Concept check Concept check Concept check Look at

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Concept check

Concept check

Concept check

Concept check

Concept check Look at the following pedigree of the family. Here the trait is

Concept check Look at the following pedigree of the family. Here the trait is expressed only in males. Do you think that the inheritance is Y linked? Explain with proper reason. What could be the alternate possible mode of inheritance? Predict the genotype of individuals A, B, C and D.

In both a & b males are affected. Are both follow same pattern of

In both a & b males are affected. Are both follow same pattern of inheritance? Or different ? Give valid justification

Concept check

Concept check

a) An enzyme X isolated from a eukaryotic cell has 192 amino acid residues

a) An enzyme X isolated from a eukaryotic cell has 192 amino acid residues and is coded by a gene with 1, 440 bp. Explain the relationship between the number of amino acid residues in the enzyme and the number of nucleotide pairs in its gene b) The b-globin gene has three exons of 140, 222 and 252 nucleotides and two introns of 130 and 850 nucleotides, using this information/data calculate the amino acids present in the polypeptide that the mature m. RNA can encode? 6

Concept check A nucleotide analog X with the following chemical structure is present in

Concept check A nucleotide analog X with the following chemical structure is present in abundance in a cell infected by HIV. This analog X blocks DNA chain elongation when it is incorporated into viral DNA synthesized by reverse transcriptase. Why does DNA synthesis stop? 7

Concept check Dr. Garrod’s finding on disease ‘Alkaptonuria’ was very significant due to the

Concept check Dr. Garrod’s finding on disease ‘Alkaptonuria’ was very significant due to the following reason. Choose most appropriate one. a. He showed that the patients are not victim of ‘black magic’ b. He showed that the ‘Alkaptonuria’ is not a major disease, patients can live with it c. He showed that genetic inheritance is connected to biochemical pathways in the body d. He proposed that first cousin marriage is wrong. No religion should encourage it. 8

The prokaryotic cells possess circular DNA as genetic material. Therefore, in the course of

The prokaryotic cells possess circular DNA as genetic material. Therefore, in the course of replication, they will not lose any genetic material as the cell divides. A eukaryotic cell having linear DNA as genetic material has a great probability of losing the information at ends of the DNA strand during replication. How has the eukaryotic cell evolved to overcome the loss of genetic matter in subsequent cycles of replication? 9

Concept check The discovery of reverse transcriptase has impact on life in and out

Concept check The discovery of reverse transcriptase has impact on life in and out of science in a myriad of ways. This enzyme has proved to be catalyzing the formation of (a) polypeptide from an RNA template (b) DNA from a polypeptide template (c) RNA from a polypeptide template (d) RNA from a DNA template (e) DNA from an RNA template

On the basis of the given sequence of DNA and the resulting protein produced

On the basis of the given sequence of DNA and the resulting protein produced in a eukaryotic cell, Determine the sequence of the m. RNA (pre-processing and postprocessing) (Note: shaded region is the promoter for the gene) DNA sequence 5’CTGAGACTGCTCCGCCTCGCCATGACTATAACTGCTATCCTACAGCAGCATCGGATATCGCCAGTCGTCGGCCTAA 3’ 3’GACTCTGACGAGGCGGAGCGGTACTGATATTGACGATAGGATGTCGTCGTAGCCTATAGCGGTCAGCAGCCGGAT T 5’ Protein Methionine. Threonine. Serine. Tyrosine. Arginine. Glutamine. Serine. Alanine- 11

On the basis of the given sequence of DNA and the resulting protein produced

On the basis of the given sequence of DNA and the resulting protein produced in a eukaryotic cell, Determine the sequence of the m. RNA (pre-processing and postprocessing) (Note: shaded region is the promoter for the gene) DNA sequence 5’CTGAGACTGCTCCGCCTCGCCATGACTATAACTGCTATCCTACAGCAGCATCGGATATCGCCAGTCGTCGGCCTAA 3’ 3’GACTCTGACGAGGCGGAGCGGTACTGATATTGACGATAGGATGTCGTCGTAGCCTATAGCGGTCAGCAGCCGGAT T 5’ m. RNA (preprocessed) 5’AUGACUAUAACUGCUAUCCUACAGCAGCAUCGGAUAUCGCCAGUCGUCGGCCUAA 3’ m. RNA (postprocessing) 5’ GMAUG ACU UCC UAU CGC CAG UCG GCC UAA AAA(n)3’ Met Thr Ser Tyr Arg Glu Ser Ala Stop 12