* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download SMU-DDE-Assignments-Scheme of Evaluation PROGRAM msc
Expression vector wikipedia , lookup
Transformation (genetics) wikipedia , lookup
Community fingerprinting wikipedia , lookup
Molecular cloning wikipedia , lookup
Real-time polymerase chain reaction wikipedia , lookup
Polyadenylation wikipedia , lookup
Endogenous retrovirus wikipedia , lookup
Two-hybrid screening wikipedia , lookup
DNA supercoil wikipedia , lookup
Messenger RNA wikipedia , lookup
Vectors in gene therapy wikipedia , lookup
Non-coding DNA wikipedia , lookup
Amino acid synthesis wikipedia , lookup
RNA polymerase II holoenzyme wikipedia , lookup
Eukaryotic transcription wikipedia , lookup
Epitranscriptome wikipedia , lookup
Gene expression wikipedia , lookup
Promoter (genetics) wikipedia , lookup
Biochemistry wikipedia , lookup
Point mutation wikipedia , lookup
Transcriptional regulation wikipedia , lookup
Deoxyribozyme wikipedia , lookup
Silencer (genetics) wikipedia , lookup
Artificial gene synthesis wikipedia , lookup
Nucleic acid analogue wikipedia , lookup
Biosynthesis wikipedia , lookup
SMU-DDE-Assignments-Scheme of Evaluation PROGRAM SEMESTER SUBJECT CODE & NAME BK ID DRIVE MARKS Q. No 1 A MSC BIOINFORMATICS 2 MBI201 MOLECULAR AND DEVELOPMENTAL BIOLOGY B1943 WINTER 2015 60 Criteria Describe the functioning of lac operon. Functioning of the lac operon: The lac operon is an operon required for the transport and metabolism of lactose in Escherichia coli and other enteric bacteria. It consists of three adjacent structural genes, lac Z, lac Y and lac A. • lacZ encodes β galactosidase (LacZ), lacY encodes β galactoside permease (LacY) and • lacA encodes β galactoside transacetylase (LacA) • There are two ways by which regulation of lac operon takes place. They are (1) Induction and (2) catabolite repression. (1) Induction:’ The lac operon is a negatively controlled inducible operon; the lac Z, lac Y, and lac A genes are expressed only in the presence of lactose. • The lac regulator gene, designated as ‘I ‘gene, encodes a repressor that is 360 amino acids long. However, the active form of lac repressor is a tetramer containing four copies of I gene product. • In the absence of inducer, the repressor binds to the lac operator, which in turn prevents RNA polymerase from catalyzing the transcription of the structural genes. • A few molecules of lac Z, lac Y, and lac A gene products are synthesized in the un induced state, providing a low background level of enzyme activity. • This background level of enzyme activity is essential for induction of the lac operon because the inducer of the operon, allolactose, is derived from lactose in a reaction catalyzed by β galactosidase. • Once formed, allolactose is bound by the repressor causing the release of repressor from the operator. In this way, allolactose induces the transcription of lac Z, lac Y, and lac A structural genes. Total Marks (Unit 5 ; Section 5.3) 10 10 Marks SMU-DDE-Assignments-Scheme of Evaluation (2) Catabolite repression: • When glucose enters an E. coli cell, it is utilized directly without the induction of any new enzymes because, the enzymes required for the glucose catabolism are always expressed in the E. coli cells. • Thus, the cell is primed to use this sugar, a central molecule in carbohydrate metabolism, above all the others. • As E. coli becomes starved for glucose, it begins synthesizing an unusual nucleotide named cyclic adenosine 3′, 5′ monophosphate, generally referred to as cyclic AMP, or cAMP. • An enzyme called adenylyl cyclase is continually active in converting ATP molecules into cAMPs. • When the cells are starved for glucose, the cell produces a pool of cAMP molecules. These cAMP molecules bind to proteins called Catabolite activator proteins (CAP) also called as cAMP receptor proteins or CRPs. • This CAP cAMP complex then binds to a site near the lac operon’s promoter called CAP site. • By themselves, both RNA polymerase and cAMP CAP complex have relatively low affinity for their respective binding sites in lac promoter DNA. Interaction between residues in CAP and α subunit of RNA polymerase forms a protein protein complex that binds much more stably to promoter DNA than either protein alone. • Thus, binding of the CAP cAMP complex to the promoter site is required for the transcription of lac operon. • This is an example of positive control, in which the binding of a protein(CAP) activates transcription. • When glucose is present in the environment and transported into the cell, the function of adenylyl cyclase is inhibited. • The result is that ATP is no longer converted into cAMP, the cAMP pool drops. • This decrease in the complex inactivates the promoter, and the lac operon is turned off. SMU-DDE-Assignments-Scheme of Evaluation • When cAMP is low, because of the lower level of glucose, cAMP CAP does not bind to the CAP site. • As a result, RNA polymerase does not bind efficiently to the lac promoter and only a little lac mRNA is synthesized. • When lactose is present and the glucose is absent, maximal transcription of the lac operon occurs. In this situation, the lac repressor does not bind to the lac operator. • The concentration of cAMP increases, and the cAMP CAP complex formed binds at the CAP site, stimulating binding of RNA polymerase strongly. This process leads to transcription of lac structural genes. • However, in the absence of lactose, no lac mRNA is formed because repressor bound to the lac operator prevents transcription. 2 A Discuss the steps involved in the processing of pre m-RNA. (Unit 3, Section 3.5, Pg. No. 81) 2.5X4 10 Following are the steps involved in the processing of pre-mRNA. 1. The addition of the 5’ cap: Almost all eukaryotic pre-mRNAs are modified at their 5’ends by the addition of a structure called a 5’cap. This capping consists of the addition of an extra nucleotide at the 5’end of the mRNA and methylation by the addition of a methyl group (CH3) to the base in the newly added nucleotide/s and to the 2’ –OH group of the sugar of one or more nucleotides at the 5’ end. Capping takes place rapidly after the initiation of transcription, the 5’ cap functions in the initiation of translation. 2. The addition of the poly (A) tail: Most mature eukaryotic mRNAs have from 50 to 250 adenine nucleotides at the 3’ end (a poly (A) tail). These nucleotides are not encoded in the DNA, but are added after transcription) in a process termed Polyadenylation. Many eukaryotic genes transcribed by RNA polymerase II are transcribed well beyond the end of the coding sequence; the extra material at the 3’ end is then cleaved and the poly (A) tail is added. 3. RNA splicing: The other major type of modification that takes place in eukaryotic pre-mRNA is the removal of introns by RNA splicing. SMU-DDE-Assignments-Scheme of Evaluation 3 A This occurs in the nucleus following transcription, but before the RNA moves to the cytoplasm. Consensus sequences and the spliceosome splicing require the presence of three sequences in the intron. A cut is made at the 3’ splice site and, simultaneously, the 3’end of exon 1 becomes covalently attached (spliced) to the 5’ end of exon 2. 4. RNA editing: A long-standing principle of molecular genetics is that genetic information ultimately resides in the nucleotide sequence of DNA, except in RNA viruses. This information is transcribed into mRNA, and mRNA is then translated into a protein. The assumption that all information about the amino acid sequence of a protein resides in DNA is violated by a process called RNA editing. In RNA editing, the coding sequence of an mRNA molecule is altered after transcription, and so the protein has an amino acid sequence that differs from that encoded by the gene. Explain any five features of genetic code. Explanation for any five of the following features of genetic code: 1. The genetic code is non-overlapping: Each nucleotide is a part of only one triplet codon. Successive codons are read in order. The translation starts at initiation codon and continues till the termination codon in a single translational frame. The nucleotide sequence is read without any punctuation because, if the reading frame is displaced by a single base, it remains shifted throughout the sequence and so the message will code for a different amino acid. 2. The codes are highly degenerate: The amino acids methionine and tryptophan have single codons, whereas, three amino acids (Leucine, Serine, Arginine) have six codons, five amino acids have four, isoleucine has three, and nine amino acids have two. Codons representing the same or related amino acids tend to be similar in sequence. Often the codons that differ in the last base will code for a single amino acid. The reduced specificity of the third base in a codon is called as third base degeneracy. Codons that specify the same amino acid are termed ‘synonyms’. 5X2 10 SMU-DDE-Assignments-Scheme of Evaluation 3. The arrangement of code table is nonrandom: Most synonyms differ only in their third nucleotide. XYU (where X and Y are two different nucleotides) and XYC always specify the same amino acid; XYA and XYG do so in all except in two cases (i.e., (case 1) UGA codes for stop codon, UGG codes for tryptophan and (case 2) AUA codes for Isoleucine whereas AUG codes for methionine). These observations indicate that the code evolved so as to minimize the deleterious effects of sudden change in nucleotide bases (i.e., mutation). 4. UAG, UAA and UGA are stop codons: These three codons (are also called nonsense codons) do not specify any amino acids, but give signal to ribosome (protein synthetic machinery) for polypeptide chain elongation. The UAG, UAA and UGA are often referred to as ‘amber’, ochre, and ‘opal’ codons respectively. 5. AUG and GUG are chain initiation codons: The codons AUG and less frequently, GUG specify the starting point for polypeptide chain synthesis. However, they also specify amino acids methionine and valine respectively at internal positions in a polypeptide chain. 6. The code is unambiguous: Each triplet codon specifies a particular amino acid and only one amino acid. 7. The standard genetic code is not universal: 4 A DNA studies in 1981 revealed that the genetic codes of certain mitochondria are variants of standard genetic codes i.e., in mammalian mitochondria, AUA, as well as standard AUG, are a methionine (initiation codon); UGA specifies tryptophan rather than stop; AGA and AGG are ‘Stop’ rather than Arginine. Thus, the standard genetic code, although very widely utilized, is not universal. Describe the phases of fertilization. Add a note on the structure of ovum. (Unit 10, Section 10.4, Page No. 268) 6 10 Explanation for the following phases of fertilization: a) Passage of sperm through corona radiata b) Penetration of zona pellucida c) Fusion of plasma membranes of the oocyte and the sperm. SMU-DDE-Assignments-Scheme of Evaluation d) Completion of second meiotic division of oocyte and formation of female pronucleus. e) Formation of male pronucleus f) Fusion of pronuclei 5 A 4 Structure of ovum: The ovum that is shed from the ovary is not fully mature. It is still a secondary oocyte, which is undergoing division to shed off the second polar body, and is surrounded by the zona pellucida. Some cells of corona radiata remain adherent to the zona pellucida. The ovum has no nucleus as it is undergoing second meiotic division, and instead has a spindle. The first polar body lies in the perivitelline space between the cell membrane and the zona pellucida. The ovum measures about 0.1 mm in size. Describe the physical properties of DNA. (Unit 1, Section 1.3, Pg. No. 16) Description for the following five physical properties of DNA: 5X2 10 1. Antiparallel orientation: The DNA strands have got directionality. The orientation of the strands is anti-parallel, with one being 5’ to 3’ and the other 3’ to 5’. The direction of the naming starts from 5’ through 3’ region. Both the chains have the same geometry but the base pairing holds them together with opposite polarity. 2. Grooves: Twin helical strands form the DNA backbone. The strand backbones are closer together on one side of the helix than on the other. The major groove occurs where the backbones are far apart, the minor groove occurs where they are close together. 3. Sense and antisense: A DNA strand is called "sense" if its sequence is the same as that of a messenger RNA copy that is translated into protein. The sequence on the opposite strand is called the "antisense" sequence. 4. Supercoiling: DNA can be twisted like a rope in a process called supercoiling With DNA in its "relaxed" state, a strand usually circles the axis of the double helix once every 10.4 base pairs, but if the DNA is twisted, the strands become more tightly or more loosely wound. SMU-DDE-Assignments-Scheme of Evaluation If the DNA is twisted in the direction of the helix, this is positive supercoiling, and the bases are held more tightly together. If they are twisted in the opposite direction, this is negative supercoiling, and the bases come apart more easily. 5. Quadruplex structures: In humans, the telomeres are the lengths of single stranded sequences, enriched with GT bases. In order to stabilize the ends, the telomeres form structures of stacked sets of four-base units, called “G-quadruplex”. In addition to these quadruplexes, the DNA also form loops called “Telomere loops” or simply “T-loops”, where in the single-stranded DNA curls around, forming a long circle. The T-loops are stabilized by telomere binding proteins. 6 A Define mutation. Discuss the three types of DNA repair mechanisms. (Unit 8, Section 8.2, Page No. 195) 1 10 Definition: Sudden heritable changes in the genetic material are called mutations. 9 Explanation for the following three types of DNA repair mechanisms: (1) Mismatch repair: Occurs immediately after DNA synthesis, uses the parental strand as a template to correct an incorrect nucleotide incorporated into the newly synthesized strand. (2) Excision repair: It entails removal of a damaged region by specialized nuclease systems and then DNA synthesis to fill the gap. (3) SOS response: A bacterial deoxyribonucleic acid (DNA) repair system in which cell division and DNA replication are blocked, and DNA repair, recombination, and mutation genes are induced. *A-Answer Note –Please provide keywords, short answer, specific terms, specific examples (wherever necessary) **********