Unit 4: Linkage, Chromosome Mapping and Bacterial Genetics - Subjective Questions
BTY551 — Genetics • Practice Questions with Detailed Answers
20 questions
Define linkage and crossing over. Explain how these two phenomena are related and why they are considered opposing forces in inheritance.
Linkage refers to the tendency of genes located on the same chromosome to be inherited together into the same gamete, because they do not assort independently.
Crossing over is the process of reciprocal exchange of segments between homologous chromosomes during Prophase I of meiosis (pachytene stage), resulting in new combinations of alleles (recombinants).
Relationship and opposing nature:
- Linkage keeps parental combinations of alleles together.
- Crossing over breaks these parental combinations and creates recombinant types.
- The greater the physical distance between two linked genes, the higher the probability of a crossover occurring between them, hence weaker linkage.
- Thus, linkage tends to reduce recombination, while crossing over tends to increase it. The observed recombination frequency is a balance between the two.
Significance: The frequency of recombinants (recombination frequency) is used as a measure of the distance between genes on a chromosome, forming the basis of genetic mapping.
Distinguish between complete linkage and incomplete linkage with suitable examples.
Complete Linkage:
- Genes are located very close together on the same chromosome.
- No crossing over occurs between them; only parental (non-recombinant) gametes are produced.
- Recombination frequency is essentially 0%.
- Classic example: Genes for body colour and wing size in male Drosophila melanogaster show complete linkage because there is no crossing over in male fruit flies.
Incomplete Linkage:
- Genes are located relatively farther apart on the same chromosome.
- Crossing over occurs between them, producing both parental and recombinant gametes.
- Recombinant types occur in a frequency less than 50% (parentals are more frequent).
- Example: Studies by Morgan on Drosophila female flies and by Bateson and Punnett on sweet peas (flower colour and pollen shape).
Comparison Table:
| Feature | Complete Linkage | Incomplete Linkage |
|---|---|---|
| Crossing over | Absent | Present |
| Gamete types | Only parental | Parental + recombinant |
| Recombination frequency | 0% | 0% < RF < 50% |
| Gene distance | Very close | Farther apart |
Describe the molecular mechanism of crossing over according to the Holliday model.
The Holliday model (proposed by Robin Holliday, 1964) explains crossing over as a form of homologous recombination at the molecular level.
Steps:
-
Alignment: Two homologous DNA duplexes align side by side.
-
Nicking: An endonuclease introduces single-strand nicks at corresponding positions in strands of the same polarity in both duplexes.
-
Strand invasion / exchange: The nicked strands leave their original partners and cross over to pair with the complementary strand of the homologous duplex.
-
Ligation: DNA ligase seals the nicks, forming a cross-shaped structure called the Holliday junction (chi structure).
-
Branch migration: The point of crossover moves along the duplexes, extending the region of heteroduplex DNA (hybrid DNA containing one strand from each parent).
-
Resolution: The Holliday junction is cut (resolved) by resolvase enzymes. Depending on the plane of cutting:
- Horizontal cut → produces non-recombinant (patch) products with only heteroduplex regions.
- Vertical cut → produces recombinant (splice) products with exchange of flanking markers.
Significance: This model explains gene conversion, heteroduplex formation, and reciprocal recombination observed during crossing over.
Explain the cytological proof of crossing over as demonstrated by Stern (in Drosophila) or Creighton and McClintock (in maize).
The cytological proof established that genetic recombination is accompanied by physical exchange of chromosome segments.
Creighton and McClintock (1931) – Maize:
- They used a chromosome 9 of maize that carried two cytological markers:
- A knob (heterochromatic thickening) at one end.
- A translocated piece of chromosome 8 at the other end.
- These physical markers were linked to genetic markers: C (coloured/colourless endosperm) and Wx (waxy/starchy).
- Plants heterozygous for both cytological and genetic markers were crossed.
- Result: Whenever the offspring showed recombinant genetic phenotypes, the chromosomes also showed a physical exchange of the cytological markers (knob and translocation).
Stern (1931) – Drosophila:
- Used X chromosomes with structural abnormalities (an L-shaped X and a fragment attached to Y) plus genetic markers.
- Recombinant flies always carried chromosomes with correspondingly exchanged physical structures.
Conclusion: Genetic crossing over corresponds to a physical exchange of chromosomal material between homologous chromosomes.
In a two-point test cross, a total of 1000 offspring were obtained: 410 parental type A, 400 parental type B, 95 recombinant type C, and 95 recombinant type D. Calculate the recombination frequency and the map distance between the two genes.
Given data:
- Parental types: 410 + 400 = 810
- Recombinant types: 95 + 95 = 190
- Total offspring = 1000
Recombination Frequency (RF):
Map Distance:
One map unit (centimorgan, cM) = 1% recombination frequency.
Conclusion: The two genes are located 19 map units (19 cM) apart on the chromosome. The predominance of parental types confirms the two genes are linked.
Explain the principle of a three-point test cross and describe how it is used to determine gene order and map distances.
A three-point test cross involves crossing a trihybrid individual (heterozygous for three linked genes) with a fully recessive (triple homozygous) individual. It allows determination of gene order, map distances, and interference in a single cross.
Principle & Procedure:
-
Cross: Trihybrid () × triple recessive ().
-
Classify offspring into 8 phenotypic classes (2^3), forming 4 reciprocal pairs:
- Parental (non-crossover) – most frequent.
- Single crossover between gene 1 and 2 (SCO I).
- Single crossover between gene 2 and 3 (SCO II).
- Double crossover (DCO) – least frequent.
-
Determine gene order: Compare the parental class with the double crossover class. The gene whose allele is switched (the one differing) in the DCO relative to parental is the middle gene.
-
Calculate map distances:
Double crossovers are added to both regions because they involve two exchanges.
- Coefficient of Coincidence (C.O.C.) and Interference (I):
Advantage: More accurate than two-point crosses since it detects double crossovers, giving true map distances.
Define interference and coefficient of coincidence. Explain their significance in chromosome mapping.
Coefficient of Coincidence (C.O.C.):
It is the ratio of the observed number of double crossovers to the expected number of double crossovers.
Expected DCO = (frequency of crossover in region I) × (frequency of crossover in region II) × total offspring.
Interference (I):
It is the phenomenon in which one crossover event inhibits the occurrence of another crossover in a nearby region. It is calculated as:
Interpretation:
- Positive interference (I > 0): One crossover reduces the chance of a second nearby crossover (most common in eukaryotes).
- No interference (I = 0, C.O.C. = 1): Crossovers are independent.
- Negative interference (I < 0): One crossover promotes another.
Significance:
- Explains why observed double crossovers are often fewer than expected.
- Essential for calculating accurate map distances.
- Reflects the physical constraints (rigidity) of chromosomes during meiosis.
What is tetrad analysis? Explain its importance and describe the types of tetrads (PD, NPD, T) obtained in fungi like Neurospora or yeast.
Tetrad analysis is a genetic technique in which all four products of a single meiosis (a tetrad) are recovered and analyzed together. It is possible in fungi such as Neurospora crassa, Saccharomyces cerevisiae (yeast), and algae, where meiotic products remain in a single structure (ascus).
Importance:
- Allows direct study of the products of a single meiotic event.
- Confirms reciprocal nature of recombination.
- Enables detection of gene conversion and first/second division segregation.
- Used to map gene-centromere distance (in ordered tetrads).
Types of Tetrads (for two genes A and B):
-
Parental Ditype (PD): All four spores are parental type. Two classes: 2 AB : 2 ab.
-
Non-Parental Ditype (NPD): All four spores are recombinant type. 2 Ab : 2 aB.
-
Tetratype (T): Contains all four types—two parental and two recombinant. 1 AB : 1 ab : 1 Ab : 1 aB.
Linkage determination:
- If PD ≈ NPD → genes are unlinked (independent assortment).
- If PD >> NPD → genes are linked.
Recombination frequency:
Explain how gene-centromere distance is calculated using ordered tetrad analysis in Neurospora.
In Neurospora crassa, the ascus contains ordered tetrads (later 8 ascospores), where the linear arrangement of spores reflects the order of meiotic divisions. This allows mapping of a gene relative to its centromere.
Concept of segregation patterns:
-
First Division Segregation (FDS): If no crossover occurs between the gene and centromere, alleles separate during Meiosis I. The spore pattern is like AAAA aaaa (4:4) — a parental arrangement.
-
Second Division Segregation (SDS): If a crossover occurs between the gene and centromere, alleles separate during Meiosis II. Patterns become AA aa AA aa or AA aa aa AA (2:2:2:2 or 2:4:2) — recombinant arrangements.
Calculation of gene-centromere distance:
Each SDS ascus arises from a single crossover involving 2 of the 4 chromatids, so only half the spores in an SDS ascus are recombinant. Therefore:
Example: If out of 100 asci, 20 show second division segregation:
Thus, the gene lies 10 cM from its centromere.
Define bacterial transformation. Describe the classical experiment of Griffith and its significance.
Bacterial Transformation is the process by which a bacterial cell takes up free (naked) DNA from its surroundings (released from dead donor cells) and incorporates it into its genome, thereby acquiring new genetic traits.
Griffith's Experiment (1928) – Streptococcus pneumoniae:
Griffith worked with two strains:
- S (smooth) strain: virulent, capsulated, causes pneumonia.
- R (rough) strain: non-virulent, non-capsulated, harmless.
Observations:
| Injection into mice | Result |
|---|---|
| Live S strain | Mice died |
| Live R strain | Mice survived |
| Heat-killed S strain | Mice survived |
| Heat-killed S + Live R strain | Mice died (live S recovered) |
Conclusion: A "transforming principle" from the dead S cells transformed the harmless R cells into virulent S cells, and this trait was heritable.
Significance:
- First demonstration of genetic transfer between bacteria.
- Later, Avery, MacLeod, and McCarty (1944) identified the transforming principle as DNA, establishing DNA as the genetic material.
Explain the process of gene mapping by transformation in bacteria. How is cotransformation frequency used to determine gene linkage?
Mapping by transformation relies on the principle that genes located close together on the donor DNA are more likely to be carried on the same DNA fragment and thus transferred together into the recipient.
Principle of Cotransformation:
- When donor DNA is fragmented (during extraction), closely linked genes tend to remain on the same fragment.
- If a recipient cell simultaneously acquires two genetic markers, they are said to be cotransformed.
- The frequency of cotransformation is proportional to the closeness of the genes.
Interpretation:
- High cotransformation frequency → genes are closely linked.
- Low cotransformation frequency → genes are far apart.
- If two genes are never cotransformed (frequency = single-transformation product), they are unlinked.
Procedure:
- Isolate DNA from a donor strain carrying markers (e.g., ).
- Add DNA to competent recipient cells ().
- Select for transformants and score for single vs double (co)transformants.
- Calculate cotransformation frequency:
Conclusion: Relative gene order and distances are deduced by comparing cotransformation frequencies among several markers.
Define transduction. Distinguish between generalized and specialized transduction with suitable examples.
Transduction is the transfer of bacterial genetic material from a donor to a recipient bacterium through the agency of a bacteriophage (virus). It was discovered by Zinder and Lederberg (1952) in Salmonella typhimurium.
Generalized Transduction:
- Occurs during the lytic cycle.
- During phage assembly, a phage head mistakenly packages random fragments of bacterial DNA instead of phage DNA.
- Any gene of the donor can be transferred.
- Carried out by phages like P22 (in Salmonella) and P1 (in E. coli).
Specialized (Restricted) Transduction:
- Occurs when a temperate (lysogenic) phage excises imprecisely from its specific integration site in the host chromosome.
- Only genes adjacent to the prophage attachment site are transferred.
- Limited/specific genes are transferred.
- Classic example: Lambda (λ) phage in E. coli transferring the gal (galactose) or bio (biotin) genes located near its att site.
Comparison Table:
| Feature | Generalized | Specialized |
|---|---|---|
| Phage cycle | Lytic | Lysogenic |
| Genes transferred | Any random gene | Only genes near att site |
| DNA packaging error | Random bacterial DNA | Faulty prophage excision |
| Example | P1, P22 | Lambda (λ) |
Describe how cotransduction frequency is used for gene mapping in bacteria. Give the formula relating cotransduction frequency to map distance.
In generalized transduction, a phage packages only a small fragment of the bacterial chromosome (~1–2% of the genome). Two genes can be cotransduced only if they are close enough to fit on the same DNA fragment inside a single phage head.
Principle:
- The closer two genes are, the higher their cotransduction frequency.
- The farther apart, the lower the frequency; genes too far apart are never cotransduced.
Wu's Formula (Cotransduction and distance):
Wu (1966) derived a relationship between cotransduction frequency and physical distance:
Where:
- = distance between the two genes (in minutes or map units).
- = length of the DNA fragment that a phage can package.
Procedure for mapping:
- Grow phage on donor ().
- Infect recipient () and select for one marker (e.g., ).
- Score the fraction that also carries the second marker ().
- Higher cotransduction = closer linkage.
Determining gene order: Three-factor cotransduction crosses help establish the correct order of three closely linked genes by comparing frequencies of different marker combinations.
Explain bacterial conjugation. Describe the role of the F (fertility) factor and distinguish between , , Hfr, and cells.
Conjugation is the transfer of genetic material from a donor to a recipient bacterium through direct cell-to-cell contact, mediated by a sex pilus. It was discovered by Lederberg and Tatum (1946) in E. coli.
F (Fertility) Factor:
- A plasmid (episome) carrying genes (tra genes) required for pilus formation and DNA transfer.
- Determines the mating type (donor vs recipient).
Types of cells:
-
cells (Donor):
- Contain the F factor as a free plasmid.
- On conjugation with , they transfer only the F plasmid, converting into . Chromosomal genes are rarely transferred.
-
cells (Recipient):
- Lack the F factor.
- Act as recipients; can become after conjugation.
-
Hfr cells (High frequency recombination):
- The F factor is integrated into the bacterial chromosome.
- Transfer chromosomal genes at high frequency in a linear, oriented manner.
- The F factor is usually not transferred completely, so recipients rarely become .
-
cells (F prime):
- Formed when an integrated F factor excises imperfectly from an Hfr chromosome, carrying adjacent chromosomal genes with it.
- Transfers both the F factor and the attached bacterial genes → leads to sexduction (F-duction).
Significance: Conjugation, especially via Hfr strains, is a powerful tool for mapping the bacterial chromosome.
Explain the interrupted mating experiment of Wollman and Jacob and how it is used to map bacterial genes in minutes.
The interrupted mating experiment (Wollman and Jacob, 1955) is used to map bacterial genes based on the time of entry of genes from an Hfr donor into an recipient during conjugation.
Principle:
- In Hfr × mating, the Hfr chromosome is transferred linearly and sequentially starting from a fixed origin (O).
- Genes closer to the origin enter the recipient earlier; distant genes enter later.
- The chromosome transfer takes about 100 minutes for the entire E. coli chromosome.
Procedure:
- Mix Hfr donor (carrying several markers, e.g., ) with recipient.
- At regular time intervals, remove samples and agitate violently in a blender to break conjugating pairs, interrupting DNA transfer.
- Plate cells on selective media to detect which donor markers have entered.
Observation:
- Each gene appears in recipients only after a specific minimum time — its time of entry.
- Plotting the time of entry vs frequency of recombinants gives the order and relative position of genes.
Mapping:
- Map distances are measured in minutes (time units) rather than percentage recombination.
- Example: If azi enters at 9 min, ton at 10 min, lac at 18 min, gal at 25 min — this gives the gene order and distances.
Significance: Established the circular nature of the E. coli chromosome and provided a time-based genetic map.
Compare the three modes of gene transfer in bacteria: Transformation, Transduction, and Conjugation.
Bacteria exchange genetic material through three main parasexual processes. Their comparison is given below:
| Feature | Transformation | Transduction | Conjugation |
|---|---|---|---|
| Definition | Uptake of free naked DNA from environment | Transfer of DNA via bacteriophage | Transfer via direct cell contact |
| Discovered by | Griffith (1928) | Zinder & Lederberg (1952) | Lederberg & Tatum (1946) |
| Vector/Agent | Free DNA fragments | Bacteriophage | Sex pilus / F factor |
| Cell contact | Not required | Not required | Required |
| DNA source | Dead donor cells | Donor packaged in phage | Living donor cell |
| Requirement | Competent recipient | Suitable phage | F factor (plasmid) |
| Amount of DNA | Small fragment | Small fragment (~1-2% genome) | Can be large (up to whole chromosome) |
| Mapping unit | Cotransformation frequency | Cotransduction frequency | Time (minutes) |
Common feature: All three are unidirectional (donor → recipient), transfer only part of the genome, and involve recombination of donor DNA into the recipient chromosome.
Describe the experiments of Bateson and Punnett on sweet peas that first indicated the phenomenon of linkage. Explain the terms coupling and repulsion.
Bateson and Punnett's Experiment (1905–1908) – Sweet Pea (Lathyrus odoratus):
They studied two traits:
- Flower colour: Purple (P) dominant over red (p).
- Pollen shape: Long (L) dominant over round (l).
Cross: Purple, long-pollen (PPLL) × Red, round-pollen (ppll).
- F1: All Purple, long (PpLl).
- F1 selfed → F2: Expected a 9:3:3:1 ratio under independent assortment.
Observation: The F2 ratio deviated from 9:3:3:1. The parental combinations (purple-long and red-round) appeared in excess, while recombinant types (purple-round, red-long) were fewer than expected.
Interpretation: The two dominant genes (P and L) tended to stay together, as did the two recessives (p and l) — indicating they were linked on the same chromosome. Bateson and Punnett called this "partial coupling" but could not explain it fully; Morgan later explained it as linkage.
Coupling and Repulsion:
- Coupling (cis) phase: Both dominant alleles are on the same chromosome (). Parentals are dominant-dominant and recessive-recessive combinations.
- Repulsion (trans) phase: Dominant allele of one gene is on the same chromosome as the recessive allele of the other ().
Morgan later showed coupling and repulsion are simply two arrangements of the same linked genes.
A three-point test cross in Drosophila gave the following results for genes , , and . Determine the gene order, map distances, and coefficient of coincidence.
| Phenotype | Number |
|---|---|
| / (Parental) | 580 + 592 |
| $a+ + $ / (SCO) | 45 + 40 |
| / (SCO) | 89 + 94 |
| / (DCO) | 3 + 5 |
Total = 1448 offspring.
Step 1: Identify the classes.
- Parental (most frequent): = 580, = 592 → Total = 1172
- DCO (least frequent): = 3, = 5 → Total = 8
- SCO region I: = 45, = 40 → Total = 85
- SCO region II: = 89, = 94 → Total = 183
Step 2: Determine gene order.
Compare parental () with DCO (). The allele that has switched position is (b moved out). Therefore is the middle gene.
Gene order:
Step 3: Calculate map distances.
Double crossovers are added to both regions.
Distance – (region containing SCO I):
Distance – (region containing SCO II):
Total map distance –: cM
Step 4: Coefficient of Coincidence (C.O.C.) and Interference.
Conclusion: Gene order is ; distances are 6.42 cM and 13.19 cM; interference of 0.35 indicates positive interference (crossovers partly inhibit each other).
Explain the chiasmatype theory and the concept of chiasma in relation to crossing over.
Chiasma (plural: chiasmata):
A chiasma is the visible X-shaped point of contact between non-sister chromatids of homologous chromosomes observed during the diplotene stage of Prophase I of meiosis. It represents the site where crossing over has occurred.
Chiasmatype Theory (Janssens, 1909):
- Proposed by F.A. Janssens, this theory states that chiasmata are the physical manifestation of crossing over.
- It suggested that at each chiasma, there is an actual breakage and reunion of non-sister chromatids, resulting in exchange of segments.
- The theory provided a cytological basis for the genetic recombination observed by Morgan.
Relationship between chiasma and crossing over:
- Each chiasma corresponds to one crossover event.
- The number and position of chiasmata influence the frequency of recombination.
- A single crossover involves only two of the four chromatids (one from each homolog); thus, maximum recombination frequency for a single crossover is 50%.
Terminalisation:
- As meiosis proceeds toward diakinesis, chiasmata appear to move toward the ends of chromosomes — a process called terminalisation.
Significance:
- Established the cytological proof and mechanism linking visible chromosome behaviour to genetic recombination.
- Confirmed that crossing over occurs at the four-strand (tetrad) stage, not the two-strand stage.
Why can the maximum recombination frequency between two genes never exceed 50%? Explain with reference to crossing over between four chromatids.
The maximum recombination frequency (RF) between two genes cannot exceed 50%, and this observation has an important cytological basis.
Explanation using four-strand (tetrad) crossing over:
At the pachytene stage, each pair of homologous chromosomes consists of four chromatids (a tetrad). A single crossover occurs between only two of these four chromatids (one non-sister chromatid from each homolog).
Consequence of a single crossover:
- Two chromatids participate → produce 2 recombinant chromatids.
- Two chromatids do not participate → remain 2 parental (non-recombinant) chromatids.
- Therefore, a single crossover yields 50% recombinants and 50% parentals among the four products.
Effect of multiple crossovers:
- When genes are far apart, multiple crossovers occur between them.
- Double crossovers involving various chromatid combinations (2-strand, 3-strand, 4-strand double crossovers) on average restore parental combinations as often as they create recombinants.
- As genes become progressively farther apart, the RF approaches but plateaus at 50%, the same value expected for independently assorting (unlinked) genes.
Conclusion:
- RF = 50% indicates either genes are on different chromosomes or are so far apart on the same chromosome that they assort independently.
- RF < 50% indicates linkage.
- Because only 2 of 4 chromatids recombine per crossover, the theoretical ceiling remains 50%. This is why very distant genes on the same chromosome must be mapped using intermediate markers.
Define linkage and crossing over. Explain how these two phenomena are related and why they are considered opposing forces in inheritance.
Linkage refers to the tendency of genes located on the same chromosome to be inherited together into the same gamete, because they do not assort independently.
Crossing over is the process of reciprocal exchange of segments between homologous chromosomes during Prophase I of meiosis (pachytene stage), resulting in new combinations of alleles (recombinants).
Relationship and opposing nature:
- Linkage keeps parental combinations of alleles together.
- Crossing over breaks these parental combinations and creates recombinant types.
- The greater the physical distance between two linked genes, the higher the probability of a crossover occurring between them, hence weaker linkage.
- Thus, linkage tends to reduce recombination, while crossing over tends to increase it. The observed recombination frequency is a balance between the two.
Significance: The frequency of recombinants (recombination frequency) is used as a measure of the distance between genes on a chromosome, forming the basis of genetic mapping.
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