Genetics-ⅱ Miscellaneous
- The nucleotide analogue used in DNA sequencing by chain termination termination method is
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DNA sequencing is the determination of the precise sequence of nucleotides nucleotides in a sample of DNA. The most popular method for doing this is called the dideoxy method or Sanger method. DNA is synthesized from four deoxynucleotide triphosphates. The dideoxy method gets its name from the critical role played by synthetic nucleotides that lack the -OH at the 3′ carbon atom. A dideoxynucleotide can be added to the growing DNA strand but when it is, chain elongation stops because there is no 3′ -OH for the next nucleotide to be attached to. For this reason, the dideoxy method is also called the chain termination method. The DNA sample is divided into four separate sequencing reactions, containing the four standard deoxynucleotides (dATP, dGTP, dCTP and dTTP) and the DNA polymerase. To each reaction is added only one of the four dideoxynucleotides (ddATP, ddGTP, ddCTP, or ddTTP). These dideoxynucleotides are the chain-terminating nucleotides, lacking a 3′ -OH group required for the formation of a phosphodiester bond between two nucleotides during DNA strand elongation. Incorporation of a dideoxynucleotide into the nascent (elongating) DNA strand therefore terminates DNA strand extension, resulting in various DNA fragments of varying length. The dideoxynucleotides are added at lower concentration than the standard deoxynucleotides to allow strand elongation sufficient for sequence
analysis.Correct Option: B
DNA sequencing is the determination of the precise sequence of nucleotides nucleotides in a sample of DNA. The most popular method for doing this is called the dideoxy method or Sanger method. DNA is synthesized from four deoxynucleotide triphosphates. The dideoxy method gets its name from the critical role played by synthetic nucleotides that lack the -OH at the 3′ carbon atom. A dideoxynucleotide can be added to the growing DNA strand but when it is, chain elongation stops because there is no 3′ -OH for the next nucleotide to be attached to. For this reason, the dideoxy method is also called the chain termination method. The DNA sample is divided into four separate sequencing reactions, containing the four standard deoxynucleotides (dATP, dGTP, dCTP and dTTP) and the DNA polymerase. To each reaction is added only one of the four dideoxynucleotides (ddATP, ddGTP, ddCTP, or ddTTP). These dideoxynucleotides are the chain-terminating nucleotides, lacking a 3′ -OH group required for the formation of a phosphodiester bond between two nucleotides during DNA strand elongation. Incorporation of a dideoxynucleotide into the nascent (elongating) DNA strand therefore terminates DNA strand extension, resulting in various DNA fragments of varying length. The dideoxynucleotides are added at lower concentration than the standard deoxynucleotides to allow strand elongation sufficient for sequence
analysis.
- The total number of fragments generated by the complete and sequential cleavage of the polypeptide given below by Trypsin followed by CNBr is _______
Phe-Trp-Met-Gly-Ala-Lys-Leu-Pro-Met-Asp-Gly-Arg-Cys-Ala-Gln
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Given sequence: Phe-Trp-Met-Gly-Ala-Lys-Leu-Pro-Met-Asp-Gly-Arg-Cys-Ala-Gln
First it is digested with Trypsin and than with Cyanogen Bromide in a process of sequential cleavage. Cyanogen bromide hydrolyzes peptide bonds at the C-terminus of methionine residues. Trypsin cleaves peptide chains mainly at the carboxyl side of the amino acids lysine or arginine, except when either is followed by proline.
After trypsin digestion, fragments generated are:
Phe-Trp-Met-Gly-Ala-Lys; Cys-Ala-Gln; Leu-Pro-Met-Asp-Gly-Arg
After CNBr digestion, the fragments generated are:
Phe-Trp-Met; Gly-Ala-Lys; Cys-Ala-Gln; Leu-Pro-Met; Asp-Gly-ArgCorrect Option: C
Given sequence: Phe-Trp-Met-Gly-Ala-Lys-Leu-Pro-Met-Asp-Gly-Arg-Cys-Ala-Gln
First it is digested with Trypsin and than with Cyanogen Bromide in a process of sequential cleavage. Cyanogen bromide hydrolyzes peptide bonds at the C-terminus of methionine residues. Trypsin cleaves peptide chains mainly at the carboxyl side of the amino acids lysine or arginine, except when either is followed by proline.
After trypsin digestion, fragments generated are:
Phe-Trp-Met-Gly-Ala-Lys; Cys-Ala-Gln; Leu-Pro-Met-Asp-Gly-Arg
After CNBr digestion, the fragments generated are:
Phe-Trp-Met; Gly-Ala-Lys; Cys-Ala-Gln; Leu-Pro-Met; Asp-Gly-Arg
- Match the entries in Group Iwith the entries in Group II.
Group I Group II P. RNAse P 1. Polyadenylation Q. RNase H 2. Splicing R. snRNAs 3. Ribozymes S. CstF 4. DNA-RNAhybrids
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Ribonuclease P (RNase P) is the endoribonuclease that generates the mature 5′-ends of tRNA by removal of the 5′-leader elements of precursor-tRNAs. Although it carries out a biochemically simple reaction, RNase P is a complex ribonucleoprotein particle composed of a single large RNA and at least one protein component. RNase H (Ribonuclease H) is an endoribonuclease that specifically hydrolyzes the phosphodiester bonds of RNA which is hybridized to DNA. The snRNAs, along with their associated proteins, form ribonucleoprotein complexes (snRNPs), which bind to specific sequences on the pre-mRNA substrate. This intricate process results in two sequential transesterification reactions. These reactions will produce a free lariat intron and ligate two exons to form a mature mRNA. This results in the splicing. CstF is recruited by cleavage and polyadenylation specificity factor (CPSF) and assembles into a protein complex on the 3’ end to promote the synthesis of a functional polyadenine tail, which results in a mature mRNA molecule ready to be exported from the cell nucleus to the cytosol for translation.
Correct Option: A
Ribonuclease P (RNase P) is the endoribonuclease that generates the mature 5′-ends of tRNA by removal of the 5′-leader elements of precursor-tRNAs. Although it carries out a biochemically simple reaction, RNase P is a complex ribonucleoprotein particle composed of a single large RNA and at least one protein component. RNase H (Ribonuclease H) is an endoribonuclease that specifically hydrolyzes the phosphodiester bonds of RNA which is hybridized to DNA. The snRNAs, along with their associated proteins, form ribonucleoprotein complexes (snRNPs), which bind to specific sequences on the pre-mRNA substrate. This intricate process results in two sequential transesterification reactions. These reactions will produce a free lariat intron and ligate two exons to form a mature mRNA. This results in the splicing. CstF is recruited by cleavage and polyadenylation specificity factor (CPSF) and assembles into a protein complex on the 3’ end to promote the synthesis of a functional polyadenine tail, which results in a mature mRNA molecule ready to be exported from the cell nucleus to the cytosol for translation.
- The RNA primer synthesized during the replication process in bacteria is removed by
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DNA Polymerase I is a template-dependent DNA polymerase which catalyzes 5′→3′ synthesis of DNA. DNA Polymerase I catalyzes the template-directed polymerization of nucleotides into duplex DNA in a 5′→3′ direction. In addition to polymerase activity, this DNA polymerase exhibits 3′ to 5′ and 5′ to 3′ exonuclease activity. It is able to utilize nicked circular duplex DNA as a template and can unwind the parental DNA strand from its template. Because of its exonuclease activity in the 3′ to 5′ direction, the polymerase can remove or excise the RNA primers synthesized during replication.
Correct Option: C
DNA Polymerase I is a template-dependent DNA polymerase which catalyzes 5′→3′ synthesis of DNA. DNA Polymerase I catalyzes the template-directed polymerization of nucleotides into duplex DNA in a 5′→3′ direction. In addition to polymerase activity, this DNA polymerase exhibits 3′ to 5′ and 5′ to 3′ exonuclease activity. It is able to utilize nicked circular duplex DNA as a template and can unwind the parental DNA strand from its template. Because of its exonuclease activity in the 3′ to 5′ direction, the polymerase can remove or excise the RNA primers synthesized during replication.
- In a genetic study, 80 people were found to have alleles for polydactyly. Only 36 of them were polydactylous. What is the extent of penetrance percentage? _______
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Given, No. of polydactylous persons = 36
No. of people having the allele for polydactyly = 80
Therefore, the extent of penetrance in the population = 36/80 x 100 % = 45 %Correct Option: C
Given, No. of polydactylous persons = 36
No. of people having the allele for polydactyly = 80
Therefore, the extent of penetrance in the population = 36/80 x 100 % = 45 %