CBSE Class 12

Here we are providing Class 12 Biology Important Extra Questions and Answers Chapter 6 Molecular Basis of Inheritance. Important Questions for Class 12 Biology are the best resource for students which helps in Class 12 board exams.

Class 12 Biology Chapter 6 Important Extra Questions Molecular Basis of Inheritance

Molecular Basis of Inheritance Important Extra Questions Very Short Answer Type

Question 1.
Name the genetic material for the majority of organisms.
Answer:
DNA (Deoxyribose nucleic acid)

Question 2.
List the function of RNA.
Answer:
RNA acts as genetic material in viruses and also functions as an adapter, structural, and in some cases as a catalytic molecule.

Question 3.
How many nucleotides are present in a bacteriophage Φ × 174?
Answer:
5386.

Question 4.
List the number of base pairs in:
(i) lambda bacteriophage
Answer:
48502 bp

(ii) E.coli and
Answer:
4.6 × 10⁶bp

(iii) haploid content of human DNA.
Answer:
3.3 × 10⁹bp.

Question 5.
Comment two chains of DNA have antiparallel polarity.
Answer:
If one chain has 5′ → 3′ polarity, the other chain has 3′ → 5′ polarity.

Question 6.
What is the difference between DNA and DNAase?
Answer:
DNAs are the number of molecules of DNA and DNAase is an enzyme that digests DNA.

Question 7.
What made DNA as genetic material?
Answer:
As RNA was unstable, it evolved further with certain chemical modifications and formed DNA. DNA became more stable.

Question 8.
What is the average rate of polymerization?
Answer:
2000 bp per second.

Question 9.
List three components of the transcription unit.
Answer:

1. A promoter,
2. The structural gene,
3. A terminator.

Question 10.
What is the term used for fully processed hn RNA?
Answer:
A messenger RNA (mRNA).

Question 11.
What is splicing?
Answer:
It is the removal of introns and the joining of exons in a definite manner.

Question 12.
Where are UTRs present in mRNA strand?
Answer:
UTRs (Untranslated region) are present at both 5′ end (before start codon) and at the 3′ end (after stop signal).

Question 13.
Write the significance of UTRs.
Answer:
They are required for the regulation of an efficient translation process.

Question 14.
Name any two non-sense codons (stop signal).
Answer:

1. UGA (Opal)
2. UAA (Ochre).

Question 15.
What are the exceptions to the general rule that DNA is the genetic material in all organisms? Give evidence that supports these exceptions.
Answer:
Some animal viruses and all plant viruses contain RNA as their genetic material.

Question 16.
What is a replication fork? (CBSE, Delhi 2011)
Answer:
When replication starts the two strands of DNA unwind to form a Y-shaped structure called the replication fork.

Question 17.
Name the technique by which Gene expression can be controlled with the help of RNA molecule. (CBSE Sample Paper 2018-19)
Answer:
Northern blotting

Question 18.
Name the parts ‘A’ and ‘B’ of the transcription unit given below. (C.B.S.E. Delhi 2008)
Answer:

A-Promoter
B-Coding strand.

Question 19.
Mention the role of the codons AUG and UGA during protein synthesis. (CBSE 2011, 2016)
Or
Mention two functions of codon AUG. (CBSE 2010)
Answer:

1. AUG acts as a start signal and also code for methionine amino acid.
2. UGA acts as a stop signal.

Question 20.
How do histones acquire positive charge? (CBSE Delhi 2011)
Answer:
A histone protein acquires a positive charge because they are rich in basic amino acid residues such as lysine, arginine, and histidine. All are positively charged in their side chains.

Question 21.
Why do DNA fragments move towards the anode during gel electrophoresis? (CBSE Delhi 2018C)
Answer:
DNA molecule is negatively charged. When placed in an electric field it moves towards a positively charged anode.

Question 22.
Name one amino acid which is coded by only one codon. (CBSE Delhi 2018 C)
Answer:
Methionine – AUG/ Tryptophan – UGG

Molecular Basis of Inheritance Important Extra Questions Short Answer Type

Question 1.
If the sequence of coding strand In a transcription unit Is written as follows: (CBSE Delhi 2011)
5′- A T G C A T G C A T G C A T G C A T G C A T G C A T G C – 3′
Write down the sequence of mRNA.
Answer:
The sequence of mRNA shall be:
5′ – U A C G U A C G U A C G U A C G U A C G U A C G U A C G – 3′.

Question 2.
What is Chargaff rule? (CBSE Delhi 2011)
Answer:
Chargaff Rules:

1. In DNA molecule, A — T base pairs equals in number to G — C base pairs.
2. A + G = T + C, i.e. purines and pyrimidines equal in amount.
3. A = T and C = G (Amount).
4. The base ratio A + T/G + C may vary from one species to another but is constant for each species. It helps in identifying the source of DNA.
5. The deoxyribose sugar and phosphate component occur in equal proportions.

Question 3.
(a) Name the component of a nucleotide responsible for giving 5’— 3′ polarity to a polynucleotide.
Answer:
A polymer has at one end a free phosphate moiety at 5′- end of ribose sugar which is referred to as 5′-end of a polynucleotide chain. Similarly, at the other end of the polymer, the ribose has a free 3-OH group which is referred to as the 3′-end of a polynucleotide chain.

(b) Where in a nucleotide is the glycosidic bond present? (CBSE Delhi 2019 C)
Answer:
A nitrogenous base is linked to pentose sugar through an N-glycosidic linkage to form nucleoside.

Question 4.
Which property of DNA double helix led Watson and Crick to hypothesize a semi-conservative model of DNA replication? Explain.
Or
Write a note on the semi-conservative mode of DNA replication. (CBSE Delhi 2008; Delhi 2011, 2014)
Answer:
The semi-conservative model of DNA replication hypothesized by Watson and Crick was based upon the property that during replication, the two strands would separate and act as a template for the synthesis of new complementary strands. After the completion of replication, each DNA molecule would have one parental and one newly synthesized strand. This scheme was termed as ‘Semi-conservative’ DNA replication.

Question 5.
RNA was the first genetic material, DNA evolved later on. Explain.
Or
Why is RNA considered the first genetic material? (CBSE, 2009 2012)
Answer:
RNA was the first genetic material. There is now enough evidence to suggest that essential life processes (such as metabolism, translation, splicing, etc.) evolved around RNA. RNA used to act as a genetic material as well as a catalyst (there are some important biochemical reactions in living systems that are catalyzed by RNA catalysts and not by protein enzymes). But, RNA being a catalyst was reactive and hence unstable. Therefore, DNA has evolved from RNA with chemical modifications that make it more stable. DNA being double-stranded and having complementary strands further resists changes by evolving a process of repair.

Question 6.
Discuss the significance of the heavy isotope of nitrogen in Meselson and Stahl’s experiment.
Answer:

1. By using the heavy isotope of nitrogen, they could find out the semiconservative nature of DNA replication as the densities of DNA having ¹⁵N in both the strands ¹⁵N/¹⁴N DNA and ¹⁴N/¹⁴N DNA were all different.
2. The hybrid DNA (¹⁵N/¹⁴N DNA) had a density intermediate between that of heavy DNA (¹⁵N/¹⁵N DNA) and that of light/normal DNA (¹⁴N/¹⁴N DNA).

Question 7.
Differentiate polycistronic mRNA and monocistronic mRNA.
Answer:
Differences between polycistronic mRNA and monocistronic mRNA:

Question 8.
You are repeating the Hershey-Chase experiment and are provided with two isotopes ³²p and ¹⁵N (in place of ³⁵S in the original experiment). How do you expect your results to be different?
Answer:

1. The use of ¹⁵N will not give any conclusive result because it is only a heavy isotope of nitrogen.
2. In the original experiment, ³⁵S was detected only in the supernatant as it was incorporated in protein only, while 15N will be incorporated into proteins as well as in DNA and hence it would appear both in the supernatant and in the sediment as well.

Question 9.
What is DNA fingerprinting? Mention its applications.
Or
List any two applications of the DNA fingerprinting technique. (CBSE Delhi 2018)
Answer:
DNA fingerprinting: The technique of using DNA fragments, resulting from restriction endonuclease enzyme cleavage to identify particular individuals, is called DNA fingerprinting.

Applications of DNA Fingerprinting:

1. Paternity disputes can be solved by DNA fingerprinting.
2. It can solve the problems of evolution.
3. It can be used to study the breeding patterns of animals facing the danger of extinction.
4. It is useful in restoring the health of the patients suffering from leukemia (blood cancer).
5. It is very useful in the detection of crime and legal pursuits.

Question 10.
What are the major enzymes of DNA replication? (CBSE 2009)
Answer:
Enzymes of DNA Replication:

1. DNA-dependent DNA polymerase. It catalyzes the polymerization of deoxynucleotides,
2. Okazaki fragments, which are then quickly joined together by an enzyme known as DNA ligase.
3. The discovery of helicases and topoisomerase enzymes explains the unwinding of the DNA helix.

Question 11.
One of the codons on mRNA is AUG. Draw the structure of the tRNA adapter molecule for this codon. Explain the uniqueness of this tRNA. (CBSE Delhi and Outside Delhi, 2008)
Answer:
Transfer RNA or soluble RNA or Adapter RNA (tRNA or sRNA). It constitutes 15% of total RNA and is the smallest out of three with only 70-85 nucleotides having a sedimentation coefficient of 45.

It is unique because it has double specificity, one for codon on mRNA strand and the other for the corresponding amino acid.

Structure of tRNA adapter

Question 12.
Briefly describe the termination of a polypeptide chain. (C8SE 2009)
Answer:
Termination of polypeptide synthesis:

1. When one of the termination codons (UAA, UAG, UGA) comes at the A-site, it does not code for any amino acid and there is no tRNA molecule for it.
2. As a result, the polypeptide synthesis (or elongation of the polypeptide) stops.
3. The polypeptide synthesized is released from the ribosome, catalyzed by a ‘release factor’.

Question 13.
Write the dual purpose served by Deoxyribonucleoside triphosphates in polymerization. (CBSE Delhi 2018)
Answer:
Deoxyribonucleoside triphosphates serve as substrates, i.e. nucleotides during replication, and also provide energy for polymerization reaction by cleavage of high energy terminal phosphates bond. Thus they serve dual purposes.

Question 14.
Although a prokaryotic cell has no defined nucleus, yet DNA is not scattered throughout the cell. Explain. (CBSE Delhi 2018)
Answer:
In the nucleoid region, DNA that is negatively charged is held with some proteins that are positively charged. The DNA in the nucleoid is organized in large loops held by proteins. And the DNA in the form of single chromosomes is attached to the mesosome at a point. 15

Question 15.
Differentiate between the genetic codes given below:
(i) Unambiguous and Universal
Answer:

• Unambiguous: The code is specific, i.e. one Condon code for onLy one amino acid.
• Universal: The code is the same in aLL organisms.

(ii) Degenerate and Initiator (CBSE Delhi 2017)
Answer:

• Degenerate: When an amino acid is coded by more than one codon, it is said to be degenerate.
• Initiator: AUG is an initiator codon, i.e. it initiates the translation process and also codes for methionine.

Question 16.
Why does the lac operon shut down sometime after the addition of lactose in the medium where E.coti was growing? Why low-level expression of the lac operon is always required? (CBSE Sample Paper 2018-19)
Answer:
After the addition of lactose, a complete breakdown of lactose to glucose and galactose takes place. Therefore, there is no more lactose to bind to the repressor protein and the lac operon shuts down.

A very low level of expression of lac operon has to be present in the cell all the time, otherwise, lactose cannot enter the cells.

Question 17.
Carefully examine structures A and B of pentose sugar given below. Which one of the two is more reactive? Give reasons. (CBSE Sample Paper 2019-20)

Answer:
The pentose sugar of figure A is more reactive.
Reasons:

1. There are two -OH groups present in the pentose sugar.
2. It makes it more labile.
3. It is easily degradable.

Molecular Basis of Inheritance Important Extra Questions Long Answer Type

Question 1.
(i) How are the following formed and involved in DNA packaging in a nucleus of a cell?
(a) Histone octamer
(b) Nucleosome
(c) Chromatin
Answer:
Packaging of DNA.
(a) Histone octamer: Five types of histone proteins (H₁, H₂A, H₂B, H_3, and H₄) are involved. Out of these, four of them H₁ A, H₂ B, H_3, and H₄ occur in pairs to produce histone octamer also called the nu body. Histones are organized in a form of a compact unit formed of 8 molecules hence called histone octamer.

(b) Nucleosome: The unit of compaction of DNA is the nucleosome. About 146 bp of DMA is wrapped
over histone octamer for 1 \(\frac{3}{4}\) turn to form nucleosome of the size 110 × 60 A.

(c) Chromatin: Linker DNA connects two adjacent nucleosomes. It bears H₁ protein. As a result, a chain is formed called chromatic. Nucleosome chain gives ‘beads on string’ appearance under the electron microscope. Chromatin are repeating units of a structure located on the nucleosome.

(ii) Differentiate between Euchromatin: 1 and Heterochromatin. (CBSE Delhi 2016)
Answer:

Question 2.
Distinguish between heterochromatin and euchromatin. Which of the two is transcriptionally active?
Answer:
Differences between heterochromatin and euchromatin:

Euchromatin is more active transcriptionally.

Question 3.
Write a note on messenger RNA.
Answer:
Messenger RNA (mRNA).
It forms only 5% of total RNA but is the longest of all. It brings instructions from DNA for the formation of a particular polypeptide. The instructions are coded in the form of a base sequence called genetic code. Three adjacent nitrogen bases specify a particular amino acid. The formation of polypeptides occurs over the ribosomes. mRNA gets attached to ribosomes.

mRNA.

It starts as a cap for attachment with the ribosome. It is followed by an initiation codon (AUG) either immediately or after a small non-coding region. It is followed by the coding region followed by the termination codon (UAA, UAG, and UGA). Then there is a small non-coding region and poly-A area at 3 termini. The mRNA may specify only a single polypeptide or a number of them called monocistronic and polycistronic respectively.

The life span of mRNA maybe a few minutes to an hour or even days in the case of RBC.

Question 4.
What is genetic code? List the properties of genetic code.
Answer:
The code language of DNA and mRNA is complementary. So, genetic code is the sequence of nucleotides in DNA and RNA that determines the amino acid sequence in proteins. Some amino acids are specified by more than one codon. The sequence of nucleotide on the tRNA molecule which complements the codon is called anticodon, e.g. one of the codons for the amino acid leucine is CUG and the anticodon is GAC. Similarly, the codon for phenylalanine is UUU, while the anticodon is AAA.

Properties of genetic code:
The following properties of genetic code have now been proved by experimental evidence.

1. The code is a triplet.
2. The code is degenerate.
3. The code is non¬overlapping.
4. The code is commaless.
5. The code is non-ambiguous.
6. The code is universal,
7. Collinearity.

Both polypeptide and DNA or mRNA have a linear arrangement of their components.

The term code letter stands for a nucleotide A, T, G, or C in DNA and A, U, G, or C in RNA. The sequence of three does not code for any amino acid, such codons are called a non-sense codon, e.g. UGA.

Question 5.
What is the role of ribosomes during translation? Ribosomes move along mRNA molecules and catalyze the assembly of amino acids into protein chambers. (CBSE Outside Delhi 2019)
Answer:
Role of ribosomes: Ribosomes usually form linear or helical groups during active protein synthesis called polyribosomes or polysomes. The mRNA strand having coded information joins along with smaller subunits of ribosomes. The adjacent ribosomes are 360 A apart.

The different parts of the ribosome connected with protein synthesis are:

1. A tunnel for mRNA.
2. A groove for the passage of newly synthesized polypeptide (larger subunit).
3. Two active sites (P-site-peptidyl transfer or donor site and A-site or aminoacyl or acceptor site).
4. A binding site for tRNA near A-site.
5. Presence of enzyme peptidyl transferase.
6. Recognition point of smaller subunit for mRNA.
7. Presence of GTP-ase, binding sites for elongation factors, and translocases.

Question 6.
Describe the steps in the sequencing of the human genome.
Answer:
Sequencing of a genome.
The method involved two major approaches:

1. Expressed Sequence Tags (ESTs): It is focused on identifying all the genes that are expressed as RNAs.
2. Sequence Annotation: It involves simply sequence the whole set of the genome that included all the coding and non-coding sequences and then assigning functions to different regions in the sequence.

HGP followed the second technique:

1. The total DNA from the cell is isolated and converted into random fragments of relatively smaller sizes.
2. These fragments are then cloned in suitable hosts using specialized vectors; the commonly used hosts are bacteria and yeast and the vectors are bacterial artificial chromosomes (BAC) and yeast artificial chromosomes (YAC).
3. The fragments are then sequenced using automated DNA sequences.
4. The sequences were then arranged on the basis of certain overlapping regions present in them; this required the generation of overlapping fragments for sequencing.
5. These sequences are annotated and assigned to the respective chromosomes.

Question 7.
(i) Why is DNA molecule a more stable genetic material than RNA? Explain.
Answer:
There is now enough evidence to suggest that essential life processes (such as metabolism, translation, splicing, etc.) revolved around RNA. RNA used to act as a genetic material as well as a catalyst (there are some important biochemical reactions in living systems that are catalyzed by RNA catalyst and not by protein enzymes). But, RNA being a catalyst was reactive and hence unstable. Therefore, DNA has evolved from RNA with chemical modifications that make it more stable. DNA being double-stranded and having complementary strands further resists changes by evolving a process of repair.

(ii) ‘Unambiguous’, ‘degenerate’ and ‘universal’ are some of the salient features of genetic code. Explain. (CBSE 2008, 2011, Delhi 2014, 2016)
Answer:
Unambiguous: Each codon codes for one amino acid, none for more than one.

Degenerate: One amino acid often has more than one triplet codon; only methionine and tryptophan have a single triplet codon.

Universal: The genetic code is universal. A given codon in DNA and m RNA specifies the same amino acid in all organisms from viruses, bacteria to human beings.

Question 8.
Explain the role of tRNA in the initiation of protein synthesis. (CBSE 2012)
Answer:
Role of tRNA’ in the initiation of protein synthesis:

1. tRNA is an adapter molecule that on one hand would bind to a specific amino acid.
2. The tRNA then called sRNA (soluble RNA).
3. tRNA has an anticodon loop that has bases complementary to the code.
4. It has an amino acid acceptor end to which it binds to an amino acid.
5. tRNAs are specific for each amino acid.
6. For initiation, there is another tRNA, that is referred to as initiator tRNA.
7. Amino acids are activated in the presence of ATP and joined to their corresponding tRNA it is called charging of tRNA or aminoacylation of tRNA.
8. Two such charged tRNAs are brought close and the formation of peptide bond occurs.

Question 9.
List the criteria that can act as genetic material must fulfill. Which one of the criteria is best fulfilled by DNA or RNA thus making one of them a better genetic material? Explain. (CBSE (Delhi) 2016)
Answer:
Essential requirements of genetic material:

1. A genetic material should be able to store and express information to confer the heritable characters of living organisms.
2. It should be capable to make its own replica.
3. Genetic material should also have the mechanism to undergo mutations that will generate variations.

Question 10.
A small stretch of DNA strand that codes for a polypeptide are shown below:
3’— — — — CAT CAT AGA TGA AAC — — — — 5’
(a) Which type of mutation could have occurred in each type resulting in the following mistakes during rep¬lication of the above original se¬quence?
(i) 3’ … … … … CAT CAT AGA TGA ATC … … … 5’
Answer:
A point mutation (single base substitution)

(ii) 3’ … … … … CAT ATA GAT GAA AC … … … 5’
Answer:
A point mutation (single base deletion)

(b) How many amino acids will be translated from each of the above strands (i) and (ii)? (CBSE Sample Paper 2019-20)
Answer:
(i) 4 amino acids
(ii) 4 amino acids

Question 11.
(i) In the human genome which one of the chromosomes has the most genes and which one has the fewest?
Answer:
Chromosome 1 has the most genes (2968) and Y-chromosome has the least gene (231).

(ii) Scientists have identified about 1.4 million single nucleotide polymorphs in the human genome. How is the information of their existence going to help the scientists? (CBSE2009)
Answer:
This information promises to revolutionize the processes of finding the chromosomal location for disease-associated sequences and tracing human history.

Question 12.
Describe the structure of an RNA polynucleotide chain having four different types of nucleotides. (CBSE Delhi 2013, 2019)
Answer:

RNA polynucleotide

RNA polynucleotide:

1. RNA is formed of polynucleotides of ribose series.
2. Each nucleotide is formed of a nitrogen base, ribose sugar (pentose), and phosphoric acid.
3. There are two types of nitrogen bases-purines and pyrimidines.
4. Adenine (A) and Guanine (G) are purines.
5. Cytosine (C) and Uracil (U) are pyrimidines.
6. The nitrogen base is linked to pentose sugar by N-glycosidic linkage to form nucleoside.
7. When a phosphate group is linked to 5′ -OH of a nucleoside through phosphodiester linkage thus forms nucleotide.
8. Nucleotides of ribose series present are AMP, GMP, CMP, and UMP.
9. Two nucleotides are linked through 3′ – 5′ phosphodiester linkage to form dinucleotides.
10. More nucleotides can be joined in such a manner to form a polynucleotide chain.

Question 13.
Describe the initiation process of transcription in bacteria. (CBSE 2010, 2016, 2019)
Answer:
Initiation of transcription in bacteria:

1. The process of copying genetic information from antisense or template strands of DNA into RNA is called transcription.
2. The segment of DNA that takes part in transcription is called the transcription unit. It has three components
   (a) a promoter,
   (b) the structural gene and
   (c) a terminator.
3. The structural gene is composed of that strand of DNA that has 3′ → 5′ polarity as transcription can occur only in the 5′ → 3′ direction.
4. Transcription requires a DNA-dependent-RNA polymerase and initiation factor.
5. Bacteria have only one type of RNA polymerase which transcribes all three types of RNAs.
6. Ribonucleotides of ribose series are activated through phosphorylation (ATP, GTP, CTP, UTP).
7. Transcription begins at the initiation site. A promotor has an RNA polymerase recognition site and it binds to the specific site.
8. Enzymes required for the unwinding of the chain are unwound and single-stranded binding proteins.
9. Nucleotides are added as per the base-pairing rule.

Question 14.
State the role of VNTRs in DNA fingerprinting. (CBSE Outside Delhi 2013)
Answer:
Role of Variable Number of Tandem Repeats (VNTRs) Short nucleotide repeats in the DNA are very specific in each individual and vary in number from person to person but are inherited. These are the ‘Variable Number Tandem Repeats’ (VNTRs). These are also called ‘minisatellites. Each individual inherits these repeats from his/her parents which are used as genetic markers in a personal identity test.

For example, a child might inherit a chromosome with six tandem repeats from the mother and the same tandem repeated four times in the homologous chromosome inherited from the father. The half of VNTR alleles of the child resemble that of the mother and half that of the father.

Question 15.
(i) List the two methodologies which were involved in the human genome project. Mention how they were used.
Answer:
(a) Expressed Sequence Tags (ESTs): This method focuses on identifying all the genes that are expressed as RNA.
(b) Sequence Annotation: It is a method of simply sequencing the whole set of the genome that contains all the coding and non-coding sequences, and then assigning different regions in the sequence with functions.

(ii) Expand ‘YAC’ and mention what it was used for. (CBSE Delhi 2017)
Answer:
‘YAC’ is an abbreviated form of ‘Yeast Artificial Chromosome’. It is used as a cloning vector for cloning DNA fragments in a suitable host so that DNA sequencing can be done.

Question 16.
Summarise the process by which the sequence of DNA bases in the Human Genome Project was determined using the method developed by Frederick Sanger. Name a free-living non-pathogenic nematode whose DNA has been completely sequenced. (CBSE Sample Paper 2019-20)
Answer:
(a) The HGP strategy:

(b) Chromosome 1
(c) Caenorhabditis elegans

Question 17.
Why is a DNA molecule considered a better hereditary material than an RNA molecule? (CBSE Delhi 2018C
Answer:
DNA is the better hereditary material compared to RNA because of the following features:

1. DNA molecule is more stable tha> RNA (as thymine is present here instead of uracil found in RNA).
2. DNA is less reactive than RNA (a- does not have 2’ OH group wl makes RNA very reactive).
3. Since DNA is less reactive so it is t easily degradable.
4. Chances of mutation are less. compared to RNA.

Question 18.
Name the three RNA polymerases found in eukaryotic cells and mention their functions. (CBSE Delhi 2018C)
Answer:

• RNA polymerase I – It transcribes ribosomal RNAs (28S, 18S, 5.8S rRNAs)
• RNA polymerase II – It transcribes precursors of mRNA – heterogeneous nuclear RNA (hnRNA)
• RNA polymerase III – It transcribes transfer RNA (tRNA), 5SrRNAand small nuclear RNAs (snRNAs)

Question 19.
Explain the post-transcriptional modifications the hn-RNA undergoes in eukaryotic cells. (CBSE Delhi 2018C)
Answer:
In eukaryotes, the primary transcript is not functional due to the presence of exons and introns. Therefore it requires certain modifications to make it functional. First of all, introns are removed and the remaining exons are joined together in order by the process called Splicing.

Then the hnRNA undergoes capping followed by tailing. In capping, methyl guanosine triphosphate is added to the 5’ end of hnRNA. In tailing 200 -300 adenylate residues are added to 3’ end in a template-independent manner. Now the hnRNA has been processed into mRNA.

Question 20.
What are the aims of bioinformatics?
Answer:
Aims of bioinformatics:

1. To spread scientifically investigated knowledge for the benefit of the research community.
2. To transform the biological polymeric sequences into sequences of digital symbols and to store them as databases.
3. To develop a variety of methods and tools of software for data analysis.

Question 21.
Describe Griffith’s experiment to demonstrate that DNA is the basic genetic material. What explanation for Griffith’s observation was given by Avery, McCarty, and MacCleod? (CBSE Delhi 2008, 2009, 2012)
Answer:
Griffith’s experiment demonstrates DNA as genetic material. Transformation experiments were initially conducted by F. Griffith in 1928.

1. He injected a mixture of two strains of Pneumococcus (Diplococcus pneumoniae) into mice. One of these two strains S III was virulent and the other strain All was non-virulent.
2. The S III type bacteria when injected into mice cause pneumonia and ultimately to death. The R II type bacteria when injected, no pneumonia occurred.
3. The S III bacteria, prior to injection are killed by heating. It is then injected into the mice. It did not cause the disease.
4. Heat killed S III and RII (non-virulent) bacteria, when injected into mice, causing pneumonia. Finally, death occurred.

Griffith’s experiment demonstration transformation in Pneumococcus.

This proved that the DNA of S III type bacteria has transformed the DNA of R II type bacteria into virulent type S III. This phenomenon of transferring characters of one strain to another by using a DNA extract of the former is called transformation. Conclusion. Griffith concluded that virulence was transferred from S-Type dead cells to R-Type living cells in the form of a capsule (some component of a cell) rather than the whole cells.

Contribution of Avery, MacCleod, and MacCarty: Avery, MacCleod, and McCarty gave the proof that the “transforming agent” is DNA. They carried out the experiments with Diplococcus and showed the transformation of type R-ll to type S-III. They found that proteases and RNases did not affect transformation but DNAses affect it. Thus they gave proof that the active component was DNA and not the RNA, proteins, or polysaccharides.

Question 22.
How did Hershey and Chase differentiate between DNA and protein in their experiment while proving that DNA is the genetic material? (CBSE 20015)
Or
Why did Hershey and Chase use 35S and 32P in their experiment? Explain. State the importance of blending and centrifugation in their experiment. Write the conclusion they arrived at after completing their experiment. (CBSE 20019 C)
Or
Hershey and Chase carried out their experiment under three steps: (a) Infection, (b) Blending, and (c) Centrifugation.
Or
Explain each one of these steps that helped them to prove that DNA is the hereditary material. (CBSE (Outside Delhi) 2019)
Answer:
1. Alfred Hershey and Martha Chase (1952) worked with viruses that infect bacteria called bacteriophages.

2. They used radioactive sulfur (35S) to identify protein and radioactive phosphorus (32P) to identify the components of nucleic acid.

3. The tadpole-shaped bacteriophage attaches to the bacteria. Its genetic material enters the bacterial cell by dissolving the cell wall of bacteria. The bacterial cell treats the viral genetic material as if it was it’s own and subsequently manufactures more virus particles.

4. Infection: They grew some viruses on a medium that contained radioactive phosphorus and others on a medium that contained radioactive sulfur.

The Hershey—Chase Experiment

5. Viruses grown in the presence of radioactive phosphorus contained radioactive DNA but not radioactive protein because DNA contains phosphorus but protein does not. Similarly, viruses grown on radioactive sulfur contained radioactive protein but not radioactive DNA because DNA does not contain sulfur.

6. Radioactive phages were allowed to attach to E. co/i bacteria. Then as the infection proceeded, the viral coats were removed from the bacteria by agitating them in a blender. The virus particles were separated from the bacteria by spinning them in a centrifuge.

7. Bacteria that were infected with viruses that had radioactive DNA were radioactive, indicating that DNA was the material that passed from the virus to the bacteria.

8. Bacteria that were infected with viruses that had radioactive proteins were not radioactive.

9. This indicates that proteins did not enter the bacteria from the viruses.

10. DNA is, therefore, the genetic material that is passed from virus to bacteria.

Question 23.
List the characteristics of DNA molecules.
Answer:
Characteristics of DNA molecule:

1. The two chains are spirally coiled about around a common axis to form a regular, right-handed double helix.
2. The double helix has a major groove and minor groove alternately.
3. The helix is 20Å wide; it’s one complete turn is 34Å long, and has 10 base pairs, and the successive base pairs are 3.4Å apart.
4. The two chains are complementary to each other with respect to base sequence.
5. The two strands are hydrogen-bonded: A on one chain is joined to T on the other chain by 2 hydrogen
   bonds; C on one chain is linked to G on the other chain by 3 hydrogen bonds.
6. The two strands run in an antiparallel direction.
7. The amount of A + G = the amount of T + C; the amount of A = the amount of T: the amount of G = the amount of C. Sugar and phosphate groups occur in equal proportion.
8. The DNA molecule is remarkably stable due to hydrogen bonding and hydrophobic interactions.
9. The DNA molecule undergoes denaturation and renaturation easily.
10. The denatured DNA strands are hyperchromic.
11. The DNA molecule can replicate and repair itself, and can also transcribe RNAs.
12. The base sequence of one chain serves as the genetic code.
13. The DNA can function in vitro.
14. The amount of DNA per nucleus is constant in all the body cells of a given species.

Question 24.
How is a long DNA molecule adjusted in a nucleus? (CBSE Delhi 2008)
Or
(i) What is this diagram representing?
(ii) Name the parts A, B, and C.
(iii) In the eukaryotes, the DNA molecules are organized within the nucleus. How is the DNA molecule organized in a bacterial cell in the ab¬sence of a nucleus? (CBSE 2009)

Answer:
The distance between two consecutive base pairs is 0.34 nm (0.34 × 10-9 m). If the length of DNA double helix in a typical mammalian cell is calculated (simply by multiplying the total number of bp with the distance between two consecutive bp, that is, 6.6 × 109 bp × 0.34 × 1(T9 m/bp), it comes out to be approximately 2.2 meters. A length that is far greater than a dimension of a typical nucleus (approx 10-6 m).

In eukaryotes, this organization is much more complex. There is a set of positively charged, basic proteins called ‘histones’. A protein acquires charge depending upon the abundance of amino acid residues with charged side chains. Histones are rich in amino acid residues lysines and arginines. Both the amino acid residues carry positive charges in their side chains. Histone organized to form a unit of eight histone molecules called ‘histone octamer’. The negatively charged DNA is wrapped around positively charged histone-octamer to form a structure called ‘Nucleosome’.

A typical nucleosome contains 200 bp of DNA helix. Nucleosomes constitute the repeating unit of a structure in the nucleus called ‘Chromatin’, thread-like stained (colored) bodies seen in the nucleus. The nucleosomes in chromatin are seen as a “beads-on-string” structure when viewed under an electron microscope (EM).
Or
(i) Nucleosome

(ii)

(iii) Organization of DNA molecule in bacteria. In prokaryotes like bacteria which lacks a nucleus, DNA is not scattered. The negatively charged DNA binds with positively charged proteins in a region called a nucleoid.

Question 25.
Describe briefly the mechanism of DNA replication. (CBSE (Delhi) 2019)
Or
Draw a labeled diagram of the replication fork. (CBSE 2011)
Or
Draw a neat labeled sketch of replicating fork of DNA.
Or
(a) Why does DNA replication occur within a replication fork and not in its entire length simultaneously?
(b) “DNA replication is continuous and discontinues on the two strands within the replication fork.” Give reasons. (CBSE (Outside Delhi) 2019)
Answer:
Mechanism of DNA replication. Watson and Crick, while explaining their model of DNA structure, suggested the semi-conservative mechanism of its replication. The quality and quantity of DNA in parent and daughter cells must be the same.

Replication is one of the most important properties of DNA and forms the very basis of life:
1.  Replication takes place during the S-phase of the interphase between two mitotic cycles.

2. Replication is a semi-conservative process in which each of the two double helices formed from the parent double strand has one old and one new strand. Repair replication is non-conservative.

3. DNA replication requires a DNA template, a primer, deoxyribonucleoside triphosphates (dATP, dGTP, dTTP, dCTP), Mg++, DNA unwinding protein, superhelix relaxing protein, a modified RNA polymerase to synthesize the RNA primer, a joining polynucleotide ligase, enzyme.

4. Watson and Crick suggested that the two strands of DNA molecules uncoil and separate, and each strand serves as a template for the synthesis of a new strand alongside it.

5. The sequence of bases which should be present in the new strands can be easily predicted because these would be complementary to the bases present in the old strands. A will pair with T, T with A, C with G, and G with C.

6. Thus, two daughter molecules are formed from the parent molecule and these are identical to the parent molecule.

7. Each daughter DNA molecule consists of one old (parent) strand and one new strand.

Continuous and discontinuous synthesis of DNA.

8. Since only one parent strand is conserved in each daughter molecule, this mode of replication is said to be semiconservative.

9. In one strand replication takes place discontinuously and short pieces called Okazaki fragments are synthesized. One strand may synthesize a continuous strand and the other Okazaki fragments. Both new strands are synthesized in the 5’ → 3’ direction. Thus one strand is synthesized forward and the other backward.

10. The Okazaki pieces are joined by polynucleotide ligase, a joining enzyme, to form continuous strands.

Question 26.
Provide experimental evidence for the semi-conservative mode of replication of DNA. (CBSE 2008 (Outside Delhi) 2012, 2016, 2019 C)
Answer:
Meselson and Stahl (1958) experimentally proved that DNA replication is semi-conservative.

1. E. coli bacterium was grown for many generations in a culture medium in which the nitrogen source contained heavy isotope N¹⁵ thus the labeling of bacterial DNA was done.
2. Later on, these bacteria were cultured in N¹⁴ non-radioactive isotope.
3. DNA was analyzed to determine the distribution of radioactivity.
4. The experiment showed that one strand of each daughter DNA molecule was radioactive whereas the other was non-radioactive.
5. During the second replication in the N¹⁴ medium, the radioactive and non-radioactive strands separated and served as the template for the synthesis of non-radioactive strands.
6. Out of four DNA molecules, two are completely non-radioactive and the other two have half of the molecule as non-radioactive.
7. This evidence shows that DNA replication is semi-conservative.
8. This biochemical evidence was supported by the direct cytological observation of duplicating DNA of E. coli.

Question 27.
(i) State the ‘Central dogma’ as proposed by Francis Crick. Are there any exceptions to it? Support your answer with a reason and an example.
Answer:
Francis Crick proposed the Central dogma in molecular biology, which states that the genetic information

Central dogma
In some viruses, central dogma is seen in the reverse direction, that is from RNA to DNA as RNA is the main genetic material. Example: Retrovirus (HIV) and the process is reverse transcription.

(ii) Explain how the biochemical characterization (nature) of the ‘Transforming Principle’ was determined, which was not defined from Griffith’s experiments. (CBSE Delhi 2018)
Answer:
Transforming principle: In 1928, Frederick Griffith, in a series of experiments with Streptococcus pneumonia which is a bacterium responsible for pneumonia, witnessed a miraculous transformation in the bacteria. During the experiment, a living organism (bacteria) changed its physical form. He concluded that the R strain bacteria had somehow been transformed by the heat-killed S strain bacteria.

Some ‘transforming principle’, transferred from the heat-killed S strain, had enabled the R strain to synthesize a smooth polysaccharide coat and become virulent. This must be due to the transfer of the genetic material. However, the biochemical nature of genetic material was not defined from his experiments.

Oswald Avery, Colin MacLeod, and Maclyn McCarty worked to determine the biochemical nature of the ‘transforming principle’ in Griffith’s experiment. They purified biochemicals (proteins, DNA, RNA, etc.) from the heat-killed S cells to see which ones could transform live R cells into S cells. They discovered that DNA alone from S bacteria caused R bacteria to become transformed.

They also discovered that protein-digesting enzymes (proteases) and RNA-digesting enzymes (RNases) did not affect transformation, so the transforming substance was not a protein or RNA. Digestion with DNase did inhibit transformation, suggesting that the DNA caused the transformation. They concluded that DNA is the hereditary material, but not all biologists were convinced.

Question 28.
(a) Write the contributions of the following scientists in deciphering the genetic code: George Gamow; Hargobind Khorana; Marshall Nirenberg; Severo Ochoa.
Answer:
Contributions of scientists:

1. George Gamow: He suggested that in order to code for all the 20 Amino acids, the code should be made up of three nucleotides. He proved that the codon is a triplet.
2. Hargobind Khorana: He provided experimental proof that genetic codon is always triplet. He was able to synthesize RNA molecules with a defined combination of bases (homopolymers and copolymers)
3. Marshall Nirenberg: He prepared a cell-free system for protein synthesis and finally deciphered the genetic code.
4. Severe Ochoa: He established that enzyme polynucleotide phosphorylase was helpful in RNA synthesis with a defined sequence in a template-independent manner.

(b) State the importance of a Genetic code in protein biosynthesis. (CBSE Delhi 2019)
Answer:
Importance of Genetic code: Genetic code codes for a specific amino acid that is required for protein synthesis. (CBSE Outside Delhi 2018)

Question 29.
Explain translation in detail.
Or
Explain the process of charging tRNA.
Answer:
Translation: During the translation process, proteins are made by the ribosomes on the mRNA strand:
The main steps are:

Continuous and discontinuous synthesis of DNA:

1. Activation of amino acid.
2. Transfer of activated amino acid to tRNA.
3. Initiation of synthesis.
4. Elongation of a polypeptide chain.
5. Termination of a chain.

Question 30.
Explain the steps involved in DNA fingerprinting. (CBSE 2009)
Answer:
Steps of DNA fingerprinting:
The steps/procedure in DNA fingerprinting include the following:

1. Extraction: DNA is extracted from the cells in a high-speed, refrigerated centrifuge.
2. Amplification: Many copies of the extracted DNA are made by a polymerase chain reaction.
3. Restriction Digestion: DNA is cut into fragments with restriction enzymes into precise reproducible sequences.
4. Separation of DNA sequences/ restriction fragments: The cut DNA fragments are introduced and passed through an electrophoresis setup containing agarose polymer gel; the separated fragments can be visualized by staining them with a dye that shows fluorescence under ultraviolet radiation.
   
   Lac operon
5. Southern Blotting: The separated DNA sequences are transferred onto a nitrocellulose or nylon membrane.
6. Hybridization: The nylon membrane is immersed in a bath and radioactive probes (DNA segments of known sequence) are added: these probes target a specific nucleotide sequence that is complementary to them.
7. Autoradiography: The nylon membrane is pressed on an X-ray film and dark bands develop at the probe sites.

Question 31.
What is the inducer in the lac operon? How does it ensure the “switching on” of genes?
(i) Draw a schematic representation of the Lac operon.
Answer:
Inducer. It is a chemical that may be a substrate, hormone, or some other metabolite which after coming in contact with the repressor, changes the latter into the non-DNA binding state so as to free the operator gene. Thus the “switch on” occurs.

(ii) Explain how does this operon gets switched ‘on’ or ‘off’.
Answer:
The expression of the genes is usually controlled to achieve maximum cellular economy. This means that the gene will be turned on or off as per requirement. A set of genes will be switched on when there is the necessity to handle and metabolize a new substrate. When these genes are turned on enzymes are produced, which metabolize the new substrate. The phenomenon is known as induction.

(iii) What is the role of a regulatory gene? (CBSE Delhi 2012, 2019 C)
Answer:
Role of the regulatory gene in a lac operon: It synthesizes a biochemical or regulator protein that can act positively as an activator and negatively as a repressor. It controls the activity of the operator gene.

Question 32.
(i) Describe the structure and function of a tRNA molecule. Why is it referred to as an adapter molecule?
Answer:
t-RNA (transfer RNA) reads the genetic code on one hand and transfers amino acids on the other hand. So it was called adapter molecule by Francis Crick. It is also called soluble RNA (SRNA).

Note: For the figure, please see the answer of Q. no. 11 short answer type question. The secondary structure of tRNA is clover leaf-like but the 3-D structure is inverted L-shaped. t-RNA has five arms or loops:

• Anticodon loop: It has bases complementary to the code.
• Amino acid acceptor end: Amino acid binds to it.
• T-loop: It helps in binding to ribosomes.
• D-loop: It helps in binding aminoacyl synthetase.
• Variable loop: Its function is not known.

(ii) Explain the process of splicing of hn-RNA in a eukaryotic cell. (CBSE Delhi 2017)
Answer:
The primary transcript formed in eukaryotes is non-functional, containing both the coding region exon and non-coding region intron in RNA and are called heterogeneous RNA or hn-RNA. hn-RNA undergoes a process where the introns are removed and exons are joined to form mRNA by the process called splicing.

Question 33.
(i) Why does replication occur in small replication forks and not in entire lengths?
Answer:
Replication occurs in a small opening of a DNA helix called replication forks for very long DNA molecules. The reason is in the case of long DNA molecules, the two strands of DNA cannot be separated in their entire length due to very high energy demand.

(ii) Why is DNA replication continuous and discontinuous in a replication fork?
Answer:
DNA-dependent DNA polymerase catalyzes DNA polymerization only in one direction, that is, 5′ → 3′. DNA strands are anti-parallel and have opposite polarity. So on template strand with polarity 3′ → 5′ DNA replication is continuous white on the template strand with polarity 5′ → 3′ replication is discontinuous.

(iii) State the importance of the origin of replication in a replication fork. (CBSE Delhi 2018C)
Answer:
Replication cannot start randomly at any point in DNA. It requires a definite region of sequences where the replication originates. This site is called the origin of replication.

Question 34.
Compare the processes of DNA replication and transcription in prokaryotes. (CBSE Delhi 2019C)
Answer:

Question 35.
Write the different components of a lac operon in E. coil. Explain its expression while in an ‘open’ state. (CBSE Delhi 2017, 2019 C)
Answer:
The Lac operon (Inducible operon): The concept of the operon was first proposed in 1961 by Jacob and Monod.
Components of an operon:

1. Structural genes: It Is the fragment of DNA that transcribes mRNA for polypeptide synthesis.
2. Promoter: It Is the sequence of DNA where RNA poLymerase binds and initiates transcription.
3. Operator: It is the sequence of DNA adjacent to the promoter.
4. Regulator gene: It is the gene that codes for repressor protein which binds to the operator due to which operon Is switched “off”.
5. Inducer: Lactose Is an Inducer that helps in switching “on” of an operon. Lac operon consists of three structural genes (z, y, a), operator (o), promoter (p), regulatory gene (r).

(a)

(b)

• Gene z codes for p-galactosidase
• Gene y codes for permease.
• Gene codes for enzymes transacetylase. When lactose is absent:

When lactose is absent, i.e. gene produces repressor protein.

This repressor protein binds to the operator and as a result, prevents RNA polymerase to bind to the operon, and the operon is switched off.

When lactose is present:

• Lactose acts as an inducer that binds to the repressor and forms an inactive repressor.
• The repressor cannot bind to the operator.
• Now the RNA polymerase binds to the operator and transcribes lac mRNA.
• Lac mRNA is polycistronic, i.e. produces all three enzymes p-galactosidase, permease, and transacetylase.
• The lac operon is switched on.

Question 36.
What is the human genome project (HGP)? Write salient features of the human genome project.
Answer:
Human Genome Project. It is a mega project involving a lot of money, the most advanced techniques, numerous computers, and scientists at work. The magnitude of the project can be imagined that if the cost of sequencing a bp is 3 dollars, sequencing of 3 × 109 bp would be a billion dollars. If the data is to be stored in books, with each book having 1000 pages and each page with 1000 letters, some 3300 books will be required. Here bioinformatics databasing and other high-speed computational devices have helped in the analysis, storage, and retrieval of information.

Salient features of the human genome:

1. The human genome contains 3164.7 million nucleotides (base pairs).
2. The size of the genes varies; an average gene consists of 3000 bases, while the largest gene, dystrophin, consists of 2.4 million bases.
3. The total number of genes is estimated to be 30000 and 99.9% of the nucleotides are the same in humans.
4. The functions of over 50% of the discovered genes are not known.
5. Only less than 2% of the genome codes for proteins.
6. Repetitive segments made up a large portion of the human genome.
7. Repetitive sequences throw light on chromosome structure and dynamics and evolution, though they are thought to have no direct coding functions.
8. (Chromosome 1 has 2968 genes and Y-chromosome has the least number (231 genes).
9. Scientists have identified about 1.4 million locations, where DNA differs in a single base in human beings. These are called single nucleotide polymorphisms (SNPs).
10. Repeated sequences make up a large portion of the human genome.

Very Important Figures:

Unidirectional repLication

BidirectionaL replication

The flow of genetic information.

Transcription in eukaryotes

Here we are providing Class 12 Biology Important Extra Questions and Answers Chapter 5 Principles of Inheritance and Variation. Important Questions for Class 12 Biology are the best resource for students which helps in Class 12 board exams.

Class 12 Biology Chapter 5 Important Extra Questions Principles of Inheritance and Variation

Principles of Inheritance and Variation Important Extra Questions Very Short Answer Type

Question 1.
Name one trait that does not blend.
Answer:
Sex does not blend.

Question 2.
Give one example of a genetic trait for each of the following in humans:
1. Lethality
Answer:
Lethality: Homozygous sickle cell anaemia.

2. Multiple allelism.
Answer:
Example of Multiple allelism: ABO blood groups.

Question 3.
What for symbols AA and Aa stand?
Answer:

• AA: Homozygous dominant,
• Aa: Heterozygous dominant.

Question 4.
Name the plant that shows incomplete dominance in respect to the colour of its flower.
Answer:
Mirabilis jalapa.

Question 5.
Write the genotypes of a man with blood group A.
Answer:
l^Al^A, l^Al⁰

Question 6.
What is Mendel’s monohybrid ratio for phenotypes?
Answer:
3:1.

Question 7.
Write down Mendel’s dihybrid ratio for phenotypes.
Answer:
9: 3: 3: 1.

Question 8.
Who were the discoverers of Mendelism?
Answer:
Hugo de Vries, Karl Correns, Erich von Tschermak were the rediscoverers of Mendelism.

Question 9.
What are the real determinants of what an organism will become?
Answer:
The complex interaction between genes and their environment really determine what an organism will become.

Question 10.
Name the disorder in humans with the following karyotype:
(a) 22 pairs of autosomes + XO
(b) 22 pairs of autosomes + 21 st chromosome + XY (CBSE Outside Delhi 2019)
Answer:

Question 11.
What wilt is the genetic makeup of an organism which suffers from sickle cell anaemia?
Answer:
Homozygous (Hb^S Hb^S).

Question 12.
Name the type of cross that would help to find the genotype of a pea plant bearing violet flowers. (CBSE Delhi 2017)
Answer:
Test cross.

Question 13.
What term is used for the two chromatids resulting from the interchange of segments during crossing over?
Answer:
Recombinants (cross overs).

Question 14.
Write one example of each of organisms exhibiting
(i) male heterogamety
Answer:
Male heterogamety: Drosophila and humans

(ii) female heterogamety. (CBS£ Delhi 2019 C)
Answer:
Female heterogamety: Birds and some reptiles

Question 15.
A geneticist is interested in study variations and pattern in living being preferred to choose an organism with the short life cycle. Provide a reason. (CBSE Delhi 2015)
Answer:
An organism with a shorter life cycle is helpful in rapid study and analysis of hereditary pattern in many generations, e.g. Drosophila, Neurospora.

Question 16.
Give an example of a human disorder that is caused due to a single gene mutation. (CBSE Delhi 2016)
Answer:
Phenylketonuria.

Question 17.
Write the sex of human having XXY chromosomes with 22 pairs of autosomes. Name the disorder this human suffers from. (CBSE Delhi 2016)
Answer:
The human is male (as Y chromosome is present). He is suffering from Klinefelter’s syndrome

Principles of Inheritance and Variation Important Extra Questions Short Answer Type

Question 1.
State the Mendelian principle which can be derived from a dihybrid cross and not from monohybrid cross.
Or
State Mendel’s Law of Independent Assortment. CBSE Sample Paper 2018-19)
Answer:
From the dihybrid cross, the law of independent assortment can be derived which states that, when two pairs of traits are combined in a hybrid, segregation of one pair of characters is independent of the other pair of characters.

Question 2.
In a cross between two tall pea plants, some of the offsprings produced pure dwarf. Show with the help of Punnett square how this is possible. (CBSE (Delhi) 2013)
Answer:

Question 3.
What is a dihybrid cross?
Answer:
Dihybrid cross. A cross in which two characters are taken into consideration during experimentation, such a cross is called dihybrid cross. A cross between a pea plant with yellow smooth seeds and a pea plant with green, wrinkled seeds is a dihybrid cross.

From the dihybrid cross, it can be derived that each gene is assorted independently of the other during its passage from one generation to the other or Law of independent assortment.

Question 4.
In order to obtain the E, generation, Mendel pollinated a pure breeding tall plant with a pure breeding dwarf plant. But forgetting the F₂ generation, he simply self-pollinated the tall F₁ plants. Why?
Answer:

1. He made crosses to study the pattern of inheritance of a few characters over generations.
2. Initially, he made pure lines.
3. To create a heterozygote or hybrid, he had to cross two different plants (pure lines).
4. To study the inheritance pattern; it is enough if the hybrids are self – pollinated, thus the segregation of factors can be studied.

Question 5.
A cross between a red flower-bearing plant and a white flower-bearing plant of Antirrhinum majus produced all plants having pink flowers. Work out across, to explain how is this possible? (CBSE Outside Delhi 2013)
Answer:

Question 6.
Differentiate gene and allele.
Answer:
Difference between gene and allele:
Allele (allelomorphs) refers to the alternate form of a gene pair present on the same loci in the homologous chromosome, whereas gene is the smallest unit of an organism capable of transmitting genetic information and expressing the same.

Question 7.
In Snapdragon, a cross between true-breeding red-flowered (RR) plants and true-breeding white-flowered (RR) plants showed a progeny of plants with all pink flowers.
(i) The appearance of pink flowers is not known as blending. Why?
Answer:
It is not a case of blending. In this case, R gene was not completely dominant over r gene and this made genotype Rr to distinguish as pink.

(ii) What is this phenomenon known as? (CBSE 2014)
Answer:
Incomplete dominance.

Question 8.
With the help of one example, explain the phenomena of co-dominance and multiple allelism in the human population. (CBSE 2014)
Answer:
In the case of co-dominance, two alleles for a trait are equally expressed.

Example: ABO blood groups are controlled by the gene I. The gene I have three alleles l^A, l^B and l^O. These alleles determine the type of sugar on the RBC surface. Alleles lA and lB are co-dominant and express the AB blood group.

Since there are three different alleles and express themselves on the basis of dominance recessiveness and co-dominance, it is a case of multiple allelism.

Question 9.
The child has a blood group of O. If the father has blood group A and mother has blood group B, work out the genotypes of the parents and the possible genotypes of the other offsprings. (CBSE Outside Delhi, 2015, 2019)
Answer:

1. Genotypes. Man (l^A l^O) Mother l^B l^O and child l^O l^O.
2. The blood group of the future offspring. A type, B type, 0 types and AB type. It is based on the following cross:

Inheritance of blood groups A, B, O, AB

Question 10.
Give examples of sex-linked inheritance in Drosophila. During his studies on genes in Drosophila that were sex-linked T.H. Morgan found F₂ population phenotypic ratios deviated from expected 9: 3: 3: 1. Explain the conclusion he arrived at. (CBSE 2010)
Answer:
Examples of sex-linked inheritance in Drosophila (Morgan’s conclusion).

1. Genes for white eye colour is located in the X-chromosome and Y-chromosome is empty carrying no normal allele for white eye colour.
2. The white-eyed female possesses a gene for white eye colour (W) on both of its X-chromosomes.
3. The white-eyed males receive X-chromosome with (W) gene from mother and (Y) from father with no gene.
4. The daughter receives one X-chromosome with (W) gene from mother and one X-chromosome with dominant (W+).

Question 11.
Briefly explain XX-XO (a type of sex determination).
Answer:
In the case of roundworms, true bugs, grasshoppers and cockroaches the females have two sex chromosomes XX, whereas the males have only one X-chromosome. The male has no second chromosome thus designated as XO. The sex ratio of 1: 1 is produced as shown in the figure below.

XX-XO determination of sex in the cockroach.

Question 12.
Explain the mechanism of ‘sex determination’ in birds. How does it differ from that of human beings? (CBSE Delhi 2018)
Answer:
Sex determination is of ZW-ZZ type in birds.
In this type, the males are homogametic and have ZZ sex chromosomes, and females are heterogametic with ZW pair of sex chromosomes.

In human beings, the chromosomal mechanism of sex determination is of XX- XY type. The human male is heterogametic and has XY sex chromosomes, whereas the human female is homogametic with XX sex chromosomes.

Question 13.
Write a note on ZO-ZZ type of sex determination.
Answer:
In case of ZO-ZZ type of sex determination the female produces two types of eggs. The one-half of eggs is with Z-chromosome and the other half without Z-chromosome. The male has homomorphic sex chromosomes and is homogametic. It forms only one kind of sperms each with Z-chromosome. On fertilisation by a sperm with Z-chromosome, the Z-containing egg gives rise to male offspring ZZ and Z- lacking egg produces female offspring ZO. Such type of sex determination is found in the case of butterflies and moths.

Question 14.
Give an example of an autosomal recessive trait in humans. Explain its pattern of inheritance with the help of a cross. (CBSE Delhi 2016)
Answer:
Autosomal recessive trait. Sickle-cell anaemia is caused by autosomal recessive trait. The disease is controlled by a single pair of alleles Hb^A and Hb^S. Only the homozygous individuals for HbsHbs show the disease. The heterozygous individuals are carriers (Hb^A Hb^s)

Question 15.
How would you find the genotype of a tall pea plant bearing white flowers? Explain with the help of a cross. Name the types of the cross you would use. (CBSE Delhi 201i 5)
Answer:
In order to find the genotype of a given plant, one has to breed it with plenty of oic recessive individual. It is called test CRC is. Tall and white plant TTww or Ttww / is crossed with dwarf white plant ttww TTww x ttww

(All tall white plants indicate that the genotype is TTww)

This ratio indicates that the genotype is Ttww

Depending on the resuLt we can determine the genotype of the flower.

Question 16.
How is polygenic inheritance different from pleiotropy? Give one example of each.
Answer:

Question 17.
Why is it not possible to study the pattern of inheritance of traits in human beings, the same way as it is done in pea plant? Name the alternate method employed for such an analysis of human traits.
Answer:

1. Control cross cannot be performed in human as in other organisms.
2. The generation time is long. Pedigree analysis is the alternative method to study the inheritance of human traits in several generations.

Question 18.
A man with blood group A married a woman with a B group. They have a son with AB blood group and a daughter with blood group 0. Work out the cross and show the possibility of such inheritance. (CBSE Delhi and Outside Delhi 2008)
Answer:

Blood groups of progeny A, B, AB and O.

Question 19.
A haemophilic father can never pass the gene for haemophilia to his son. Explain. (CBSE Delhi 2016)
Answer:
Haemophilia is a sex-linked recessive disorder where X chromosome carries the defective haemophilic gene and Y chromosome is healthy. And son inherits only the Y chromosome from his father which is not carrying the gene for haemophilia. Therefore, the haemophilic father can never pass haemophilia to his son.

Principles of Inheritance and Variation Important Extra Questions Long Answer Type

Question 1.
Study the given pedigree chart and answer the questions that follow:
1. Is the trait recessive or dominant?
Answer:
Dominant.

2. Is the trait sex-linked or autosomal?
Answer:
Autosomal.

3. Give the genotypes of the parents shown in generation I and their third child is shown in generation II and the first grandchild shown in generation III. (CBSE Sample Paper 2018-19)

Answer:
The genotype of parents in generation I – Female: aa and Male: Aa
The genotype of a third child in generation II-Aa Genotype of the first grandchild in generation III – Aa

Question 2.
Haemophilia is a sex-linked recessive disorder of humans. The pedigree chart given below shows the inheritance of Haemophilia in one family.

Study the pattern of inheritance and answer the questions given.
1.Give all the possible genotypes of the members 4, 5 and 6 in the pedigree chart.
2. A blood test shows that the individual 14 is a carrier of haemophilia. The member numbered 15 has recently married the member numbered 14. What is the probability that their first child will be a haemophilic male?
(CBSE 2009, CBSE Sample Paper 2018-19)

Answer:

1. Genotypes of member 4 – XX or XX^h Genotypes of member 5 – X^hY Genotypes of member 6 – XY
2. The probability of first child to be a haemophilic male is 50%.

Question 3.
Mention the advantages of selecting a pea plant for the experiment by Mendel.
Answer:
Advantages of selecting pea plant as experimental material:

Mendel selected pea plant (Pisum sativum) because:

1. Many varieties were available with observable alternative forms for a trait or a characteristic.
2. Peas normally self-pollinate; as their corolla completely encloses the reproductive organs until pollination is complete.
3. It was easily available.
4. It has pure lines for experimental purpose, i.e. they always breed true.
5. It has contrasting characters. The traits were seed colour, pod colour, pod shape, flower shape, the position of flower, seed shape and plant height.
6. Its life cycle was short and produced a large number of offsprings.
7. The plant can be grown easily and does not require care except at the time of pollination.

Question 4.
Differentiate between the following:
(i) Dominance and recessiveness
Answer:
Differences between dominance and recessiveness:

(ii) Homozygous and heterozygous
Answer:
Differences between homozygous individual and a heterozygous individual:

Question 5.
Explain the law of dominance using a monohybrid cross.
Answer:
Law of Dominance. According to this law, when two factors of a character are unlike, one of them will manifest in the body and is called dominant while the other remains hidden and is termed recessive factor.

The law can be well explained by the monohybrid cross by studying the following crosses:
(i) Pure tall = TT, Hybrid tall = Tt

Gametes of TT parent = \(\frac { 1 }{ 2 }\)T + \(\frac { 1 }{ 2 }\)T
Gametes of Tt parent = \(\frac { 1 }{ 2 }\)T + \(\frac { 1 }{ 2 }\)t

The 50% are pure tall and 50% hybrid tall. Then pure tall plants will produce 100% tall in F₂ generation and hybrid plants will produce in the ratio of 1: 2: 1 in the F2 generation.

(ii) When the cross is made between pure tall and pure dwarf, we get results as follows (Fig.).

A Punnett square used to understand a typical monohybrid cross conducted by Mendes between true-breeding tall plants and true-breeding dwarf plants

Question 6.
Define and design a test cross.
Answer:
Test Cross: It is a cross between an organism of an unknown genotype and a homozygous recessive organism.

Results of a Test Cross: If the test cross yields offspring of which 50% show the dominant character and 50% show the recessive character, i.e. F1 ratio is 1: 1, the individual under test is heterozygous (see fig.). This is so because the individual showing the recessive trait (say white coat colour in the guinea pig, dwarf size in pea plant) must have received one recessive allele (b in a guinea pig, t in pea plant) from each parent.

Genetics of a test cross

If all the offspring of the test cross show the dominant trait, the individual being tested is homozygous dominant with genotype BB for a guinea pig and TT for pea plant (fig.).

Question 7.
When a cross is made between a tall plant with yellow seeds (TtYy) and tall plant with green seeds (Ttyy), what proportions of phenotype In the offspring could be expected to be
(i) tall and green
(ii) dwarf and green?
Answer:

Cross between a tall plant with yellow seeds

Question 8.
Differentiate back cross and test cross.
Answer:
Differences between the back cross and test cross:

Question 9.
Differentiate incomplete dominance and codominance.
or
Explain the following terms with an example:
(i) Codominance
(ii) Incomplete dominance (CBSE Outside Delhi 2008; Delhi 2011, 2019)
Answer:
Differences between incomplete dominance and codominance:

Question 10.
(i) Why is the human ABO blood group gene considered a good example of multiple alleles?
Answer:
A B O blood groups are controlled by a single gene: (I) The plasma membrane of the red blood cells has sugar polymers that protrude from its surface and is controlled by the gene. The gene (I) has three alleles l^A, l^Band l^O. Presence of more than two types of alleles at the same locus governing the same character is called multiple alleles.

(ii) Work out across up to F₂ generation only, between a mother with blood group A (Homozygous) and the father with blood group B (Homozygous). Explain the pattern of inheritance exhibited. (CBSE (Delhi) 2013)
Or
Describe the mechanism of a pattern of inheritance of ABO blood groups in human. (CBSE 2011)
Answer:
Patterns of inheritance of the ABO blood group.

F₁ generation. All with blood group AB. It is a case of co-dominance.

Question 11.
(i) What is polygenic inheritance? Explain with the help of a suitable example, (ii) How are pleiotropy and Mendelian pattern of inheritance different from the polygenic pattern of inheritance? (CBSE Outside Delhi 2016)
Or
Certain phenotypes in the human population are spread over a gradient and reflect the contribution of more than two genes. Mention the term used for the type of inheritance. Describe it with the help of an example in the human population. (CBSE Sample Paper 2019-20)
Answer:
(i) Polygenic inheritance. It is a type of inheritance controlled by three or more genes in which the dominant alleles have a cumulative effect. Each dominant allele expresses a part or unit of a trait. It is also called quantitative inheritance or multiple factor inheritance. The genes involved in such kind of inheritance are termed polygenes.

Examples:

1. Kernel colour in wheat
2. Cob length in maize
3. Skin colour in human
4. Human intelligence Human Skin Colour. It is caused by the pigment melanin.

The quantity of melanin is due to three pairs of polygenes (A, B and C). The genotype of black will be (AA BB CC) and white will have (aa bb cc). Marriage between two such persons will show variations. In progeny (Aa Bb CC) there will be 7 types of phenotypes i.e. very dark, 6 dark, 15 fairly dark, 20 intermediates, 15 family light, 6 light and have very light.

(ii) How are pleiotropy and Mendelian pattern of inheritance different from the polygenic pattern of inheritance? (CBSE Outside Delhi 2016)
Answer:

1. Pleiotropy and Mendelian pattern of inheritance: In the case of pleiotropy one gene has an effect on two on more traits. One effect is more evident in the case of one trait (major effect) and less evident in the case of others (secondary effect). The mendelian pattern of inheritance is monogenic.

2. In pleiotropism and monogenic inheritance no intermediates are produced and show discontinuous variations in the expression of a trait. Intermediates are quite common in polygenic inheritance and produce continuous variations in the expression of a trait.

Question 12.
(i) Write the scientific name of the organism Thomas Hunt Morgan and his colleagues worked with for their experiments. Explain the correlation between linkage and recombination with respect to genes as studied by them.
Answer:
Thomas Hunt Morgan and his colleagues worked with Drosophila melanogaster. They carried out several dihybrid crosses in Drosophila to study gens that were sex-linked.

Morgan and his group knew that the genes were located on the X chromosome and noticed that when the two genes in a dihybrid cross were situated on the same chromosome, the proportion of parental gene combinations were much higher than the non-parental type.

Morgan attributed this due to the physical association or linkage of th< two genes. He coined the term links to describe this physical associate of genes on a chromosome and t/ term recombination to describe t generation of non-parental age combination. Morgan and his also found that even when genes w grouped on the same chromos or some genes were very tightly linked, they showed very low recombine while others were loosely linked.

(ii) How did Sturtevant explain gene mapping while working with Morgan? (CBSE Delhi 2018)
Answer:
Alfred Sturtevant was Morgan’s student. He used the frequency of recombination between gene pairs on the same chromosome as a measure of the distance between genes and ‘mapped’ their position on the chromosome. Today genetic mappings are extensively used as a starting point in the sequencing of the whole genomes.

Question 13.
What is recombination? Discuss the applications of recombination from the point of view of genetic engineering.
Answer:
Recombination refers to the generation of a new combination of genes which is different from the parental types. It is produced due to crossing over that occurs during meiosis prior to gamete formation.

Applications of recombination:

1. It is a means of introducing new combinations of genes and hence new traits.
2. It increases variability which is useful for natural selection under changing environment.
3. It is used for preparing linkage chromosome maps.
4. It has proved that genes lie in a linear fashion in the chromosome.
5. Breeders have to select small or large population for obtaining the required cross-overs. For obtaining cross-overs between closely linked genes, a very large population is required.
6. Useful recombinations produced by crossing over are picked up by breeders to produce useful new varieties of crop plants and animals. Green revolution and white revolution were implemented using the selective recombination technique.

Question 14.
How is sex determined? (CBSE Delhi 2015)
Answer:
Determination of the sex of the child. Sex chromosomes determine sex in human beings. In males, there are 44+XY chromosomes, whereas in female there are 44+XX chromosomes. Here X and Y chromosomes determine sex in human beings.

Two types of gametes are formed in male, one type is having 50% X-chromosome, whereas another type is having Y-chromosome. In a female, gametes are of one type and contain X-chromosome. Thus females are homogametic and males are heterogametic. If male gamete having Y-chromosome (endosperm) undergoes fusion with female gamete having X-chromosome, the zygote will have XY chromosome and this gives rise to a male child.

If male gamete having X-chromosome (gymnosperm) undergoes fusion with female gamete having X-chromosome, the zygote will be having XX-chromosome and this gives rise to the female child.

Genetics of sex in human beings. The letter A represents autosomes.

Question 15.
Both haemophilia and thalassemia are blood-related disorders in human. Write their causes and the difference between the two. Name the category of genetic disorder they both come under. (CBSE Delhi 2017)
Answer:
Both haemophilia and thalassemia are Mendelian disorders:

• Haemophilia is a sex-linked recessive disorder. The gene for haemophilia is located on X-chromosome. The gene passes from a carrier female to her son.
• Thalassemia is an autosomal-linked recessive disease.
• It occurs due to either mutation or deletion resulting in the reduced rate of synthesis of one of the globin chains of haemoglobin.
• Difference between Haemophilia and Thalassemia. In haemophilia, clotting is affected, i.e. there can be a non-stop bleeding even after a minor cut.
• In Thalassemia anaemia is the characteristic of this disease. It is caused by faulty haemoglobin synthesis.

Question 16.
Differentiate male and female heterogamety. (CBSE Delhi 2015, 2019 C)
Answer:
Differences between male and female heterogamety:

Question 17.
Why Drosophila has been used extensively for genetical studies? (CBSE 2014, 2019 C)
Answer:
Advantages of using Drosophila as genetic material:
Drosophila is a very useful organism for genetical experiments because:

1. A very large number of offsprings are produced after each mating.
2. It can be cultured in large number in laboratory and animals can be easily examined under a hand lens.
3. Its life cycle is very short and is completed in 10-12 days. A new generation can be obtained every two weeks.
4. It has four pairs of chromosomes all different in size and easily distinguishable.
5. They produce numerous variants.
6. It has heteromorphic (XY) chromosomes in the male.
7. Female Drosophila flies can be easily differentiated from the males by the large body size and presence of ovipositor in the abdomen.

Question 18.
Mendel published his work on the inheritance of characters in 1865, but it remained unrecognised till 1900. Give three reasons for the delay in accepting his work. (CBSE Delhi 2014)
Answer:
Mendel’s work published as “Experiments on plant hybridisation” remained unnoticed and unappreciated for some 34 years due to:

1. Limited circulation of the “Proceedings of Brunn Natural Science Society” in which it was published.
2. He could not convince himself about his conclusions being universal since Mendel failed to reproduce the results on Hawkweed (Hieracium) undertaken on the suggestion of Naegeli. It was due to non-availability of pure lines.
3. Absence of aggressiveness in his personality.
4. The scientific world was being rocket^ at that time by Darwin’s theory c}f Natural Selection.

Question 19.
Compare in any three ways the chromosomal theory of inheritance as proposed by Sutton and Boveri with that of experimental results on pea pie int presented by Mendel. (CBSE Delhi 2019)
Answer:

Question 20.
(a) Explain linkage and recombination as put forth by T.H. Morgan based on his observations with Drosophila melanogaster crossing experiment.
(b) Write the basis on which Alfred Sturtevant explained gene mapping. (CBSE Delhi 2019)
Answer:
(a) Linkage and recombination:

• Morgan is called the father of experimental genetics.
• Morgan used Drosophila for experiments of genetics.
• Linkage: It is the phenomenon of certain genes staying together during inheritance through several generations without any change or separation of these being present on the same chromosome. The two genes do not segregate independently of each other. So, F₂ generation deviates significantly from 9:3:3:1.
• Recombination: Loosely linked genes show a higher frequency of recombinant frequency which is around 37.2%. Tightly linked genes tend to show fewer recombinant frequency which is around 1.3%.

(b) Morgan’s student Alfred Sturtevant used the frequency of recombination between gene pairs on the same chromosome as a measure of the distance between genes and mapped their position on a chromosome.

Question 21.
ExplaIn how a test cross can be conducted to distinguish between a homozygous and heterozygous dominant genotype. What is the test cross? How can it decipher heterozygosity of a plant? (CBSE Delhi 2016)
Or
How will you find out whether a given plant is homozygous dominant? (CBSE 2008)
Or
You are given a tall pea plant and asked to find its genotype. How will you find it? (CBSE Outside Delhi 2019)
Answer:
Test Cross. When an individual is crossed to recessive parent it is called a test cross. The results can be easily analysed. If you follow the monohybrid cross where the FT is test crossed, a ratio of 1:1 will be obtained. On the same basis, you can work out that in a dihybrid case, the test cross ratio will be 1:1:1:1. Test cross can also be used for another purpose.

You must have understood by now that the homo and heterozygous genotypes for a dominant trait cannot be differentiated because they show the same phenotype. If we put them through a test cross, you will see that all homozygous dominant combinations will breed true but heterozygous genotypes will follow the segregation.

A test cross can be conducted to differentiate between a homozygous and heterozygous dominant genotype.

Question 22.
Explain the law of independent assortment with a dihybrid cross. (CBSE Outside Delhi, 2013, 2014)
Answer:
Law of independent assortment: According to this law, the factors of different pairs of contrasting characters do not influence each other. They are independent of one another in their assortment to form a new combination during gamete formation. Dihybrid cross. A cross in which two characters are taken into consideration during experimentation, such a cross is called dihybrid cross.

A cross between a pea plant with yellow smooth and a pea plant with green, wrinkled seeds is considered. Explanation. When a cross is made between pea plant having yellow smooth seeds (YYSS) and a pea plant with green wrinkled seeds (yyss). At the time of cross-pollination, yellow smooth (YYSS) produces gametes with genes (YS) and green wrinkled will produce gametes with gene (ys). Gametes unite at random. The seeds obtained when placed in the soil will grow to form plants and produce seeds which are yellow smooth (YySs) because yellow and smooth characters are dominant over green and wrinkled. These are called plants of F₁ generation.

When plants of F₁ generation are allowed to self-pollinate gametes formed YS, Ys, yS and ys by meiosis, they unite at random forming seeds. The plants thus obtained are called F₂ generation. They are Yellow smooth (YYSS, YySS, YySs, YYSs); yellow wrinkled (YYss, Yyss), green smooth (yySS, yySs) and green wrinkled (yyss) in the ratio of 9: 3: 3: 1. The result of a dihybrid cross can be shown in Fig. on the chequerboard.

Result of a dihybrid cross.

From the above dihybrid cross, it can be derived that each gene is assorted independently of the other during its passage from one generation to the other or law of independent assortment is justified.

Question 23.
In four o’clock plants, red colour (R) is incompletely dominant over white (r), the heterozygous having pink colour. What will be the offspring in a cross between a red flower and a pink flower? (CBSE Outside Delhi, 2013)
Answer:

1. In the monohybrid cross, red is incompletely dominant over white.
2. Red flowered plants have genotype RR and white-flowered plants have genotype rr.
3. Pink flowers have a genotype Rr.
4. Red flowering plants will form gametes with R genes and pink flowers will produce two types of gametes with R gene and r gene.
5. Arrangement of gametes in chequerboard.

Question 24.
Explain the pattern of inheritance of haemophilia in humans. Why is the possibility of a human female becoming a haemophilic extremely rare? Explain. (CBSE Delhi, 2008; Outside Delhi 2011)
Or
Why is human female rarely haemophilic? Explain how do haemophilic patients suffer? (CBSE Outside Delhi 2013)
Answer:
A pattern of inheritance of haemophilia:

1. It is the sex-linked recessive trait which is known as bleeder’s disease because the exposed blood does not readily clot due to deficiency of plasma thromboplastin (haemophilia B/ Christmas disease) or Antihaemophilia globulin (haemophilia A)
2. The defect has been inherited in the family of British Crown through Queen Victoria.
3. In females, haemophilia appears when both the sex chromosomes carry its recessive gene, X^h X^h. Such females die before birth.
4. A woman having a single allele of the trait appears normal but is a carrier of the disease XX^h
   
5. For sex-linked genes, human males are hemizygous. Therefore, X^h Y is haemophilic.
6. Marriage between haemophilic male and carrier female produces haemophilic sons (X^hY, 50%), normal sons (XY, 50%), carrier daughters (XX^h, 50%) and haemophilic daughters (X^hX^h, 50%, die before birth).
7. Haemophilic man (X^hY) and normal woman (XX) produce carrier girls (XX^h) and normal boys (XY).
   
8. Marriage between carrier woman and normal man produce 50% carrier girls (XX^h), 50% normal girls (XX), 50% normal boys (XY) and 50% haemophilic boys (X^hY).
   
   Sons- 50% normaL 50% heamophitic
   Daughters- 50% normaL 50% camer
9. The possibility of a female becoming haemophilic is very rare because the mother of such a female has to be at least a carrier and father should be haemophilic.

Question 25.
A colourblind child is born to a normal couple. Work out a cross to show how is it possible. Mention the sex of this child. (CBSE Delhi 2014, 2016)
Answer:
(a) Colourblindness is an X-linked recessive disease

So, the sex of the child is male.

Question 26.
1. How does a chromosomal disorder differ from a Mendelian disorder?
2. Name any two chromosomal aberra¬tion associated disorders.
3. List the characteristics of the disorders mentioned above that help in their diagnosis. (CBSE 2010)
Or
How does gain or loss of chromosome(s) take place in humans? Describe one example each of chromosomal disorder along with the symptoms involving an autosome and a sex chromosome. (CBSE Sample Paper 2019-20)
Answer:
1. Mendelian disorders are mainly determined by alteration or mutation in a single gene. These disorders are transmitted to the offspring on the basis of Mendelian inheritance, e.g. haemophilia, sickle cell anaemia. Chromosomal disorders are caused due to absence or excess or abnormal arrangement of one or more chromosomes. They are caused due to failure of segregation of chromatids during cell division or due to polyploidy. e.g. Down’s syndrome, Klinefelter syndrome.

2. Chromosomal aberration associated disorders.
(a) Down’s syndrome
(b) Klinefelter syndrome.

3. (a) Down’s syndrome. It is caused due to an additional copy of chromosome number 21 (Trisomy).
Symptoms: Short statured body, small rounded head, furrowed tongue, partially open mouth.

(b) Klinefelter syndrome. It is caused due to an additional copy of X-chromosomes (47 chromosome XXY).
Symptoms. Overall masculine development but the development of breast also occurs. These individuals are sterile.

Question 27.
A true-breeding pea plant, homozygous for inflated green pods (FFGG) is crossed with another pea plant with constricted yellow pods (ffgg). What would be the phenotype and genotype  F₁ and F₂ genotype? Give the phenotype ratio of F₂ generation. (CBSE Delhi 2008)
Answer:

Question 28.
A true-breeding pea plant homozygous for axial violet flowers (AAW) crossed with another pea plant with terminal white flowers (aaw).
(i) What would be phenotype and genotype of F₁ and F₂ generations
Answer:

(ii) Give the phenotype ratio of F₂ generations. (CBSE Delhi. 2008)
Answer:

Question 29.
A child suffering from Thalassemia is born to a normal couple. But the mother is being blamed by the family for delivering a sick baby.
(i) What is Thalassemia?
Answer:
Thalassemia: It is an autosomal recessive blood disease that appears in children of two unaffected carriers, heterozygote parents. The defect occurs due to mutation or deletion of the genes controlling the formation of globin chain (commonly a and P) of haemoglobin. Imbalanced synthesis of globin chains of haemoglobin causes anaemia. Thalassemia is of three types a, p, and 8.

(ii) How would you counsel the family not to blame the mother for delivering a child suffering from this disease? Explain.
Answer:
I would explain to the people around that this disease can be caused due to the presence of a defective gene in both the parents or it may be caused due to certain changes with the genetic setup.

(iii) List the values your counselling can propagate in the families. (CBSE Delhi 2013)
Answer:
People had a good understanding and had to realize the situation. They become supportive and made joint efforts to help the patients.

Question 30.
In a dihybrid cross, white-eyed, yellow-bodied female Drosophila was crossed with red-eyed, brown-bodied male Drosophila. The cross produced 1.3 per cent recombinants and 98.7 progeny with parental type combinations in the F2 generation. Analyse the above observation and compare with the Mendelian dihybrid cross. (CBSE Sample Paper 2018-19)
Answer:
Morgan observed that the two genes did not segregate independently of each other and the F2 ratio deviated vary significantly from the 9:3:3:1 ratio.

He attributed this to physical association or linkage of two genes and coined the term linkage and the term recombination to describe the generation of non-parental gene combinations.

Morgan and his group found that even when the genes are grouped on the same chromosome, some genes are very tightly linked (show very low recombination) while others were loosely linked (showed higher recombination). in the Mendelian dihybrid cross, the phenotypes round, yellow; wrinkled, yellow; round, green and wrinkled, green appeared in the ratio 9:3:3:1.

Wrinkled, yellow and round, green is possible because the distance between two genes is more. Therefore, recombination of parental type is possible.

Question 31.
Aneuploidy of chromosomes in human beings results in certain disorders. Draw out the possibilities of the karyotype in common disorders of this kind in human beings and its consequences in individuals. (CBSE Sample Paper 2018-19)
Or
A doctor after conducting certain tests on a pregnant woman advised her to undergo M.T.P., as the foetus she was carrying showed trisomy of 21st chromosome.
(a) State the cause of trisomy of the 21 st chromosome.
Answer:
(a) Down’s syndrome, Turner’s syndrome, Klinefelter’s syndrome are common examples of Aneuploidy of chromosomes in human beings.

• Down’s syndrome results in the gain of the extra copy of chromosome 21- trisomy.
• Turner’s syndrome results due to the loss of an X chromosome in human females- XO monosomy.
• Klinefelter’s syndrome is caused due to the presence of an additional copy of X- chromosome resulting in XXY condition.

(b) Why was the pregnant woman advised to undergo M.T.P. and not to complete the full term of her pregnancy? Explain. (CBSE Delhi 2019 C)
Answer:
Down’s Syndrome: The affected individual is

• short statured with small round head furrowed tongue and partially open mouth
• Palm is broad with characteristic palm crease
• Physical, psychomotor, and mental development is retarded.

Klinefelter’s Syndrome: The affected individual is

• a male with development of breast, i.e. Gynecomastia
• Such individuals are sterile.

Turner’s Syndrome: The affected individual shows the following characters:

• Females are sterile as ovaries are rudimentary
• lack of other secondary sexual characters

Very Important Figures:

XX-XO determination of sex in the cockroach.

Formation of recombinant as well as non-recombinant (parental type) gametes.
Forms of chromosomal mutations.

Answer:

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