This page contains the AQA AS Biology B8 Dna Genes And Synthesis Questions and kerboodle answers for revision and understanding.This page also contains the link to the notes and video for the revision of this topic.
8.1 Genes And Tripple Code AQA AS B8 DNA Genes And Synthesis Kerboodle Answers
Ans 1 A gene is the basic physical and functional unit of heredity. Genes, which are made up of DNA and located on chromosome, act as instructions to make molecules called proteins.
Ans 2 Six consecutive amino acids will be coded by 3X6 = 18 bases.
Ans 3 A change in one base may lead to a change in the amino acid coded for. If this amino acid happens to be part of the active site of the enzyme, then the active site may no longer exist and so the enzyme won’t work. Another way the enzyme might be rendered non-functional is if the amino acid change is somewhere in the enzyme that causes a change in the general shape of the enzyme; enzymes generally have a very specific shape necessary to their function, and if the shape changes then the enzyme might not work.
4.aAns total 5 amino acids including the start code (methionine amino acid) will be coded by this section of bases.
b.Ans ATG codes for methionine.
C,Ans Different codons can produce the same amino acid due to the way they bind to transfer RNA (tRNA). Different organisms use different codons. So in theory, as you say, arginine can be coded for by 6 different codons. However, in practice, not all of those are used in a single organism – it depends on the tRNA they express. Codon usage databases exist online which will allow you to find the known codon usage from any given organism.
Interpreting the genetic code
Ans 1 To amino acids that have only one codon are Methionine (START Codon; ATG/AUG)
And Tryptophan (Trp; TGG/ UGG).
Ans 2 a. CUC= Leucine (CTC)
- AAA= Lysine
- GAU= Aspartic acid (GAT)
Ans a. Metheonine; Arginine; Stop Codon; Glycine; Serine
- Alanine; Lysine; Phenylalanine; Proline; Aspartic acid.
8.2 DNA and chromosomes AQA AS B8 DNA Genes And Synthesis Kerboodle Answers
Ans 1 There is two types of cells: prokaryotic cells and eukaryotic cells. These types of cells show difference at their DNA level also.
| Prokaryotic Cell | Eukaryotic Cell | |
| Size of DNA molecule | Shorter | Longer |
| Form of DNA | Circular | Linear |
| Associated with proteins | Not associated | Associated with histone proteins |
| Type of DNA | Chromosome is not there | Chromosomes are there. |
| Organelle DNA | Not present | Present in mitochondria and chloroplast. Organelle DNA is like prokaryotic DNA, short, circular and not associated protein. |
Ans 2 Protein found on chromosome is known as histone protein. In biology, histones are highly alkaline proteins found in eukaryotic cell nuclei that package and order the DNA into structural units called nucleosomes. They are the chief protein components of chromatin, acting as spools around which DNA winds, and playing a role in gene regulation. Without histones, the unwound DNA in chromosomes would be very long (a length to width ratio of more than 10 million to 1 in human DNA).
3.Ans Chromosome packaging is facilitated by certain proteins known as histones.Footnote Histones have a positive charge because they have a high proportion of amino acids with basic side chains. The positively charged histones associate with DNA, which has a negative charge because of its phosphate groups, to form structures called nucleosomes. The fundamental unit of each nucleosome consists of a beadlike structure with 146 base pairs of DNA wrapped around a disc-shaped core of eight histone molecules (two each of four different histone types). Although the nucleosome was originally defined as a bead plus a DNA segment that links it to an adjacent bead, today the term more commonly refers only to the bead itself (i.e., the eight histones and the DNA wrapped around them).
4Ans a. Total 46 chromosomes are there that means 2.3 m/ 230 cm/ 2300 mm length of DNA will be equally divided into all 46 chromosomes. i.e. 230/ 46= 5 cm. Mean length of DNA in each chromosome will 5 cm or 50 mm.
- 2.3 m/ 2300 mm will the length of DNA present is a human brain cell because every cell has same length of DNA.
8.3 Structures of ribonucleic acid AQA AS B8 DNA Genes And Synthesis Kerboodle Answers
Ans 1 Messenger RNA (mRNA) functions as a carrier for genetic information from the DNA to target recipients, like ribosome for protein synthesis and production. Transfer RNA (tRNA) attaches and transports amino acids into growing chains to form proteins.
Structural difference between mRNA and tRNA
Structure of tRNA
Transfer RNA brings out the amino acids to a growing polypeptide chain at the ribosomal site of protein synthesis during translation. All tRNA’s have a similiar sequences of 73-93 nucleotides. The image of tRNA is in a compact “L” shape. Transfer RNA has three structures; A primary, secondary and a tertiary. Each one contains a specific anticodon triplet sequence that can be a base pair to one or more codons for an amino acid.
Structure of mRNA
Messenger RNA allows the genetic coding in DNA. mRNA is always singly stranded. It contains bases Adenine, Guanine, Cytosine and Uracil. There is random coiling in messenger RNA and there is no specific base pairing. In fact Base Pairing actually destroys its biological activity.
Figure: Structure of mRNA (1) and tRNA (2)
Ans 2 DNA and RNA differ in their sugar content, their nucleobase content and their three-dimensional structure.
| DNA | RNA | |
| Sugar content | Deoxy-ribose sugar | Ribose sugar |
| Nucleobase content | Adenine; Thymine; Guanine and cytosine | Adenine; Uracil; guanine and cytosine |
| Three dimensional structure | Double helix linear | Single stranded linear or clover leaf shape. |
Ans 3 A codon is a three-base sequence (three nitrogen bases in a row) on mRNA. It calls for a specific amino acid to be brought to the growing polypeptide.
An anticodon is a three-base sequence on tRNA. It matches the codon. That’s how the right amino acid is put onto the polypeptide next. The tRNA must fit its anticodon onto the mRNA. Each tRNA can only bring one kind of amino acid.
Comparison of DNA, messenger RNA and transfer RNA
Ans 1 a. Gametes are formed after meiosis thus the DNA content becomes half of the parent cell. That means parents are diploid and the gametes are haploid and the DNA content is restored after the fusion of gametes.
Ans 1 b. DNA content is different in parent and gametes. By this way DNA content is maintained from one generation to another generation.
Ans 2 a The genetic material must be extremely stable so that sequence information can be passed on from generation to generation without degradation. DNA is more stable than RNA thus it is a better genetic information resource.
- Ans 2 b mRNA is broken down easily.
- It’s part of regulation of gene expression. A cell can make a protein when the protein is needed, and then it can stop making that protein when the protein is not needed.
It’s part of the defence against positive strand RNA viruses. If mRNA is being broken down relatively quickly, then there’s a better chance for a +strand RNA virus to get chopped up before it does damage.
8.4 Polypeptide synthesis- transcription and splicing AQA AS B8 DNA Genes And Synthesis Kerboodle Answers
Ans 1 RNA polymerase is an enzyme that is responsible for copying a DNA sequence into an RNA sequence, duyring the process of transcription. As complex molecule composed of protein subunits, RNA polymerase controls the process of transcription, during which the information stored in a molecule of DNA is copied into a new molecule of messenger RNA.
Ans 2 In the starting of the transcription an enzyme is needed to open the DNA strands. This enzyme is helicase enzyme. It unwind the DNA and makes it compatible for transcription and RNA polymerase binding.
Ans 3. mRNA sequence is read in batches of 3 nucleotides. pre-mRNA occur in eukaryotic cells only. It consist of coding regiosn(exons) and non-coding regions (introns). Introns are removed during splicing and remaining exons are joined together. This will correct reading frame for the protein synthesis. Consider the following sentence:
THEHGECATYYUATEHPORATTHANDTLRAN
By reading the sentence as a whole it does not make any sense. However if we read the triplets after removing the junk parts or “introns” then it reads:
THE CAT ATE RAT AND RAN
So what we have done is remove the necessary parts so that the sequence is the correct one for the mRNA and thus producing the right polypeptide. If these introns were left in, then the final product would contain extra unnecessary nucleotides therefore giving the wrong message.
Ans 4 a. Sequence of pre-mRNA is UAGCUUCAGGUC (Complementary to the template DNA strand)
- It will code for 4 amino acids.
Ans 5 mRNA is obtained after splicing. In a gene there are both coding and non coding sequences there. The part which expresses in gene is exon thus transcribed mRNA is of less base pair length than the its actual length.
8.5 Polypeptide synthesis-translation AQA AS B8 DNA Genes And Synthesis Kerboodle Answers
Ans 1 Ribosome is the cell organelle which is involved in translation.
Ans 2 a. tRNA anticodon for AUC is UAG.
- DNA base sequence is TAG for mRNA codon AUC.
Ans 3 A transfer RNA (abbreviated tRNA and formerly referred to as sRNA, for soluble RNA) is an adaptor molecule composed of RNA, typically 76 to 90 nucleotides in length, that serves as the physical link between the mRNA and the amino acid sequence of proteins. tRNA does this by carrying an amino acid to the protein synthetic machinery of a cell (ribosome) as directed by a three-nucleotide sequence (codon) in a messenger RNA (mRNA). As such, tRNAs are a necessary component of translation, the biological synthesis of new proteins in accordance with the genetic code.
Ans 4 mRNA is of 64 codons but it produced only 63 amino acids. It is because of the stop codon at the end of the mRNA. Stop codon indicates the end of polypeptide synthesis.
Stop codons do not code for any amino acids so there is 1 less amino acid than there are codons.
Protein Synthesis:
Figure 5. It shows the information of a part of a polypeptide along a section of eight codons. Codon 4 and 5 have been left blank. Using figure 5 and table 1 answer the following questions:
Ans 1 X= Ribosome; Y= mRNA.
Ans 2 At the N-terminal of polypeptide chain amino group (NH2) is there.
Ans 3 The anticodon sequence on tRNA molecule 4 will be AUG.
| DNA codon | mRNA codon | Amino acid |
| CAA | GUU | Glutamine |
| TGG | ACC | Tryptophane |
| GCT | CGA | Alanine |
| GAC | CUG | Aspartic acid |
| AGT | UCA | Serine |
Ans 5
| DNA codon | mRNA codon |
| CAA | GUU |
| TGG | ACC |
| GCT | CGA |
Ans 6 CAG is converted into UAG. CAG was coding for Glutamine but after mutation it become UAG that is a stop codon. As a result of this mutation the polypeptide chain will get terminate at codon 8 position.
Ans 7 a. Inversion of code CAG will be GAC. CAG was coding for Glutamine whereas GAC will code for aspartic acid. Aspartic acid is a substitute of Glutamine. Both are polar in nature. Both are of negative charge. Thus the nature of polypeptide will not get changed.
- If the polypeptide formed from this mutant gene forms part of an enzyme than it could be possible that enzyme may not function. But there are less chances of it because glutamine can be replaced by aspartic acid due to similar nature and structure. Enzyme substrate reactions are very specific thus it might be possible the enzyme will not function.
Cracking the code:
Ans DNase is added to cell extract to avoid DNA contamination and to obtain RNA only.
- In one experiment the radioactive amino acid was phenylalanine and four mixtures, differing only in their mRNA, were set up as follows:
- mRNA made up of a chain of nucleotides containing only the base adenine=poly A
- mRNA made up of a chain of nucleotides containing only the base Uracil=poly U
- mRNA made up of a chain of nucleotides containing only the base cytosine=poly c
- no mRNA was present.
The results are shown in Table 2.
Ans 2 The codon UUU – because the very radioactive polypeptide (39800 counts min) was only produced from the mixture containing poly U. this polypeptide must be made up of phenylalanine because this is the only radioactive amino acid present. As the synthetic mRNA contains only the bases sequence UUUUUUU, etc., one codon for phenylalanine must be UUU.
Ans 3 As a control experiment to show that the radioactivity was due to the labelled phenylalanine rather than some other factor such as background radiation.
By this method Nirenberg deciphered 47 of the 64 possible codons in the genetic code. The remaining 17 codons gave ambiguous results. Khorana devised another technique. Khorana made very long mRNA molecules with repeating sequence of nucleotide bases such as GUGUGUGUGUGUGUG. The polypeptide will have two amino acid cysteine and valine in alternation.
Ans 4. It is not possible to say what the codon is for each amino acid because there are 20 amino acids which are coded by 64 codons. One amino acid will be coded by one or more codon.
From Nirenberg’s earlier experiments, it was known that UGU was a codon for cysteine. This meant that GUG was a codon for valine. By seeing the results of similar experiments using specific sequences of mRNA he Dr. Khorana was able to decipher the complete genetic code and shown that the code was degenerate. Nirenberg’s experiments confirmed the results.
Ans 5. The genetic code can be described as degenrate (more than one codon may specify a particular amino acid) but not ambiguous; no codon specifies more than one amino acid.
Ans 6. It is not easy to find amino acids coded for by certain codons because genetic codes are redundant, and degenerate.
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Practice questions:
Ans (i) 9 amino acids this sequence of DNA bases can code for.
Ans The number of amino acids coded for are fewer than the 9. It is because of the fact that 6th codon is stop codon thus termination of polypeptide will take place.
Ans b. DNA base sequence guided the amino acids sequence on the polypeptide chain. Change in DNA base sequence will directly affect the amino acid sequence. Amino acid sequence in protein will determine its structure and function. Any abnormality in amino acid sequence due to DNA base sequence change will result in abnormal, non-functional protein.
Ans
| Number of bases | ||||
| C | G | A | T | |
| Strand A | 26 | 19 | 20 | 9 |
| Strand B | 19 | 26 | 9 | 20 |
Ans (i) R= Deoxyribonucleic sugar
(ii) Q= Phosphate
Ans b. Hydrogen bond
Ans Minimum number of DNA bases needed to code for ribonuclease will be {START CODON}{GENE}{STOP CODON} i.e. 3+127*3+3=387.
Ans The mRNA will be coded from complementary strand. The sequence of complementary strand is CAAATGATGAGAAGAAGAAAT. Thus the amino acids are as following:
| Gln | Met | Met | Arg | Arg | Arg | Asn |
. Ans e. DNA base sequence guided the amino acids sequence on the polypeptide chain. Change in DNA base sequence will directly affect the amino acid sequence. Amino acid sequence in protein that could be an enzyme, will determine its structure and function. Any abnormality in amino acid sequence due to DNA base sequence change will result in abnormal, non-functional enzyme.
Ans 3 (a) introns are the non coding region of gene.
Ans (b) Isoleucine; glycine; Valnine; Serine
Ans (i) Mutation 1 will not cuase any affects of the amino acid sequence because after mutation the code is becoming GGA instead or GGC. Both are coding for Glycine only.
Ans (ii) In this mutation the amino acid Valnine changes into leucine. Thus is will affect the enzyme functioning ans structure.
i)Ans Gene mutation occurs during S phase of cell cycle.
(ii)Ans Gene mutation has to do with the duplication of the DNA, and if something wrong goes with this duplication, and then the new DNA will not be exactly like the original one. This duplication happens in the S phase.
(A) (i) Ans two substances are ribonucleic sugar and phosphate backbone.
(ii) Ans TAGGCA
(b) (i) Ans mRNA are linear whereas tRNA is of clover leaf shape.
(ii) Ans pre-mRNA is a newly created mRNA that contains introns (non-coding) and exons (coding regiosn) of gene. Whereas mRNA contains only the information needed for protein synthesis (exons)
(c) (i) Ans 100-(38+20+24)= 18% Uracil for the middle part of chromosome.
For the END part 100-(31+22+26)= 21% Uracil.
(ii) Ans Well, different parts of the chromosome contain different genes. What makes a gene code for a specific protein is the sequence of bases. Thus the sequence of bases of genes in the middle will be different to those at the end.
Q 5 (a) Ans Translation
(b) Ans Molecule Q is tRAN.
(c) Ans Complementary DNA base sequence: TAC (ATG)
Anticodon sequence: UAC
(d) Ans
| Aspartic acid | D | Asp | Behaves similarly to glutamic acid. Carries a hydrophilic acidic group with strong negative charge. Usually is located on the outer surface of the protein, making it water-soluble. Binds to positively-charged molecules and ions, often used in enzymes to fix the metal ion. When located inside of the protein, aspartate and glutamate are usually paired with arginine and lysine. |
| Proline | P | Pro | Contains an unusual ring to the N-end amine group, which forces the CO-NH amide sequence into a fixed conformation. Can disrupt protein folding structures like α helix or β sheet, forcing the desired kink in the protein chain. Common in collagen, where it often undergoes a posttranslational modification to hydroxyproline. |
Aspartic acid Proline
(e)
Ans After deletion the Base sequence will be AUGCCUACCGACU. AUG (methionine); CCU (proline). Before mutation the codon was CCG coding for proline. But due to deletion the amino acid sequence after that is changed. Thus the function and structure of protein get changed.
6 (a) Ans Mean number of base pair per chromosome in sweet rush is 21.7 Mbp.
(b)Ans Approximately 21000 DNA fragments need to be made to sequence the DNA of monkey flower plant.
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