3. Restriction Enzymes 1) Consider the sequence of DNA given below and answer the following questions 5’ ATTGAGGATCCGTAATGTGTCCTGATCACGCTCCACG 3’ 3’ TAACTCCTAGGCATTACACAGGACTAGTGCGAGGTGC 5’
Gene Interactions
When the expression of a single trait is influenced by two or more different non-allelic genes, it is termed as genetic interaction. According to Mendel's law of inheritance, each gene functions in its own way and does not depend on the function of another gene, i.e., a single gene controls each of seven characteristics considered, but the complex contribution of many different genes determine many traits of an organism.
Gene Expression
Gene expression is a process by which the instructions present in deoxyribonucleic acid (DNA) are converted into useful molecules such as proteins, and functional messenger ribonucleic (mRNA) molecules in the case of non-protein-coding genes.
Part 3. Restriction Enzymes
1) Consider the sequence of DNA given below and answer the following questions
5’ ATTGAGGATCCGTAATGTGTCCTGATCACGCTCCACG 3’
3’ TAACTCCTAGGCATTACACAGGACTAGTGCGAGGTGC 5’
- a) You cut the sequence of DNA shown above using BamHI (see table 19.1 from the text). How many fragments of DNA would you expect to result from this restriction digest?
- b) If you cut the sequence of DNA shown above using BclI (recognition sequence = 5’ TGATCA 3’, enzyme cuts after the first T) instead of BamHI how many fragments do you expect?
2) For each given sequence/restriction enzyme pair, determine how many pieces of DNA would result form the digest and indicate whether those pieces would have blunt or sticky ends. NOTE: in the given recognition sites, the dash represents where the cut is made.
a) HpaI, recognizes 5’ GTT – AAC 3’
5’ GGATGTTAACAATCTCTACGGGTTAACACCCTTGGGTTAACATCCGCGG 3’
3’ CCTACAATTGTTAGAGATGCCCAATTGTGGGAACCCAATTGTAGGCGCC 5’
Number of fragments of DNA: Type of end:
- b) EcoRI, recognizes 5’ G – AATTC 3’
5’ ACGACGTATTAGAATTCTTATCCGCCGCCGGAATTCTCATCA 3’
3’ TGCTGCATAATCTTAAGAATAGGCGGCGGCCTTAAGAGTAGT 5’
Number of fragments of DNA: Type of end:
- c) HpaI, recognizes 5’ GTT – AAC 3’ AND SspI, recognizes 5’ AAT – ATT 3’
5’ AGTTAACCGACAATATTGTATTATATCCGCCGCCGTCGTTAACATCA 3’
3’ TCAATTGGCTGTTATAACATAATATAGGCGGCGGCAGCAATTGTAGT 5’
Number of fragments of DNA: Type of end:
- d) HindII, recognizes 5’ GTC – GAC 3’, AND HaeIII, recognizes 5’ CC – GG 3’, AND BamHI, recognizes 5’ CCTAG – G 3’
5’ ACGCCGGACGTACCTAGGTTTAGTCGACTCCGCCGCCCCTAGGGTCATCA 3’
3’ TGCGGCCTGCATGGATCCAAATCAGCTGAGGCGGCGGGATCCCAGTAGT 5’
Number of fragments of DNA: Type of end:
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