Molecular analysis of given plant DNA using DNA based markers
Easy
A.Protein Coding Region
B.Primer Coupled Replication
C.Plasmid Cloning Reaction
D.Polymerase Chain Reaction
Correct Answer: Polymerase Chain Reaction
Explanation:
PCR stands for Polymerase Chain Reaction, a technique used to amplify specific DNA sequences in vitro.
Incorrect! Try again.
2Which enzyme is commonly used in PCR to synthesize new DNA strands?
Molecular analysis of given plant DNA using DNA based markers
Easy
A.Restriction endonuclease
B.Taq DNA polymerase
C.DNA ligase
D.RNA polymerase
Correct Answer: Taq DNA polymerase
Explanation:
Taq DNA polymerase is a heat-stable enzyme isolated from Thermus aquaticus, widely used in PCR because it withstands the high denaturation temperatures.
Incorrect! Try again.
3What is the correct order of the three main steps in a single PCR cycle?
Molecular analysis of given plant DNA using DNA based markers
Easy
dNTPs (dATP, dTTP, dGTP, dCTP) are the nucleotide building blocks used by DNA polymerase to synthesize new DNA strands.
Incorrect! Try again.
14Which instrument is used to carry out PCR by rapidly cycling through different temperatures?
Molecular analysis of given plant DNA using DNA based markers
Easy
A.Spectrophotometer
B.Autoclave
C.Centrifuge
D.Thermal cycler
Correct Answer: Thermal cycler
Explanation:
A thermal cycler (PCR machine) automatically changes temperatures to perform the denaturation, annealing, and extension steps.
Incorrect! Try again.
15A DNA molecular marker is best described as:
Molecular analysis of given plant DNA using DNA based markers
Easy
A.A dye that labels RNA
B.An enzyme that cuts DNA
C.A DNA sequence used to identify genetic variation
D.A protein used to stain DNA
Correct Answer: A DNA sequence used to identify genetic variation
Explanation:
A molecular marker is a known DNA sequence or fragment used to detect differences (polymorphisms) among individuals or species.
Incorrect! Try again.
16Which molecular marker technique requires prior knowledge of the DNA sequence to design specific primers?
Molecular analysis of given plant DNA using DNA based markers
Easy
A.RAPD
B.SSR (microsatellites)
C.AFLP
D.RFLP
Correct Answer: SSR (microsatellites)
Explanation:
SSR markers require known flanking sequences to design specific primers, unlike RAPD which uses arbitrary primers.
Incorrect! Try again.
17The annealing step in PCR allows which of the following to occur?
Molecular analysis of given plant DNA using DNA based markers
Easy
A.Primers bind to the template DNA
B.DNA strands separate
C.New strands are fully extended
D.Enzymes are denatured
Correct Answer: Primers bind to the template DNA
Explanation:
During annealing, the temperature is lowered so primers can bind (hybridize) to their complementary sequences on the single-stranded template.
Incorrect! Try again.
18What is loaded alongside DNA samples on a gel to indicate fragment sizes?
Molecular analysis of given plant DNA using DNA based markers
Easy
A.Restriction enzyme
B.Loading dye only
C.DNA ladder (molecular weight marker)
D.Primer mix
Correct Answer: DNA ladder (molecular weight marker)
Explanation:
A DNA ladder contains fragments of known sizes and serves as a reference to estimate the size of sample DNA bands.
Incorrect! Try again.
19The term polymorphism in molecular marker analysis refers to:
Molecular analysis of given plant DNA using DNA based markers
Easy
A.Variation in DNA sequence among individuals
B.The number of PCR cycles
C.The color of a gel band
D.The size of a single gene
Correct Answer: Variation in DNA sequence among individuals
Explanation:
Polymorphism means the presence of different DNA sequence variants at a locus among individuals or populations.
Incorrect! Try again.
20Which of the following is a PCR-based molecular marker?
Molecular analysis of given plant DNA using DNA based markers
Easy
A.RFLP
B.Karyotyping
C.RAPD
D.Southern blot
Correct Answer: RAPD
Explanation:
RAPD is a PCR-based marker technique, whereas RFLP relies on restriction digestion and hybridization rather than amplification.
Incorrect! Try again.
21A researcher uses RAPD markers to analyze genetic diversity among plant accessions. Which feature makes RAPD suitable when no prior sequence information is available?
Molecular analysis of given plant DNA using DNA based markers
Medium
A.Use of short arbitrary primers of about 10 bases
B.Need for radioactive probes
C.Use of two specific long primers per locus
D.Requirement of Southern blotting
Correct Answer: Use of short arbitrary primers of about 10 bases
Explanation:
RAPD uses single short arbitrary primers (~10 nucleotides) that bind at random sites, so no prior sequence knowledge of the plant genome is needed.
Incorrect! Try again.
22During RFLP analysis of plant DNA, a point mutation removes a restriction site. What is the expected effect on the resulting banding pattern?
Molecular analysis of given plant DNA using DNA based markers
Medium
A.The number of bands stays exactly the same
B.All bands disappear completely
C.One larger fragment replaces two smaller fragments
D.Two smaller fragments replace one larger fragment
Correct Answer: One larger fragment replaces two smaller fragments
Explanation:
Loss of a restriction site means the enzyme no longer cuts there, so the two adjacent fragments merge into one larger fragment, altering the RFLP pattern.
Incorrect! Try again.
23Two plant varieties show SSR (microsatellite) alleles of 180 bp and 192 bp at the same locus. If the repeat unit is a dinucleotide, what is the difference in repeat number between them?
Molecular analysis of given plant DNA using DNA based markers
Medium
A.3 repeats
B.24 repeats
C.6 repeats
D.12 repeats
Correct Answer: 6 repeats
Explanation:
A dinucleotide repeat adds 2 bp per unit. The size difference is bp, so repeat units differ.
Incorrect! Try again.
24Which property of SSR (microsatellite) markers makes them preferred for cultivar identification over RAPD?
Molecular analysis of given plant DNA using DNA based markers
Medium
A.Requirement of no primers
B.Codominant inheritance and high reproducibility
C.Detection only of methylation changes
D.Dominant inheritance and low cost
Correct Answer: Codominant inheritance and high reproducibility
Explanation:
SSRs are codominant (heterozygotes can be distinguished) and highly reproducible due to locus-specific primers, making them reliable for cultivar fingerprinting.
Incorrect! Try again.
25In AFLP analysis, plant genomic DNA is first digested and then adaptors are ligated. What is the immediate purpose of ligating adaptors to the restriction fragments?
Molecular analysis of given plant DNA using DNA based markers
Medium
A.To provide known primer-binding sites for selective amplification
B.To circularize the fragments
C.To remove all non-coding regions
D.To fluorescently stain the DNA
Correct Answer: To provide known primer-binding sites for selective amplification
Explanation:
Adaptors add known sequences at fragment ends so that AFLP primers (adaptor + selective bases) can anneal and selectively amplify a subset of fragments.
Incorrect! Try again.
26A dominant marker such as RAPD cannot distinguish between which two genotypes at a locus?
Molecular analysis of given plant DNA using DNA based markers
Medium
A.Two unrelated loci
B.Homozygous recessive and heterozygous
C.Two different homozygous recessives
D.Homozygous dominant and heterozygous
Correct Answer: Homozygous dominant and heterozygous
Explanation:
Dominant markers score only presence/absence of a band. Both homozygous dominant (AA) and heterozygous (Aa) show the band, so they cannot be distinguished.
Incorrect! Try again.
27While designing a PCR to amplify a plant marker region, the melting temperatures of the two primers differ by . What is the most likely consequence?
Molecular analysis of given plant DNA using DNA based markers
Medium
A.Increased Taq polymerase activity
B.Formation of a longer product than expected
C.Poor amplification due to unequal primer annealing
D.Complete failure of DNA denaturation
Correct Answer: Poor amplification due to unequal primer annealing
Explanation:
A large difference means one primer anneals poorly at the chosen annealing temperature, reducing amplification efficiency and yielding weak or no product.
Incorrect! Try again.
28A plant DNA sample gives smeared, non-specific bands in RAPD even after repeating the reaction. Which adjustment is most appropriate to improve specificity?
Molecular analysis of given plant DNA using DNA based markers
Medium
A.Lower the denaturation temperature
B.Remove the primer entirely
C.Increase the annealing temperature
D.Add more template DNA
Correct Answer: Increase the annealing temperature
Explanation:
Raising the annealing temperature increases primer binding stringency, reducing non-specific priming and cleaning up the banding pattern.
Incorrect! Try again.
29SCAR markers are often developed from RAPD markers. What is the main advantage of converting a RAPD band into a SCAR marker?
Molecular analysis of given plant DNA using DNA based markers
Medium
A.Removal of any need for gel electrophoresis
B.Elimination of the need for PCR
C.Ability to detect methylation patterns
D.Greater reproducibility using longer specific primers
Correct Answer: Greater reproducibility using longer specific primers
Explanation:
SCAR markers use longer locus-specific primers designed from the sequenced RAPD fragment, giving far more reproducible and robust amplification.
Incorrect! Try again.
30In agarose gel electrophoresis of amplified plant DNA markers, smaller DNA fragments migrate:
Molecular analysis of given plant DNA using DNA based markers
Medium
A.Faster and farther toward the anode
B.Slower and remain near the well
C.Only if stained with EtBr
D.Toward the cathode
Correct Answer: Faster and farther toward the anode
Explanation:
DNA is negatively charged and moves toward the positive anode; smaller fragments move through the gel matrix faster, travelling farther.
Incorrect! Try again.
31A study compares 50 plant genotypes and requires markers spread across the whole genome, generated quickly and cheaply without prior sequence data. Which marker system best fits these constraints?
Molecular analysis of given plant DNA using DNA based markers
Medium
A.SSR
B.SNP arrays
C.AFLP
D.RFLP
Correct Answer: AFLP
Explanation:
AFLP generates many polymorphic bands per reaction across the genome, needs no prior sequence information, and is relatively rapid and cost-effective for diversity screening.
Incorrect! Try again.
32If a marker is described as codominant, what can be directly inferred from its banding pattern in a plant?
Molecular analysis of given plant DNA using DNA based markers
Medium
A.Only the dominant allele is visible
B.The marker detects only deletions
C.Both alleles at a locus can be distinguished
D.The locus is always homozygous
Correct Answer: Both alleles at a locus can be distinguished
Explanation:
Codominant markers reveal both alleles, so heterozygotes (two bands) can be told apart from homozygotes (one band), unlike dominant markers.
Incorrect! Try again.
33A PCR reaction for a plant SSR marker produces no band, but the positive control works. The template DNA has an ratio of 1.4. What is the most likely problem?
Molecular analysis of given plant DNA using DNA based markers
Medium
A.Too high DNA purity
B.Protein contamination in the DNA sample
C.Excess RNA in the sample
D.Primer concentration too low
Correct Answer: Protein contamination in the DNA sample
Explanation:
A pure DNA sample has an ratio near 1.8. A ratio of 1.4 indicates protein or phenol contamination, which can inhibit PCR.
Incorrect! Try again.
34Which step of the PCR cycle is directly responsible for separating the double-stranded plant template DNA?
Molecular analysis of given plant DNA using DNA based markers
Medium
A.Denaturation at about
B.Annealing at about
C.Extension at about
D.Final hold at
Correct Answer: Denaturation at about
Explanation:
Denaturation (~) breaks hydrogen bonds between complementary strands, separating the double helix into single strands for priming.
Incorrect! Try again.
35For phylogenetic analysis of plant genetic distance, marker data are often converted into a similarity matrix. A Jaccard coefficient of 0 between two accessions indicates:
Molecular analysis of given plant DNA using DNA based markers
Medium
A.No shared bands between them
B.Identical banding patterns
C.A sequencing error
D.Complete heterozygosity
Correct Answer: No shared bands between them
Explanation:
The Jaccard coefficient measures shared presence of bands; a value of 0 means the two accessions share no common bands, indicating maximum dissimilarity.
Incorrect! Try again.
36Why are SNP markers increasingly favored for high-throughput genotyping of plant DNA?
Molecular analysis of given plant DNA using DNA based markers
Medium
A.They are abundant and amenable to automated detection
B.They require Southern blotting for scoring
C.They occur only in chloroplast DNA
D.They are always dominant markers
Correct Answer: They are abundant and amenable to automated detection
Explanation:
SNPs are the most abundant type of variation in genomes and can be scored on automated, high-throughput platforms, making them ideal for large-scale genotyping.
Incorrect! Try again.
37In a RAPD gel, one primer amplifies 8 bands across the sample, of which 3 are polymorphic among genotypes. What is the percentage polymorphism for this primer?
Molecular analysis of given plant DNA using DNA based markers
Medium
A.62.5\%
B.37.5\%
C.26.7\%
D.50\%
Correct Answer: 37.5\%
Explanation:
Percentage polymorphism .
Incorrect! Try again.
38A researcher wants to fine-map a disease-resistance gene and needs a marker tightly linked to it and easily transferable across labs. Which marker type is most appropriate?
Molecular analysis of given plant DNA using DNA based markers
Medium
A.AFLP
B.RAPD
C.SCAR
D.Random RFLP
Correct Answer: SCAR
Explanation:
SCAR markers use specific, sequence-defined primers, making them highly reproducible and easily transferable between laboratories for marker-assisted selection.
Incorrect! Try again.
39During AFLP, the use of two restriction enzymes (a rare cutter and a frequent cutter) primarily helps to:
Molecular analysis of given plant DNA using DNA based markers
Medium
A.Increase DNA methylation
B.Eliminate the need for adaptors
C.Denature the DNA permanently
D.Generate fragments of manageable size and number
Correct Answer: Generate fragments of manageable size and number
Explanation:
A rare cutter plus a frequent cutter produce a controlled number of fragments in a suitable size range, optimizing the pattern for selective amplification and resolution.
Incorrect! Try again.
40A dendrogram built from plant marker data groups two accessions on the same terminal branch with a very short branch length. This indicates that the two accessions are:
Molecular analysis of given plant DNA using DNA based markers
Medium
A.From different species always
B.Completely unrelated
C.Genetically most distant
D.Genetically very similar
Correct Answer: Genetically very similar
Explanation:
Short branch lengths and clustering together on a dendrogram reflect low genetic distance, meaning the accessions share a high proportion of marker alleles.
Incorrect! Try again.
41A researcher performs RAPD analysis on two plant cultivars using a single 10-mer primer. Cultivar A shows a band at 800 bp that is absent in Cultivar B. Upon repeating the reaction with a slightly lower annealing temperature, several new faint bands appear in both cultivars. What is the most likely explanation for the new bands?
Molecular analysis of given plant DNA using DNA based markers
Hard
A.The template DNA of both cultivars became partially degraded during storage
B.The Taq polymerase gained processivity at the lower temperature
C.Reduced annealing stringency permits mismatched primer binding at additional loci
D.Primer dimers migrated as high-molecular-weight products
Correct Answer: Reduced annealing stringency permits mismatched primer binding at additional loci
Explanation:
RAPD is highly sensitive to annealing temperature. Lowering it reduces stringency, allowing the short arbitrary primer to bind imperfectly matched sites, producing additional (often faint, non-reproducible) amplicons. This is why RAPD reproducibility is a known limitation.
Incorrect! Try again.
42In an RFLP analysis of a plant population, a restriction enzyme recognition site is lost in some individuals due to a point mutation. On a Southern blot with a probe spanning the region, what banding change would you expect in individuals homozygous for the mutation compared to wild type?
Molecular analysis of given plant DNA using DNA based markers
Hard
A.A single larger fragment replaces two smaller fragments
B.Two smaller fragments replace one larger fragment
C.No hybridization signal is detected at all
D.The band intensity doubles but sizes remain identical
Correct Answer: A single larger fragment replaces two smaller fragments
Explanation:
Loss of a restriction site means the enzyme no longer cuts there, so the two adjacent fragments remain joined as one larger fragment. Homozygotes show only this larger band, while heterozygotes would show all three bands.
Incorrect! Try again.
43A codominant SSR (microsatellite) marker is scored in an F2 plant population. A given locus in the parents shows alleles of 150 bp and 160 bp. Which banding pattern in an F2 individual indicates heterozygosity?
Molecular analysis of given plant DNA using DNA based markers
Hard
A.A smear between 150 and 160 bp
B.Two bands, one at 150 bp and one at 160 bp
C.A single band at 150 bp only
D.A single band at 155 bp
Correct Answer: Two bands, one at 150 bp and one at 160 bp
Explanation:
SSR markers are codominant, so both alleles are visualized. A heterozygote carries one allele from each parent and thus displays both the 150 bp and 160 bp bands simultaneously.
Incorrect! Try again.
44During AFLP analysis, genomic DNA is digested with a rare cutter (EcoRI) and a frequent cutter (MseI), followed by adapter ligation and selective amplification. Why are two enzymes of differing cutting frequency used together?
Molecular analysis of given plant DNA using DNA based markers
Hard
A.To generate a manageable number of fragments suitable for selective amplification
B.To increase the average fragment size beyond 5 kb
C.To ensure complete methylation of all restriction sites
D.To eliminate the need for adapter ligation
Correct Answer: To generate a manageable number of fragments suitable for selective amplification
Explanation:
The rare cutter limits the total number of fragments while the frequent cutter produces small fragments ideal for PCR. This combination yields a moderate, resolvable number of fragments for selective amplification and gel display.
Incorrect! Try again.
45A dominant marker (e.g., RAPD) and a codominant marker (e.g., SSR) are compared for mapping a recessive disease-resistance gene in an F2 population. Why is the codominant marker generally more informative?
Molecular analysis of given plant DNA using DNA based markers
Hard
A.It is unaffected by DNA template quality
B.It distinguishes homozygotes from heterozygotes, extracting full genotypic information
C.It requires no polyacrylamide gel for scoring
D.It amplifies a larger region of the genome per reaction
Correct Answer: It distinguishes homozygotes from heterozygotes, extracting full genotypic information
Explanation:
Codominant markers reveal both alleles, allowing heterozygotes to be told apart from homozygotes. Dominant markers only score presence/absence, so heterozygotes and dominant homozygotes look identical, reducing mapping information in F2 populations.
Incorrect! Try again.
46You genotype 4 SSR loci that are all unlinked, each with 2 alleles at equal frequency in a random-mating plant population. What is the probability that two unrelated individuals share the identical genotype at all four loci? (Assume each locus genotype probability follows simplification with match probability of per locus.)
Molecular analysis of given plant DNA using DNA based markers
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
With independent (unlinked) loci, match probabilities multiply. For a biallelic locus at equal frequency, the probability two individuals share a genotype is . Across four loci: .
Incorrect! Try again.
47A SCAR marker is developed from a polymorphic RAPD band. What is the primary advantage of converting a RAPD marker into a SCAR marker?
Molecular analysis of given plant DNA using DNA based markers
Hard
A.It increases the number of amplified loci per reaction
B.It converts a dominant marker into a mitochondrial marker
C.Longer, locus-specific primers greatly improve reproducibility and specificity
D.It removes the need for a thermal cycler
Correct Answer: Longer, locus-specific primers greatly improve reproducibility and specificity
Explanation:
SCARs use longer sequence-specific primers designed from the ends of a cloned RAPD fragment. Because they anneal at high stringency to a single locus, they overcome the poor reproducibility of the original arbitrary RAPD primers.
Incorrect! Try again.
48In a CAPS (Cleaved Amplified Polymorphic Sequence) analysis, a 500 bp region is PCR-amplified from two alleles and then digested with a restriction enzyme. Allele 1 is cut into 200 bp + 300 bp fragments; allele 2 is uncut. This pattern indicates that the polymorphism is:
Molecular analysis of given plant DNA using DNA based markers
Hard
A.A methylation difference at the primer sites
B.A tandem repeat expansion in allele 1
C.A large insertion of 300 bp in allele 2
D.A SNP that creates or destroys the restriction site between the two alleles
Correct Answer: A SNP that creates or destroys the restriction site between the two alleles
Explanation:
CAPS detects sequence differences that alter a restriction site. Since both alleles amplify at 500 bp but only allele 1 is cut, the two alleles differ by a SNP that gives allele 1 an enzyme recognition site absent in allele 2.
Incorrect! Try again.
49A researcher observes that an SSR locus amplifies a main allele band accompanied by a ladder of smaller bands (stutter bands). What is the molecular cause of these stutter bands?
Molecular analysis of given plant DNA using DNA based markers
Hard
A.Incomplete denaturation of the template
B.Non-specific primer binding to ribosomal DNA
C.Contamination with genomic DNA from another species
D.Slippage of Taq polymerase across the repeat units during amplification
Correct Answer: Slippage of Taq polymerase across the repeat units during amplification
Explanation:
Stutter bands arise from polymerase slippage on the tandem repeat template, adding or removing repeat units. This produces minor bands differing by one repeat, which can complicate allele scoring in SSR analysis.
Incorrect! Try again.
50Two plant DNA samples give identical RAPD profiles across 15 primers. A colleague concludes they are genetically identical clones. What is the most valid criticism of this conclusion?
Molecular analysis of given plant DNA using DNA based markers
Hard
A.RAPD samples only a limited fraction of the genome, so identical profiles do not prove complete identity
B.RAPD detects only mitochondrial DNA and misses nuclear differences
C.RAPD requires sequencing confirmation for every band
D.RAPD cannot be used on plant DNA at all
Correct Answer: RAPD samples only a limited fraction of the genome, so identical profiles do not prove complete identity
Explanation:
RAPD interrogates only the loci where primers happen to bind, a small genome fraction. Identical banding is consistent with clonal identity but cannot rule out differences elsewhere; more markers or sequencing would be needed for firm conclusions.
Incorrect! Try again.
51In marker-assisted selection, a molecular marker is 5 cM from the target gene. What is the approximate probability that a recombination event separates the marker from the gene in a single meiosis?
Molecular analysis of given plant DNA using DNA based markers
Hard
A.About 0.5%
B.About 5%
C.About 25%
D.About 50%
Correct Answer: About 5%
Explanation:
By definition, 1 cM corresponds to roughly a 1% recombination frequency over short distances. A marker 5 cM from the gene has approximately a 5% chance of being separated by recombination per meiosis, making tighter markers preferable.
Incorrect! Try again.
52A dominant AFLP marker segregates in an F2 population. In a 3:1 (band present : band absent) ratio, what fraction of the band-present individuals are expected to be heterozygous?
Molecular analysis of given plant DNA using DNA based markers
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
In a 3:1 ratio the band-present class (3 parts) consists of 1 homozygous dominant and 2 heterozygous individuals. Thus of band-present plants are heterozygous, and dominant markers cannot distinguish these two genotypes directly.
Incorrect! Try again.
53When analyzing plant DNA, ISSR markers use primers targeting the region between two microsatellites. Why do ISSR primers not require prior sequence knowledge of the target genome, unlike SSR markers?
Molecular analysis of given plant DNA using DNA based markers
Hard
A.They anchor to ubiquitous, conserved microsatellite repeat motifs rather than unique flanking sequences
B.They bind only to chloroplast origins of replication
C.They amplify only single-copy genes
D.They rely on restriction digestion instead of amplification
Correct Answer: They anchor to ubiquitous, conserved microsatellite repeat motifs rather than unique flanking sequences
Explanation:
ISSR primers are complementary to common repeat motifs (e.g., (GA)n) with a short anchor. Because such repeats are abundant across genomes, no locus-specific flanking sequence data are needed, unlike SSRs which require designing primers from unique flanks.
Incorrect! Try again.
54A genetic diversity study computes Jaccard similarity from binary marker data. Why is the Jaccard coefficient often preferred over simple matching for dominant markers like RAPD and AFLP?
Molecular analysis of given plant DNA using DNA based markers
Hard
A.It ignores shared band absences, which are ambiguous for dominant markers
B.It requires codominant scoring to be valid
C.It weights band absences more heavily than presences
D.It corrects for polymerase slippage artifacts
Correct Answer: It ignores shared band absences, which are ambiguous for dominant markers
Explanation:
For dominant markers a shared absence (0,0) is ambiguous—it may reflect different underlying causes. Jaccard similarity excludes joint absences from the calculation, making it more appropriate than simple matching, which counts them as similarity.
Incorrect! Try again.
55A CAPS marker fails to distinguish two alleles even though a SNP is known to exist between them. The SNP does not fall within any available enzyme's recognition site. Which technique best rescues this situation without sequencing every sample?
Molecular analysis of given plant DNA using DNA based markers
Hard
A.Switching to a longer AFLP selective primer
B.Running the products on a native agarose gel at higher voltage
C.Increasing the RAPD primer concentration
D.dCAPS, which introduces a mismatch in the primer to create an enzyme site around the SNP
Correct Answer: dCAPS, which introduces a mismatch in the primer to create an enzyme site around the SNP
Explanation:
Derived CAPS (dCAPS) engineers a deliberate mismatch in one primer so that, combined with the SNP, a restriction site is created in one allele only. This allows enzyme-based discrimination of SNPs that do not naturally lie in a recognition sequence.
Incorrect! Try again.
56In a bulked segregant analysis (BSA) for a plant disease resistance gene, DNA from resistant and susceptible individuals is pooled into two bulks and screened with markers. A marker showing a band only in the resistant bulk most likely indicates the marker is:
Molecular analysis of given plant DNA using DNA based markers
Hard
A.An artifact of unequal DNA pooling
B.A neutral polymorphism unlinked to resistance
C.Located on the chloroplast genome
D.Linked to the resistance locus
Correct Answer: Linked to the resistance locus
Explanation:
In BSA, the two bulks differ (ideally) only around the target locus because other regions are randomized by pooling. A marker polymorphic between the bulks is therefore inferred to be linked to the resistance gene.
Incorrect! Try again.
57A polymorphism information content (PIC) value is calculated for two SSR loci. Locus X has 2 alleles at frequencies 0.5/0.5; Locus Y has 4 alleles at 0.25 each. Which locus is more informative and why?
Molecular analysis of given plant DNA using DNA based markers
Hard
A.Locus X, because fewer alleles reduce scoring errors
B.Locus Y, because more alleles at even frequencies raise heterozygosity and PIC
C.Both are equal because total allele number does not affect PIC
D.Locus X, because biallelic loci always have maximal PIC
Correct Answer: Locus Y, because more alleles at even frequencies raise heterozygosity and PIC
Explanation:
PIC increases with the number of alleles and their evenness of frequency. Locus Y with four equally frequent alleles yields higher expected heterozygosity and PIC than the biallelic Locus X, making it more informative for diversity and mapping.
Incorrect! Try again.
58During PCR of plant DNA for marker analysis, no amplification occurs despite good template quality. Adding PVP or BSA to subsequent reactions restores amplification. The original failure was most likely caused by:
Molecular analysis of given plant DNA using DNA based markers
Hard
A.Too high a denaturation temperature destroying dNTPs
B.Excessive primer concentration degrading the template
C.Polyphenols and polysaccharides co-purified with plant DNA inhibiting Taq polymerase
D.Absence of magnesium ions in the buffer
Correct Answer: Polyphenols and polysaccharides co-purified with plant DNA inhibiting Taq polymerase
Explanation:
Plant extracts commonly carry polyphenols and polysaccharides that inhibit Taq. Additives like PVP (binds polyphenols) and BSA (sequesters inhibitors) relieve this inhibition, restoring amplification—hence the diagnostic recovery.
Incorrect! Try again.
59A phylogenetic dendrogram built from AFLP data shows one accession clustering far from morphologically similar accessions. Before concluding it is genetically distinct, what is the most important check?
Molecular analysis of given plant DNA using DNA based markers
Hard
A.Increase the annealing temperature and rebuild the tree
B.Convert all AFLP markers to mitochondrial markers
C.Assume the morphology data are always wrong
D.Verify sample identity and rule out DNA contamination or scoring errors
Correct Answer: Verify sample identity and rule out DNA contamination or scoring errors
Explanation:
An unexpected outlier often reflects a technical problem—mislabeled sample, contamination, or misscored bands—rather than true divergence. Verifying identity and data quality is essential before drawing biological conclusions from the dendrogram.
Incorrect! Try again.
60SNP genotyping is increasingly favored over SSRs for large-scale plant analysis. Which combination of properties best explains this shift?
Molecular analysis of given plant DNA using DNA based markers
Hard
A.High genome abundance, bi-allelic simplicity, and amenability to high-throughput automation
B.Higher per-locus allele number and manual gel scoring
C.Requirement for restriction digestion and radioactive probes
D.Dominant inheritance and low genome coverage
Correct Answer: High genome abundance, bi-allelic simplicity, and amenability to high-throughput automation
Explanation:
SNPs are extremely abundant genome-wide and their simple bi-allelic nature suits automated, high-throughput array and sequencing platforms. Although individually less polymorphic than SSRs, their sheer number and scalability make them preferred for large studies.
Incorrect! Try again.
Did this save you a night before the exam?
LPU Notes is free, and it stays free. Ads cover part of the server bill.
The rest comes out of a student's own pocket: the domain, the storage,
and keeping the site up through the weeks everyone needs it at once.
The payment button didn't load. An ad blocker or a filtered network is the usual reason.
to try again.
Nothing here is ever locked, and nothing unlocks. Chip in only if it was worth it.
What it pays for →