CBSE Class 12 Biology: 10 Most Asked Questions On Genetics And Evolution In Boards

Unit-VII Genetics and Evolution (20 marks)

This unit carries 29 per cent of total theory marks in the CBSE board exam. There are three chapters included in this unit.

Chapter-5: Principles of Inheritance and Variation

Chapter-6: Molecular Basis of Inheritance (1)

Chapter-7: Evolution

CBSE Class 12 Boards Biology: Top 10 Probable Questions

Question 1: Describe an operon. How lac operon controls gene expression in the bacteria?

Tips for solution:

• Explain what is an operon with the help of a diagram.
• Components of an operon
• Then show how lac operon controls gene expression in bacteria with the help of a diagram.

Answer: An operon is a functional unit of DNA containing a set of genes including a structural gene, and a regulatory gene under the control of a single operator gene (promotor). These genes are transcribed together under the control of the promotor into mRNA strands and may be translated or undergo a splicing process and create monocistronic mRNA which is translated separately resulting in the expression of the gene together. Several genes must be transcribed together to define an operon. The expression of the prokaryotic operon generates polycistronic mRNA while the eukaryotic operon leads to the generation of monocistronic mRNAs. In prokaryotic organisms such as bacteria, viruses, and algae operons are found.

The lac operon (lactose operon) is a group of genes or operons with a single promoter (transcribed as a single mRNA) required for the metabolism and transport of lactose in E. coli and other bacteria.

In bacteria, certain regulatory molecules control whether a gene will be transcribed into RNA from one promoter site or RNA polymerase binding site, which is found in clusters on chromosomes. This cluster of certain regulatory molecules (genes) under the control of the promotor is called an operon. These molecules bind to DNA close to the gene and help or block the transcription enzyme (RNA polymerase). The lac operon consists of genes that code for that protein which is involved in the metabolism and uptake of lactose sugar. The cell is allowed to express the set of the gene by operons whose products are required at the same time.

Question 2: How did SL Miller create the conditions that existed before the origin of any life on Earth?

Answer: In a lab, SL Miller built an environment resembling the one that existed before life began. In a closed flask containing CH4, H2, NH3 and water vapour at 800∘C, electric discharge was created.

The conditions were similar to those in primitive atmospheres. After a week, he observed the presence of amino acids and complex molecules like sugars, nitrogen bases, pigments, and fats in the flask. This provides experimental evidence for the theory of chemical origin.

Question 3: Outline the steps in the PCR technique and describe the effects of temperature change.

Answer: Polymerase chain reaction is a technique used to make millions of copies rapidly of a DNA sample. It allows the scientist to amplify a very small DNA sample to a large amount for a detailed study. This technique was invented by Kary Mullis. It is a common and indispensable technique used in the clinical laboratory and medical laboratory for a variety of applications including criminal forensic and biomedical research. The PCR technique majorly relies on thermal cycling.

The steps in the PCR technique include -

1. Denaturation – In this process, the two strands of the DNA is separated by raising the temperature of the mixture which causes the breakage of hydrogen bond between complementary strands.

2. Annealing – one primer binds to each strand of the target DNA sequence and polymerization is initiated. It is done by lowering the temperature.

3. Extension – new strands are synthesised by using the original strand as a template. Often Taq polymerase enzymes are used to join the free DNA nucleotides. According to the sequence of the template strand, the sequence of the free nucleotide is determined.

The result of 1 cycle of PCR is 2 double-stranded sequences of the target DNA, in which 1 is a newly made strand and 1 original strand. The cycle is repeated many times to get millions of copies.

Effects of temperature change – when the sample is heated between 94-98°C, it disrupts the hydrogen bonds as well as the base stacking interaction that holds the DNA strands together. This process is known as denaturation. After the separation of the strands, the temperature is decreased for the process of annealing. The annealing occurs between 48-72°C and it is related to the Tm (melting temperature) of the primers. The extension step, in which the primer is extended by the polymerase to form a nascent DNA strand and occurs at 68-72°C. After the completion of PCR, the thermal cycler is set to 4°-10°C to maintain the product integrity.

Question 4: Explain the Hardy-Weinberg principle with the help of an algebraic equation. List the five factors that influence the law.

Answer: According to the Hardy-Weinberg principle, allele frequencies in a population are stable and constant from generation to generation and the complete gene pool will remain constant.

The Hardy-Weinberg principle is stated by using an algebraic equation p+q=1, where p is the allele frequency of one allele (example ‘A’) and q is the frequency of another allele (example ‘a’) which is always equal to one. It is expanded using the binomial theorem into

p2+2pq+q2=1

In a population of diploid organisms.

If the frequency of allele A = p

And the frequency of allele a = q, then the expected genotype frequency under random mating is -

AA=P2(for AA homozygotes)

aa=q2 (for aa homozygotes)

Aa =2pq (for Aa heterozygotes)

In the absence of selection, mutation, genetic drift or other forces, p and q are constant through generations.

Therefore,

p2+2pq+q2=1

The five factors influencing the law are:

• Genetic drift

• Mutation

• Gene flow

• Genetic Recombination

• Natural Selection

Question 5: Explain Griffith’s ‘transforming principle’ experiment.

Answer: Frederick Griffith observed a miraculous change in Streptococcus pneumoniae, the bacterium that causes pneumonia, during a series of experiments he conducted with it in 1928. A living organism (bacteria) had undergone physical change as a result of his experiment. Some Streptococcus pneumonia (pneumococcus) bacteria grow as smooth shiny colonies (S) while others grow as rough colonies (R) on a culture plate. This is because, in contrast to the R strain, the bacteria of the S strain have a mucous (polysaccharide) coat. Mice infected with the S strain (virulent) succumb to pneumonia but mice infected with the R strain do not.

S strain —----> injected into mice —--------> mice die

R strain —----> injected into mice —--------> mice live

By heating the bacteria, Griffith was able to destroy them. He noticed that injecting mice with heat-killed S-strain bacteria did not result in their death. The mice died after receiving an injection of a mixture of heat-killed S and live R bacteria. Additionally, he extracted live S bacteria from the dead mice.

S strain (heat-killed ) —----> injected into mice —--------> mice live

S strain (heat-killed ) + R strain (live) —----> injected into mice —--------> mice die

He came to the conclusion that the heat-killed S strain bacteria had somehow transformed the R strain bacteria. The R strain was able to create a smooth polysaccharide coat and develop virulence thanks to some sort of "transforming principle" that was passed from the heat-killed S strain. Genetic material must have been transferred in order for this to occur. His experiments, however, did not specify the biochemical nature of genetic material.

Question 6: What is adaptive radiation?

Answer: Darwin visited the Galapagos Islands while on his travels. He saw an incredible variety of creatures there. Small, black birds that would later be known as Darwin's Finches fascinated him in particular. He came to understand that the same island was home to a variety of finches. He hypothesised that all the varieties originated on the island itself.

Many other forms with modified beaks developed from the original seed-eating traits, enabling them to become insectivorous and vegetarian finches. Adaptive radiation is the term used to describe the process by which various species in a particular geographic area evolve, literally radiating from a point to other geographic areas (habitats).

One of the best examples of this phenomenon is Darwin's finches. Australian marsupials are another illustration.

A variety of marsupials, each distinct from the others (Figure 7.6), developed from a common ancestor on the island continent of Australia. One can refer to this as convergent evolution when it appears that more than one adaptive radiation (representing various habitats) occurred in a solitary geographic location. In Australia, placental mammals have also undergone adaptive radiation, developing into different species that each resemble their corresponding marsupial in appearance (such as the Tasmanian wolf-marsupial and placental wolf).

Question 7: Explain - Divergent vs. convergent evolution."

Answer: Comparative anatomy and morphology highlight the similarities and differences between modern and extinct organisms. These similarities can be used to infer whether or not common ancestors were shared. For instance, the arrangement of the bones in the forelimbs of mammals such as whales, bats, cheetahs, and humans is similar.

Although the forelimbs of these animals serve various purposes, they all share the same anatomical structures, including the humerus, radius, ulna, carpals, metacarpals, and phalanges. Consequently, due to adaptations to various needs, the same structure took different forms in these animals. These structures exhibit homology and divergent evolution. Common ancestry is indicated by homology. Vertebrate brains or hearts are two more examples. The thorn and tendrils of the Bougainvillaea and Cucurbita plants show homology in plants as well. Analogy refers to a completely different situation than homology, which is based on divergent evolution. The wings of butterflies and birds are similar. Despite performing similar functions, their anatomical structures are not the same. Thus, analogous structures are the outcome of convergent evolution, wherein various structures develop for the same function and subsequently share similarities. Other analogies include the eyes of mammals and octopuses, as well as the flippers of penguins and dolphins.

Question 8: Explain the semi-conservative replication model and the evidence that supports this model.

Answer: While proposing the double helix structure for DNA, Watson and Crick immediately proposed a scheme for the replication of DNA. To quote their original statement that is as follows: “it has not escaped our notice that the specific pairing we have postulated immediately suggests a possible copying mechanism for the genetic material” (Watson and Crick, 1953).

According to the proposed mechanism, the two strands would separate and serve as a template for the synthesis of fresh complementary strands. Each DNA molecule would have one parental and one newly synthesised strand after replication was finished. Semi-conservative DNA replication is the term used to describe this system.

The experimental proof for the semi-conservative replication model:

Matthew Meselson and Franklin Stahl performed experiments to prove that DNA replicates semi-conservatively in E. coli and in higher organisms such as human cells, and plants.

• E. coli was grown in a medium containing 15NH4Cl. The result showed that this heavy isotope of nitrogen was integrated into newly synthesised DNA. This heavy molecule of DNA could be separated in a CsCl density gradient on the basis of densities from the 14N.

• Thereafter, these cells were transferred by them into a medium with normal 14NH4Cl and taken sample at different certain time intervals as cells multiplied, then extracted the DNA still remained as double-stranded. On the CsCl gradient samples were separated for measuring the density of DNAs.

• From the culture when the DNA was extracted after one generation (E. coli divides in 20 minutes so after 20 minutes) when transferred from 15N to 14N. this generation had the intermediate density or hybrid. The DNA extracted from the second generation (after 40 minutes) from the culture was a combination of this hybrid and light DNA in equal amounts.

Thus, the experiment proved that the DNA replicates semi-conservatively.

In conservative replication, after 1.0 generation, half of the molecules would be 15N-15N, the other half 14N-14N. After 2.0 generations, one-quarter of the molecules would be 15N-15N, and the other three-quarters 14N-14N. Hybrid 14N-15N molecules would not be observed in conservative replication.

Question 9: How does an RNA nucleotide differ from a DNA nucleotide?

Answer: Both ribonucleic acid (RNA) and deoxyribonucleic acid (DNA) are composed of nucleotides. And a nucleotide is a composition of 3 smaller molecules that are a 5-carbon sugar, a nitrogenous base, and a phosphate group.

RNA and DNA nucleotides differ from each other according to which five-carbon sugar is present, and whether the nitrogenous base thymine or uracil is present.

A DNA nucleotide is composed of a five-carbon sugar called deoxyribose, a phosphate group, and one of the four nitrogenous bases that are adenine (A), guanine (G), cytosine (C), and thymine (T). In deoxyribonucleotides, the 2' carbon has two hydrogens bonded to it

While an RNA nucleotide is composed of five-carbon sugar called ribose, a phosphate group, and one of the four nitrogenous bases; adenine (A), guanine (G), cytosine (C), and uracil (U). Uracil takes place of thymine in RNA.

In ribonucleotides, the 2' carbon has a hydroxy group and hydrogen bonded to it.

Question 10: Explain the law of dominance using a monohybrid cross. In a monohybrid cross of two heterozygous parents (Pp), what would the expected genotypes of the offspring be? Show your solution.

Answer: Monohybrid crosses – when we cross two parents for a gene controlling only one character (for example, colour) only is called a monohybrid cross

A Punnett square is used to understand a typical monohybrid and dihybrid cross conducted by the Mendel between true-breeding tall plants and true breeding dwarf plant.

It is logical to use the capital and lowercase of an alphabetical symbol to define the concept of dominance and recessive. The dominant allele is represented by the capital alphabet while the recessive allele is represented by the small letter.

Alleles of a gene can be similar (homozygous) as PP or dissimilar (heterozygous) Pp.

By using punnet square for 2 of two heterozygous parents (Pp)

We get: 1 homozygous dominant = PP (genotype)

2 heterozygous = Pp

1 homozygous recessive = pp

Question 11: What genotypes would be expected for the offspring from the following matings? What are the phenotypic ratios for each cross? Explain how you came up with this ratio.

1. AABB x aaBB

2. AaBB x AABb

3. AaBb x aabb

Answer:

Step 1: In the given question, offspring with different genotypes is given for the cross. The test cross is used to determine the genotype of individuals by crossing it with another individual of known genotype or for the prediction of the genotype in offspring. zygosity can be either heterozygous or homozygous. Heterozygous has one recessive allele and one dominant allele. Homozygous have both dominant alleles or recessive alleles.

Step 2:

1. Test cross between AABB and aaBB. The gametes of AABB will be AB and the gametes of the aaBB will be aB.

Hence, all progeny will have the same genotype and consequently the same phenotype.

2. Now we are going to test the cross between AaBB and AABb. The gametes of genotype AaBB will be AB and aB, and the gametes of the AABb will be AB and Ab

The resultant genotypes of the offspring are

AABB, AaBB, AABb and AaBb. Each has a different genotype but the phenotype of all the progeny will be the same for each offspring. The capital letter in the genotype shows the dominant character and the small letter shows the recessive character. In the genotype, if there is one dominant allele and one recessive allele controlling the one character then the dominant allele will exhibit its character in its phenotype, If both alleles are recessive only in that condition the allele will exhibit its trait in its phenotype. The AaBB and AABb are controlling two traits in an individual with its four alleles, Aa is controlling one trait and BB controls another trait

All genotypes of progeny that we have after cross will have the same phenotype because, in each genotype, there is one dominant allele present. In the case of the first trait, there is one dominant A is present in each genotype and in the second trait there Is one B is present in each genotype.

The phenotype of all the progeny will be the same.

(C)

Now we are going to cross between AaBb and aabb. The gametes produced by the AaBb will be AB, Ab, aB, and ab. Each gamete of the aabb will be ab.

Test cross between both:

AaBb x aabb

The genotype of the offspring we got by the cross are of four types:

four offspring will have - AaBb,

four offspring will have Aabb,

four offspring will have aaBb,

and four offspring will have - aabb

and each genotype will exhibit different phenotypes as they have dominant alleles as well as both recessive alleles.

So the phenotypic ratio will be in the ratio of 1:1:1:1

Hope these questions help you in your preparation for your board exam. All the best!