
Last updated on Jun 19, 2025
Latest Biophysical Method MCQ Objective Questions
Biophysical Method Question 1:
A purified 150 kDa species obtained from a gel filtration column was run on a 2- dimensional SDS-PAGE as shown below:
Answer (Detailed Solution Below)
Biophysical Method Question 1 Detailed Solution
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.
Biophysical Method Question 2:
A purified 150 kDa protein species from a gel filtration column was run on a 2-dimensional SDS-PAGE as shown below.
What is the likely form of the 150 kDa protein species from this observation?
Answer (Detailed Solution Below)
Biophysical Method Question 2 Detailed Solution
The correct answer is There are at least two proteins in the complex that are linked through covalent bonds.
Concept:
- Gel filtration chromatography separates proteins based on their size. A 150 kDa protein species identified through this method suggests it is either a single protein or a complex of proteins of combined molecular weight 150 kDa.
- 2-dimensional SDS-PAGE: In this method, proteins are separated based on their isoelectric point (1st dimension: isoelectric focusing) and molecular weight (2nd dimension: SDS-PAGE). If covalent bonds (like disulfide bonds) exist, they often hold protein subunits together unless a reducing agent is used to break these bonds.
- When analyzing the results from 2D SDS-PAGE, the presence of multiple spots or bands for the same molecular weight can provide insights into the protein’s structure or complex composition.
- DTT (Dithiothreitol) is a reducing agent that breaks disulfide bonds (covalent bonds) between cysteine residues.
- When run without DTT, disulfide-linked protein complexes remain intact.
- When run with DTT, disulfide bonds are reduced, causing the complex to dissociate into individual protein subunits.
Explanation:
- In the observation described, the 150 kDa protein species from gel filtration shows multiple distinct spots on the 2D SDS-PAGE under non-reducing conditions. This indicates the presence of at least two proteins linked through covalent bonds.
- The covalent bonds between these proteins are likely disulfide bonds, as these are common in protein complexes and contribute to their stability.
- Without a reducing agent in SDS-PAGE, disulfide bonds are not broken, allowing the proteins in the complex to stay linked. This explains why multiple spots corresponding to different components of the complex are visible on the gel.
Other Options:
Option 1: "There are at least two proteins that are linked through non-covalent interactions."
- This is incorrect because non-covalent interactions (e.g., hydrogen bonding or ionic interactions) are usually disrupted by SDS during electrophoresis. Therefore, if the proteins were linked only through non-covalent interactions, they would separate completely, and we wouldn't see multiple spots corresponding to covalently bonded subunits.
Option 3: "There are two proteins in the mixture without forming a complex."
- This is incorrect because gel filtration already purified the 150 kDa species as a single entity, indicating it is a complex rather than a simple mixture. If the proteins were not part of a complex, the gel filtration step would have separated them based on size, and the 2D SDS-PAGE would not show any interaction between them.
Option 4: "There is only one protein that has a disulfide bond."
- This is incorrect because the presence of multiple spots on the 2D SDS-PAGE indicates there are at least two distinct proteins in the 150 kDa species. If it were a single protein with a disulfide bond, it would appear as a single spot unless reduced..
Biophysical Method Question 3:
When the molecular weight of human immunoglobulin light chain is 24 kDa , the total molecular weight of human IgG iskDa .
Answer (Detailed Solution Below) 140 - 160
Biophysical Method Question 3 Detailed Solution
The correct answer is: 148 kDa.
Explanation:
- Human IgG is composed of two identical heavy chains and two identical light chains.
- Given that the molecular weight of one human immunoglobulin light chain is 24 kDa, the total contribution from the light chains is: 2 × 24 kDa = 48 kDa
- The molecular weight of one heavy chain in human IgG is typically around 50 kDa. Thus, for two heavy chains, the contribution is: 2 × 50 kDa = 100 kDa
- Therefore, the total molecular weight of human IgG can be calculated as follows: Total molecular weight of IgG = 48 kDa + 100 kDa = 148 kDa
Key Points:
-
Structure of IgG:
- Composition: IgG is a Y-shaped molecule composed of four polypeptide chains: two identical heavy chains and two identical light chains.
- Domains: Each heavy chain consists of three constant regions (CH1, CH2, CH3) and one variable region (VH), while each light chain has one variable region (VL) and one constant region (CL).
- Fab and Fc Regions: The two arms of the Y-shaped structure are known as the Fab (Fragment antigen-binding) regions, which bind to antigens, while the stem is called the Fc (Fragment crystallizable) region, which interacts with other components of the immune system.
-
Function of IgG:
- Neutralization: IgG can neutralize toxins and viruses by binding to them, preventing their interaction with host cells.
- Opsonization: It enhances phagocytosis by marking pathogens for destruction by immune cells. The Fc region of IgG binds to Fc receptors on phagocytic cells like macrophages and neutrophils.
- Complement Activation: IgG can activate the complement system, a group of proteins that aid in the destruction of pathogens.
- Long-term Immunity: IgG is the most abundant antibody in serum and provides long-term immunity against infections. It can persist for months or years after an infection or vaccination.
Biophysical Method Question 4:
In rate-zonal centrifugation, a mixture is spun just long enough to separate molecules that differ in mass but may be similar in shape and density. Which one of the following options represents the component that is typically used to create a gradient in this technique?
Answer (Detailed Solution Below)
Biophysical Method Question 4 Detailed Solution
The correct answer is Sucrose.
Explanation:
Rate-zonal centrifugation is a technique used to separate biomolecules based on their size and mass, typically using a gradient to allow for differential sedimentation.
-
Phospholipids: While phospholipids are important components of cell membranes, they are not typically used to create a gradient in centrifugation.
-
Sucrose: Sucrose is commonly used to create density gradients in rate-zonal centrifugation. It is advantageous because it is biochemically inert, soluble in water, and can form gradients of varying densities. This allows for the effective separation of macromolecules based on their sedimentation rates.
-
KCl: Potassium chloride is used to adjust ionic strength in solutions but is not typically used to create density gradients for separation in centrifugation.
-
Starch: While starch can be used in some biochemical applications, it is not commonly used to create density gradients for rate-zonal centrifugation.
Conclusion: Sucrose is widely recognized for creating density gradients in rate-zonal centrifugation, facilitating the separation of molecules based on their mass. Therefore, the correct answer is Sucrose.
Biophysical Method Question 5:
You have purified a protein X and observed the following‐
A. When run on a native PAGE, it gives rise to a single band.
B. When run on a non‐reducing SDS‐PAGE, you obtain two bands corresponding to 40kDa and 60kDa.
C. When run on a reducing SDS page you get three bands‐ corresponding to 60kDa, 30kDa, and 10kDa.
What can you conclude about the purified protein X?
Answer (Detailed Solution Below)
Biophysical Method Question 5 Detailed Solution
The correct answer is X contains 2 polypeptide chains that form a complex
Concept:
- Polyacrylamide Gel Electrophoresis (PAGE) is a technique used to separate proteins based on their size and charge.
- Native PAGE maintains the protein's native structure, while SDS-PAGE (sodium dodecyl sulfate-PAGE) denatures proteins, with reducing SDS-PAGE further breaking disulfide bonds.
- Disulfide bonds are covalent bonds that can form between cysteine residues within or between polypeptide chains, contributing to the stabilization of protein structure.
- Non-reducing SDS-PAGE keeps disulfide bonds intact, whereas reducing SDS-PAGE breaks these bonds, allowing for the analysis of subunit composition and the presence of disulfide linkages.
Explanation:
- Native PAGE: Observing a single band suggests that protein X exists as a single complex in its native form, without being denatured or separated into subunits.
- Non-reducing SDS-PAGE: The appearance of two bands (40 kDa and 60 kDa) indicates that protein X is composed of at least two different polypeptide chains, linked by non-covalent interactions or disulfide bonds, as their separation occurs when the protein is denatured.
- Reducing SDS-PAGE: The appearance of three bands (60 kDa, 30 kDa, and 10 kDa) when disulfide bonds are broken indicates the presence of smaller polypeptide chains. This suggests that the 40 kDa band from the non-reducing SDS-PAGE is composed of a 30 kDa and a 10 kDa subunit linked by disulfide bonds.
Conclusion: Given this information, protein X contains two distinct polypeptide chains (60 kDa and a combined 40 kDa, which splits into 30 kDa and 10 kDa under reducing conditions), forming a complex where the 40 kDa chain is further composed of two smaller subunits linked by disulfide bonds. Thus, the correct answer is that X contains 2 polypeptide chains that form a complex.
Top Biophysical Method MCQ Objective Questions
Consider the four results that were obtained from immunophenotyping of human breast cancer cells.
Which one of the following options correctly depicts the above results?
Answer (Detailed Solution Below)
Biophysical Method Question 6 Detailed Solution
Download Solution PDFThe 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 molecular ion peak [M].+ of an analyte as measured by Electron Ionization Mass Spectrometry has an m/z of 149 and a relative abundance of 100%. The [M].+ has a relative abundance of 6.7% and the [M + 2].+peak has a relative abundance of 5%. The abundance of the major isotope of H, C, N, O, and S are 1H-100%, 12C-98.9%, 13C-1.1% , 14N-99.6%, 15N-0.4%, 16O, 99.8%, 18O-0.2%, 32S-95.0%, 33S-0.75% and 34S-4.2%. The most probable molecular formula of the compound is:
Answer (Detailed Solution Below)
Biophysical Method Question 7 Detailed Solution
Download Solution PDFConcept:
- 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
A protein solution (0.2 ml) of unknown concentration was diluted with 0.8 ml of water. To 0.5 ml of this diluted solution 4.5 ml of biuret reagent was added and the color allowed to develop. The absorbance of this mixture taken in a test tube of 1cm diameter at 540 nm was observed to be 0.20.
0.5 ml of BSA (4 mg/ml) solution plus 4.5 ml of biuret gave an absorbance of 0.20 when measured as above. What is the protein concentration (mg/ml) in the undiluted unknown solution?
Answer (Detailed Solution Below)
Biophysical Method Question 8 Detailed Solution
Download Solution PDFConcept:
- Absorption spectrophotometry is a technique that measures light absorbance of the particular chemical solution. It is used to determine the amount of light passing through a solution (transmittance) or the amount of light absorbed by a solution (absorbance).
- When electromagnetic radiation passed through the solution some part of it is absorbed while the rest passes through the prism which is called the absorption spectrum.
- The absorption spectrum of a substance does not change with varying concentrations.
- This is based on Beer-Lambert's law
- Lambert law states that the intensity of the transmitted monochromatic light decreases exponentially as the thickness of absorbing material increases.
- Beer Law states that the intensity of the transmitted monochromatic light decreases exponentially as the concentration of the absorbing substance increases.
- Beer’s law states that A = εbc, where A=absorbance, ε = molar extinction coefficient, b = path length of the cuvette and c = concentration of the protein.
- Mathematical expression of Beer-Lambert's law \(A=log\frac{I_0}{I} = \epsilon cl\)
- So, By calculating the slope of the absorbance vs. concentration plot, we can obtain the molar extinction coefficient, as in most instances the path length (b) of the cuvette is equal to 1 cm, and the slope is the same value as ε.
Explanation:
Volume of undiluted unknown protein solution = 0.2ml
Volume of water added to the protein solution = 0.8ml
The dilution factor of the protein sample \(DF=\frac{V_f}{V_i} = \frac{0.2+0.8}{0.2} = \frac{1.0}{0.2} = 5\)
Absorbance of diluted unknown protein (\(A_{540}\)) = 0.2
Concentration of BSA standard = 4mg/ml
Absorbance of BSA (\(A_{540}\)) = 0.2
- Here, BSA is used as the standard, if the absorbance of 4mg/ml BSA standard solution and unknown diluted protein is 0.2, then we can conclude that the concentration of diluted unknown protein is also 4mg/ml.
- Now, we have to estimate the concentration of the unknown undiluted protein sample.
- The protein was diluted ⅕ so the dilution factor is 5.
So, the undiluted unknown protein \(= 4 \times dilution factor = 4 \times20 = 20mg/ml\)
So, the concentration of the undiluted unknown solution is 20mg/ml.
Hence, the correct answer is option 1.
In rate-zonal centrifugation, a mixture is spun just long enough to separate molecules that differ in mass but may be similar in shape and density. Which one of the following options represents the component that is typically used to create a gradient in this technique?
Answer (Detailed Solution Below)
Biophysical Method Question 9 Detailed Solution
Download Solution PDFThe correct answer is Sucrose.
Explanation:
Rate-zonal centrifugation is a technique used to separate biomolecules based on their size and mass, typically using a gradient to allow for differential sedimentation.
-
Phospholipids: While phospholipids are important components of cell membranes, they are not typically used to create a gradient in centrifugation.
-
Sucrose: Sucrose is commonly used to create density gradients in rate-zonal centrifugation. It is advantageous because it is biochemically inert, soluble in water, and can form gradients of varying densities. This allows for the effective separation of macromolecules based on their sedimentation rates.
-
KCl: Potassium chloride is used to adjust ionic strength in solutions but is not typically used to create density gradients for separation in centrifugation.
-
Starch: While starch can be used in some biochemical applications, it is not commonly used to create density gradients for rate-zonal centrifugation.
Conclusion: Sucrose is widely recognized for creating density gradients in rate-zonal centrifugation, facilitating the separation of molecules based on their mass. Therefore, the correct answer is Sucrose.
For a nuclear spin of spin quantum no. \(\left(I=\frac{1}{2}\right)\), precessing in a magnetic field at a Larmor frequency of 300 MHz, the wavelength of incident radiation required to excite the nuclear spins must be approximately
Answer (Detailed Solution Below)
Biophysical Method Question 10 Detailed Solution
Download Solution PDFConcept:
- Nuclear magnetic resonance, or NMR, is a physical phenomenon that occurs when atomic nuclei are exposed to a specific frequency of electromagnetic radiation and a magnetic energy level resonance transition.
- One can obtain the NMR spectrum by identifying the absorption signals.
- People can investigate the architectures of molecules quantitatively based on the locations, intensities, and fine structure of resonance peaks.
- When an external magnetic field is provided, energy can be transferred from lower energy levels to higher energy levels.
- Energy is transferred at a wavelength that matches the radio frequency.
- Additionally, when the spin returns to its initial base level, energy is released at the same frequency.
Explanation:
- The wavelength of the incident radiation can be calculated using the equation:
- wavelength = c / frequency
- where c is the speed of light (approximately 299,792,458 meters per second) and frequency is the Larmor frequency.
- Substituting the given values:
- wavelength = 299,792,458 m/s / 300,000,000 Hz = 0.99964 m = 999.64 mm
- So, the wavelength of the incident radiation required to excite the nuclear spins must be approximately 999.64 mm, which is approx 1nm.
Therefore, the correct answer is option 1 i.e. 1 nm.
Biophysical Method Question 11:
The conformation of a 30-residue peptide is studied by NMR spectroscopy. The JNH for most of the amide protons is 4Hz. The 2D NOESY spectrum shows prominent Ni-Ni+1 connectivities. The conformation of the peptide is
Answer (Detailed Solution Below)
Biophysical Method Question 11 Detailed Solution
Explanation:
Using Nuclear Magnetic Resonance (NMR) spectroscopy we identify atoms in the molecules that are close together in space and from this information about the distance between atoms we derive indirectly a 3-D model of proteins. It is also possible to refine protein structures based on chemical shifts and even with insufficient information for a structure determination. Chemical shifts can be used to distinguish between alpha helix and beta sheet via the chemical shift index. NOES are the by far most important source of information for structure calculation. They are usually observed for protons separated by less than 5 Å. For most values of J more than one solution exists. Scalar couplings may be used to define dihedral angle restraint ranges, which sometimes help to improve the convergence of the structure calculation.
| Secondary structure | Theoretical 3 JHN-(Hz) Value |
| Alpha helix | 3.9 |
| Antiparallel beta sheet | 7.9 |
| Parallel bets sheet | 8.0 |
Biophysical Method Question 12:
The structure of a protein is known from X-ray diffraction studies which gave 30% α- helix, 50% β-sheet and 20% random coil. Circular dichroism (CD) measurements gave 50% α-helix, 40% β-sheet and 10% random coil. What could not be a possible explanation for these observations.
Answer (Detailed Solution Below)
Biophysical Method Question 12 Detailed Solution
Concept:
- Various biological processes depend heavily on proteins, which are significant molecules.
- Sequence and structural homology serve as the main foundation for the prediction of protein structure.
- Since a protein's activity primarily depends on its three-dimensional structure, protein structure prediction or modelling is crucial.
- Similar to this, a protein's amino acid makeup determines its 3D structure.
- A slight change in the protein's sequence can result in significant structural changes in the protein's native structure.
- Although it is critical to have a thorough understanding of protein 3D structure, it can be challenging to determine a protein's natural structure when it exists in the physiological environment of the body.
Explanation:
- Currently, the main techniques used to determine protein 3D structure are X-ray crystallography and nuclear magnetic resonance (NMR).
- The atomic and molecular structure of a crystal can be ascertained via X-ray crystallography.
- The fundamental idea is that crystallized atoms cause an X-ray beam to diffract in a variety of distinct directions.
- An image of the density of electrons within the crystal can be created in three dimensions by measuring the angles and intensities of these diffracted beams.
- The mean locations, chemical bonds, disorder, and several other properties of the atoms in the crystal can all be calculated from this electron density image.
- Many biological substances, such as vitamins, medications, proteins, and nucleic acids like DNA, as well as their structures and functions, were made known through this technique.
The following are some benefits of this method:
- It produces results with great atomic resolution and is not constrained by the protein's molecular weight.
- This method can be used to learn more about how a protein detects and binds to ligands.
The following are this method's drawbacks:
- Crystallization of the protein is required for this approach but large proteins, however, may have trouble crystallizing because of their bulk and poor solubility.
- The protein's dynamic state cannot be determined; only its static state can.
- It is very difficult to determine the location of individual atoms in many protein crystal structures because they are very poorly resolved.
- Comparable size chains cannot always be distinguished from one another.
Circular Dichriosm:
- Circular Dichroism, an absorption spectroscopy, uses circularly polarized light to investigate structural aspects of optically active chiral media.
- Finding the secondary structure of proteins with CD is a very effective technique.
- Due to "exciton" interactions, when the chromophores of the amides of proteins are arranged in arrays, their optical transitions are either displaced or split into many transitions.
- As a result, various structural elements have distinctive CD spectra.
- For instance, α-helical proteins have positive and negative bands at 193 nm and 222 nm, respectively.
- Positive bands at 195 nm and negative bands at 218 nm are found in proteins with well-defined antiparallel ß-pleated sheets (ß-helices), whereas disordered proteins show very low ellipticity above 210 nm and negative bands close to 195 nm.
Option 1: Protein structure in the crystal is different from that in the solution.
- Consider the explanation above, thus this option is true.
- Some protein crystal structures, are very poorly resolved in X-ray crystallography technique, whereas same proteins secondary structure can be elucidated well using CD technique.
Option 2: CD analysis for structural components is not appropriate for this protein.
- Consider the explanation above thus this option is not true
- For elucidating the secondary structure of proteins with CD is a very effective technique.
Option 3: Contributions from other chromophores also contribute to the CD spectrum of the protein
- The three aromatic side chains that are found in proteins—the phenyl group in Phe, the phenolic group in Tyr, and the indole group in Trp—also contain ultraviolet absorption bands.
- However, proteins often contribute very little to the CD spectra in the near UV, which is where secondary structural information is found.
- Consider the explanation above thus this option is not true
Option 4: Protein contains high content of disulphide bonds.
- Disulphide bonds are prevalent in secreted proteins, however be CD or X ray crystallography technique both can elucidate it.
- thus this option is not true.
Hence the correct answer is option 1,
Biophysical Method Question 13:
Consider the four results that were obtained from immunophenotyping of human breast cancer cells.
Which one of the following options correctly depicts the above results?
Answer (Detailed Solution Below)
Biophysical Method Question 13 Detailed Solution
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.
Biophysical Method Question 14:
Which one of the following strategies will generate the most precise mutation at the predetermined location of a plant genome?
Answer (Detailed Solution Below)
Biophysical Method Question 14 Detailed Solution
The correct answer is Option 1 i.e. CRISPR/Cas9 editing.
Explanation-
The CRISPR/Cas bacterial immune system is used as a simple, RNA-guided platform for highly efficient and targeted genome editing. CRISPR genome editing technology takes advantage of DNA repair. crRNAs guide Cas9 proteins to target sequences, which are recognized through their complementarity to the crRNA. The Cas9 nuclease then introduces a double-strand break in the target DNA.
CRISPR/Cas9 (Clustered Regularly Interspaced Short Palindromic Repeats/CRISPR-associated protein 9) is a technology that has revolutionized the field of genetic engineering by providing a simple, efficient method for making precise, targeted changes to the genome of almost any organism.
CRISPR/Cas9 is based on a natural defense mechanism found in bacteria, which uses CRISPR sequences and Cas proteins like Cas9 to defend against viral attacks by slicing up the invading DNA. The Cas9 protein is combined with a guide RNA (gRNA) that has been engineered to match the sequence of the targeted gene. This system can be used for a variety of genetic modifications, such as knockouts, where the targeted gene is disabled, as well as more precise edits, like changing a single base pair.
In contrast to CRISPR/Cas9:-
- T-DNA insertion mutagenesis is a method used to introduce DNA into the genome of an organism, most commonly used in plants. However, the integration of T-DNA into the plant genome is random, which means it's not ideal when precision is required at a specific location.
- Transposon mutagenesis is also a powerful tool in genetics, as transposons can cause mutation when they insert themselves at random positions within genes. Again, the key drawback is that this is a random process.
- TILLING (Targeting Induced Local Lesions in Genomes) are techniques that allow the identification of point mutations in specific genes, but they require a large-scale mutagenesis and screening effort.
Conclusion- CRISPR/Cas9 offers the most precision and control over where in the genome modifications occur, and it also allows for more types of edits beyond just insertions or deletions. This is why, CRISPR/Cas9 is the most suited for generating precise mutations at predetermined locations of a plant genome.
Biophysical Method Question 15:
What can you infer if the correlation coefficient, [Pearson correlation (r)], is close to -1 (minus 1) for two set of variables?
Answer (Detailed Solution Below)
Biophysical Method Question 15 Detailed Solution
The correct answer is Option 4 i.e.There is a linear relationship in which, when there is an increase in one variable, there is a decrease in the second variable.
Explanation-
The Pearson correlation coefficient (r) ranges from -1 to 1. The sign of the correlation coefficient indicates the direction of the relationship, while the magnitude indicates the strength of the relationship.
- If r= -1, it indicates a perfect negative (inverse) linear relationship. In this case, as one variable increases, the other variable decreases in a perfectly straight-line fashion.
- If r = 1, it indicates a perfect positive linear relationship. In this case, as one variable increases, the other variable also increases in a perfectly straight-line fashion.
- If r=0, it indicates no linear relationship between the variables.
Conclusion:-
In summary, when the Pearson correlation coefficient is close to -1, it suggests a strong negative linear relationship between the two variables.