Gene mapping methods MCQ Quiz in தமிழ் - Objective Question with Answer for Gene mapping methods - இலவச PDF ஐப் பதிவிறக்கவும்

The correct answer is Phenocopy

Explanation:

A phenocopy is an individual whose phenotype, induced by environmental factors, mimics a specific genotype other than its own. This means that despite having a normal genotype for a given trait, the phenotype exhibited is similar to that of a known mutant due to environmental influences, rather than genetic changes.

In the scenario with the Drosophila larvae:

• The environmental factor (exposure to 37 °C during growth) induced the crossveinless phenotype in an individual that presumably had a normal genotype with respect to the genes controlling vein formation in the wings.
• When this environmentally induced crossveinless fly was crossed to a genetic crossveinless mutant, all the progeny had a normal phenotype, which indicates that the parental fly with the induced phenotype did not genetically carry the mutation for the crossveinless trait. Instead, its phenotype was a copy (phenocopy) of the mutation induced by the environmental condition (high temperature).

This distinguishes the observed phenomenon from the other options because:

• Conditional Mutant refers to a mutant that exhibits a particular phenotype only under certain (specific) environmental conditions. While this seems close, the key difference is that in a phenocopy, the organism does not have a mutant genotype, whereas conditional mutants do.
• Penetrance refers to the proportion of individuals carrying a particular variant of a gene (allele) that also express an associated trait (phenotype). In this case, the trait was not genetically inherited, hence penetrance is not the correct explanation.
• Pleiotropy occurs when one gene influences two or more seemingly unrelated phenotypic traits. This concept doesn't apply here since the observed change in phenotype was due to an environmental factor, not the multiple effects of a single gene.

Therefore, Phenocopy (Option 2) is the correct explanation for the observed phenotype, as it accounts for the change in phenotype due to environmental factors, rather than genetic mutations.

The correct answer is Option 3 i.e.b c d e a f g

Explanation-

• The key principle here is that point mutations can recombine with deletions that do not extend past the mutation, but they cannot recombine to yield wild-type phages with deletions that do extend past the mutation.
• In a simpler way we can find the sequence by observing the length of mutation and which gene functions are getting affected by that and then comparing the overlapping regions we can predict the correct sequence.
• In this question if we focus on the last gene in the map, it is affected by only deletion 5.
• In the table function of g gene is affected by only deletion 5. So g must be last in the sequence.
• Similarly by overlapping the deletions and comparing the loss of function of genes we can predict the sequence which is option 3 here.
• The correct order is bcdeafg

The correct answer is Option 1 i.e. 20.

Explanation-

Given that the genes w⁺ and cv are located 20 cM apart, we can use the concept of genetic recombination to estimate the frequency of recombination between these genes.

Parental Genotypes:

• The homozygous wild-type female (w⁺ w⁺ cv⁺ cv⁺) is crossed to a white-eyed, crossveinless male (w w cv cv).
• Female gametes: w⁺ cv⁺
• Male gametes: w cv

Possible Gamete Combinations in F1:

• w⁺ cv⁺ (Parental)
• w cv (Parental)
• w⁺ cv (Recombinant)
• w cv⁺ (Recombinant)

Cross Between F1 Individuals:- F1 female (w⁺ cv⁺) x F1 male (w cv)
Possible genotypes in F2:-

• w⁺ w⁺ cv⁺ cv⁺
• w w cv cv
• w⁺ w cv cv⁺ (recombinant)
• w w⁺ cv⁺ cv (recombinant)

Phenotypic Analysis:- The white-eyed, crossveinless phenotype is associated with the double mutant genotype (w w cv cv).
Calculation:-

• The recombinant genotypes (w⁺ w cv cv⁺ and w w⁺ cv⁺ cv) represent the individuals with the white-eyed, crossveinless phenotype.
• The frequency of recombinant gametes is proportional to the recombination frequency (20 cM). Therefore, the frequency of recombinant progeny is half of the recombination frequency.

If the recombination frequency is 20 cM, the frequency of recombinant progeny is\(\frac{80}{100} \times \frac{1}{2} \times \frac{1}{2} \)
Therefore, 20% of the F2 progeny will be white-eyed and crossveinless.

The correct answer is C, E, F

Concept:

• Genetics of Drosophila: Drosophila melanogaster, commonly known as the fruit fly, is a model organism in genetics. It has four pairs of chromosomes, one of which is the sex chromosome (X and Y). Traits such as eye color and wing formation can be linked to specific genes on these chromosomes.
• X-linked Recessive Traits: Traits that are linked to the X chromosome and are recessive are expressed in males (XY) if they inherit the X chromosome carrying the mutation from their mother. Females (XX) need two copies of the recessive allele to express the trait.
• Genetic Mapping: The distance between genes on a chromosome can be measured in centimorgans (CM), where 1 CM represents a 1% chance of recombination occurring between two genes during meiosis.

Explanation:

• Statement A: F1 females were also of four types as that of males.
  • This is incorrect because F1 females will be heterozygous for both traits (white eye and crossveinless) and will display the wild-type phenotype if these traits are recessive.
• Statement B: The white-eyed crossveinless male flies appeared due to independent assortment.
  • This is incorrect because these traits are X-linked, and their appearance is due to recombination, not independent assortment.
• Statement C: The map distance between the genes for white eye and crossveinless is estimated to be 12 CM.
  • This is correct. The number of recombinants (white-eyed and crossveinless males) is 6 out of 100, giving a recombination frequency of 6%. Therefore, the map distance is 12 CM (since 1% recombination = 1 CM).
• Statement D: The map distance between white eye and crossveinless is estimated to be 6 CM.
  • This is incorrect because the correct map distance is 12 CM as calculated from the recombination frequency.
• Statement E: All F1 females are expected to be wild type.
  • This is correct. F1 females will inherit a wild-type allele from the wild-type male and a mutant allele from the virgin female, making them heterozygous and thus displaying the wild-type phenotype.
• Statement F: The F1 wild-type males appeared due to crossing over.
  • This is correct. The wild-type males are a result of recombination between the X chromosomes during gamete formation in the heterozygous females.

The correct answer is Option 2 i.e. A, B, D

Explanation:

A) Recombination could be identified by genotyping parents and offsprings for a pair of loci.

• Genotyping involves determining the genetic makeup of an organism by examining its DNA sequence at specific loci. By comparing the genotype of parents and their offspring, scientists can identify instances of recombination
• Recombination refers to the process by which alleles (variants of a gene) are shuffled during meiosis, the type of cell division that produces gametes (sperm and eggs).
• When alleles from different loci on homologous chromosomes are assorted into gametes in new combinations, this genetic shuffling can be discerned by noting allele combinations in offspring that differ from those in the parents, providing evidence of crossing-over events.

B) Recombination frequency does not exceed 0.5, and therefore, 50cM would be the maximum distance between two loci.

• The recombination frequency is a measure of genetic linkage and is used to estimate the physical distance between two loci on a chromosome.
• The frequency ranges from 0 (completely linked, with no recombination observed between the loci) to 0.5 (unlinked, with loci either far apart on the same chromosome or on different chromosomes, appearing to assort independently). Because of this, the genetic map distance is capped at 50 centimorgans (cM), as a frequency of 0.5 (or 50%) implies random assortment.
• However, the actual physical distance between loci can be larger than what 50cM might suggest, particularly for loci on different chromosomes or far-separated on the same chromosome.
• The reason for this upper limit is due to the way genetic linkage is detected and quantified, as well as the mechanism of crossing over during meiosis, which does not increase linearly with physical distance beyond a certain point.

C) Recombination is usually a reciprocal process involving the exchange of genetic material of equal value between the participating chromosomes.

• Recombination, particularly during meiosis, is indeed often reciprocal, meaning it involves the mutual exchange of genetic segments between homologous chromosomes. This process contributes to genetic diversity among gametes (sperm and egg cells).The term "equal value" in the statement can be misleading because it implies a strict equivalence in the exchanged genetic material's content or function, which is not always the case from a functional or evolutionary perspective.
• Sometimes, recombination can be non-reciprocal, leading to "gene conversion" events.
• In gene conversion, a sequence from one chromosome is copied to the other, replacing the original sequence. This does not involve an equal exchange but rather a unidirectional transfer of genetic material. Gene conversion can result in alleles being copied or overwritten, which affects genetic diversity and allele frequencies in populations without a reciprocal exchange.

D) Occasionally non-homologous recombination happens and this functions as a source of chromosomal rearrangement.

• Normally, recombination occurs between homologous sequences, meaning between similar or identical sequences on paired chromosomes during meiosis, ensuring proper chromosome segregation and contributing to genetic diversity.
• However, errors can occur, leading to non-homologous recombination - the exchange of genetic material between non-identical loci. This can result in chromosomal rearrangements such as deletions (loss of a chromosome segment), duplications (gain of extra copies of a segment), inversions (reversal of a segment orientation), and translocations (movement of segments between non-homologous chromosomes).
• These rearrangements can have significant evolutionary implications by creating new genetic variants. They can also lead to diseases if they disrupt gene function or regulation.

Conclusion:

Therefore, the correct statements are A,B and D

Concept:

Recombination

• Recombination is a process that occurs during the formation of sex cells (sperm and eggs) in a process called meiosis, enabling the exchange of genetic material between homologous chromosomes.
• This genetic exchange contributes to the genetic diversity of offspring, crucial for evolution and adaptation.
• The points at which these exchange events occur are called 'crossing over' points.
• Recombination has several key characteristics.
• It is reciprocal, meaning that the exchange of DNA occurs between homologous chromosomes.
• However, scenarios exist where non-reciprocal exchange occurs, leading to gene conversion.
• Occasionally, non-homologous recombination occurs, such as during repair of double-strand breaks in DNA, which can lead to chromosomal rearrangements—a significant source of genetic variation and potentially diseases.
• Recombination frequency is a measure used to indicate the relative distance between two genes on a chromosome based on the likelihood of a recombination event occurring between them.
• If the genes are close together (linked), they are likely to be inherited together and not typically separated by recombination (low recombination frequency).
• Conversely, if the genes are far apart or on different chromosomes (unlinked), the likelihood of a recombination event between them is high (high recombination frequency).
• It's worth noting that the recombination frequency is often expressed in units called centimorgans (cM), where 1 cM equates to a 1% chance that a recombination event will occur between two genes per generation.
• The maximum typical value for recombination frequency is 0.5 or 50 cM, which implies that the genes assort independently and are treated as if they are on separate chromosomes.

Explanation:

Statement A is incorrect 

• Because recombination can indeed be identified by genotyping parents and offspring for a pair of loci.
• Detecting recombination events depends on observing changes in the arrangement of specific genes or genetic markers between parents and their offspring.

Statement B is correct 

• The recombination frequency in genetic mapping is often measured in centimorgans (cM), and a maxim distance of 50 cM tends to be the point at which recombination is as likely as no recombination.
• Therefore, 50 cM represents a sort of "maximum" genetic distance in these cases.

Statement C is incorrect 

• Generally, recombination is a reciprocal process where segments of DNA are exchanged between homologous chromosomes during meiosis.
• Non-reciprocal recombination - where one DNA sequence replaces another without exchange - is less common, and is typically referred to as gene conversion, rather than general recombination.

Statement D is correct 

• Occasionally, non-homologous recombination does occur and can lead to chromosomal rearrangements.
• This process involves breaking and rejoining of non-homologous (genetically dissimilar) chromosomes.
• Errors in this process can lead to duplications, deletions, inversions, and translocations.
• These changes contribute to genomic diversity, evolution, and can sometimes lead to certain genetic diseases.

Therefore, the combination of all correct statements would be B and D.

Hence the correct answer is option 3

The correct answer is Option 4 i.e. C and D  

Key Points

Genetic linkage and mapping of genes:

• This question is related to genetic linkage and the mapping of genes on a chromosome based on the frequency of recombination events.
• Bacteriophages that have two mutations causing different plaque phenotypes, and the genes responsible for these mutations are 9 cM (centimorgans) apart.
• Genetic linkage refers to the tendency of genes that are physically close on a chromosome to be inherited together more frequently.
• The closer two genes are, the less likely they are to undergo recombination during gamete formation.
• Recombination is the exchange of genetic material between homologous chromosomes during meiosis.
• Based on the provided information, we have to predict the outcome of a genetic cross involving bacteriophages with specific genotypes.
• The expected proportions of different plaque types are determined by the distances between the genes and the likelihood of recombination events occurring during the mixing and infecting process.

Explaination:

• Cross given c⁺ m^- x c^- m^+. Since the genes are 9 cM apart means the the genes are linked.
• Therefore the proportion of parental types (c⁺ m^- and c^- m⁺) will be much larger than the recombinant types (c⁺ m⁺ and c^- m^-).
• Based on this we can eliminate the proportions given in A and B.
• Distance between the genes is recombination frequency percentage.
• 9 cM distance means recombination frequency of 9%.
• Therefore, proportion of each recombinants (c^₊ m⁺ and cm^-) = 0.09/ 2 = 0.045.
• So, the proportion of each parentals (c⁺ m^- and c^- m⁺) = (1-0.09)/ 2 = 0.91/2 = 0.455.
• For statement C, Total = 45+455+455+45 = 1000.
• Number of each recombinants (c⁺ m⁺ and c^- m^-) = 0.045 x 1000 = 45; Number of each parentals (c⁺ m^- and c^- m⁺) = 0.455 x 1000 = 455.
• So C is correct.
• For statement D, Total = 65+ 680+685+70 = 1500. Number of each recombinants (c⁺ m⁺ and c^- m^-) = 0.045 x 1500 = 67.5; Number of each parentals (c⁺ m^- and c^- m⁺) = 0.455 x 1500 = 682.
• So, D is also possible. 

Hence the correct answer is Option 4

Concept:

• Environment involves both living organisms and the non-living physical conditions.
• These two are inseparable but inter-related.
• For food, shelter, growth and development, all life systems interact with the environment.
• Environment is a life supporting system. In the subject of ecology, the term ecosystem refers to the environment of life. It is a self-sustaining, structural and functional unit of biosphere.
• An ecosystem may be natural or artificial, land-based or water-based.
• Artificial systems may include a cropland, a garden, a park or an aquarium. 

​Explanation:

Option 1: Ecosystem is an open system

• Because energy and nutrients are exchanged between the physical world and living things in an ecosystem, an ecosystem is an open system.
• In a closed environment, other organisms, such as decomposers, use an organism's waste product.

Option 2:  In an artificial ecosystem flow of energy is not unidirectional

• Manufactured ecosystems are man-made systems in which biotic and abiotic elements are designed to coexist in harmony.
• It can die without human assistance because it is not self-sustaining.
• A few examples of artificial ecosystems are zoos, farming areas, and aquariums.
• Energy flow is always unidirectional in any ecosystem.

Option 3:  Ecosystem is self sustaining and dynamic structure 

• In a balanced condition, ecosystem functioning is self- regulating and self-sustaining.
• This dynamic nature of ecosystem is dependent upon a number of factors including flow of energy, cycling of materials and perturbations, both intrinsic and extrinsic.

Option 4: Sun  is the ultimate source of energy for any ecosystem

• Because it is the main source of energy for the planet, the sun is referred to as the ultimate source of energy.
• All green plants, or producers, use solar energy to carry out the process of photosynthesis, which yields sustenance.
• Animals then consume plants to obtain the same chemical energy for all of their actions.

​

hence the correct answer is option 2

The correct answer is Option 3 i.e. Sepia and tubby flies in a ratio of 1 ∶ 2

Explanation-

• In Drosophila genetics, balancer chromosomes are used to maintain specific genetic combinations and prevent recombination between homologous chromosomes.
• Balancer chromosomes typically carry multiple inversions, which suppress recombination between the wild-type chromosome and the chromosome carrying the desired mutation.
• Additionally, balancers often carry dominant markers to facilitate their identification in crosses.
• In this scenario, a Drosophila male with sepia eye color (se/se) is crossed with a female carrying a third chromosome balancer called TM6B. The TM6B balancer chromosome carries a dominant marker for a tubby phenotype (+/TM6B), and homozygosity for the balancer allele (TM6B/TM6B) is lethal.

The F1 progeny resulting from this cross will be heterozygous for both the sepia allele and the TM6B balancer allele (+/se; +/TM6B). When these F1 individuals are sib-mated, the possible genotypic combinations in the F2 generation are as follows:

Tt x Tt = TT:Tt:tt

As mentioned, an individual homozygous for TM6B balancer does not survive so TT individual will die.

Therefore, Tt : 2, tt :1

• The phenotypic outcome for the heterozygous (Tt) individuals express a "tubby" phenotype 
• The homozygous recessive (tt) individuals will have the "sepia" phenotype.

This ratio is consistent with the expected Mendelian inheritance pattern for a cross involving a recessive trait (sepia eye color) and a dominant marker (tubby phenotype) carried on a balancer chromosome.

The correct answer is Option 2 

Explanation-

Bands labeled (b) and (d) are allelic:

This conclusion is drawn from the observation that bands (b) and (d) are present in both mating types P1 and P2. This indicates that they represent alleles of the same gene because both parental strains contribute genetic material to the diploid ascus cell formed after fertilization.

This means that during the second meiotic division, the homologous chromosomes separated, leading to the distribution of alleles into different daughter cells and it would support the concept of Second Division Segregation (SDS). 
Segregation occurred during meiosis II:

Meiosis II involves the separation of sister chromatids, leading to the distribution of alleles into different ascospores. The observation of bands (b) and (d) in different combinations in the octads suggests that segregation of these alleles occurred during meiosis II, as the chromatids carrying these alleles were separated into different daughter cells.

Additional InformationFirst division segregation and second division segregation are terms used in genetics, particularly in the study of genetic recombination and gene mapping. They refer to different patterns of segregation of alleles during meiosis. Here's an explanation of each:

First Division Segregation (FDS):

• First division segregation occurs during the first meiotic division (meiosis I).
• In first division segregation, homologous chromosomes pair and undergo crossing over, exchanging genetic material between chromatids.
• During anaphase I of meiosis I, homologous chromosomes separate and move to opposite poles of the cell. However, in first division segregation, the centromeres of the chromosomes are pulled to opposite poles independently of the alleles they carry. As a result, each daughter cell receives a mixture of maternal and paternal chromosomes.
• This process leads to the formation of two types of gametes: those containing both parental alleles (non-recombinant or parental gametes) and those containing a combination of parental alleles (recombinant gametes).

• The ratio observed in FDS is typically 2:2 or 1:1, indicating an equal distribution of the two different alleles into the resulting spores. This equal distribution occurs because each pair of homologous chromosomes segregates independently during meiosis I.

• For example, if an individual heterozygous for a gene (Aa) undergoes meiosis, the resulting spores would contain equal numbers of two different types of gametes: one containing the A allele and the other containing the a allele. Therefore, the ratio of the resulting spores would be 1:1.

Second Division Segregation (SDS):

• Second division segregation occurs during the second meiotic division (meiosis II), specifically during anaphase II.
• In second division segregation, sister chromatids separate, resulting in the distribution of different alleles into different gametes.
• During anaphase II, the sister chromatids of each chromosome are pulled apart, with each chromatid going to a different daughter cell.
• If crossing over occurred during meiosis I, the alleles on the sister chromatids may be different due to recombination. As a result, different combinations of alleles are distributed into different gametes during second division segregation.
• This process can lead to the production of gametes with various combinations of alleles, including those with both parental alleles (non-recombinant gametes) and those with a mixture of parental alleles (recombinant gametes).

• The ratio observed in SDS depends on the specific genetic configuration and the number of alleles involved. Unlike First Division Segregation (FDS), where the ratio is typically 2:2 or 1:1, the ratio in SDS can vary depending on the genetic makeup of the organism and the specific gene being studied.

• For example, if an individual is heterozygous for two genes (AaBb) and undergoes SDS, the resulting spores or daughter cells may exhibit various ratios depending on the assortment of alleles during meiosis. The possible ratios in SDS can range from 4:0 to 2:2 to 0:4, depending on how the alleles segregate during the second meiotic division or mitosis.

In summary, first division segregation involves the independent assortment of chromosomes during meiosis I, while second division segregation involves the separation of sister chromatids during meiosis II. Both processes contribute to genetic variation by generating gametes with different combinations of alleles.