1. a) 1/64 b) 3/64 c) 64/3
3. a). 126 are synonymous mutations…look at the genetic code (Table 10.3)…for each codon that corresponds to a particular Amino Acid…if that last nucleotide can be changed to form the same amino acid that is one syn mutation
ex:
CUU-Ser-The last nucleotide (U) can be changed to C, A, or G and still be Serine so this codon has 3 synonomous mutations
CUC-Ser-The last nucleotide (C) can be changed to U,A, or G and still be serine so this codon has 3 synonomous mutations
CUA-Ser-etc
CUG-Ser-etc
So for this set of Serine codons…there are a total of 12 synonomous mutations…do this for all codons in the table and you should get 126.
b. 126/183
5. Ex: 5'-CAA AAA AAA AAA AAC-3'
so CAA and AAC correspond to Gln and Asp respectively.
I think the AAC codon actually corresponds to Asn. Anyone else?
AAC does correspond to Asn, pretty sure anyways.
13. a) Trp-Tyr-Val
Trp=(UGG), Tyr=UAU, UAC, Val=GUU, GUC, GUA, GUG
so to figure out the # of different combinations, look at how many different codons correspond to one amino acid and times them all together….so 1x2x4=8 different combinations
and the generic combination is UGG UAY GUN (like an AK with a banana clip(not the 1980's hairstyle clip))
b) Pro-Arg-Lys
Pro=CCU, CCC, CCA, CCG, Arg=CGU, CGC, CGA, CGG, Lys=AAA, AAG
so 4x4x2=32 possible combinations
You seemed to miss two combinations for Arg = AGA AGG which changes the possible combinations… 4x6x2=48 possible combinations
Generic combination is CCN CGN AAR
10.20 1-5x10^-4 = 0.9995 so……0.9995^300 = .861 = 86.1%
10.21 7 interons (one for every loop)…each loop represents ssDNA that did not anneal to corresponding mRNA. the interons are spliced out of mRNA(not NMR hoa) so there are gaps where mRNA is attempted to match of with corresponding DNA strand.
22.
DNA-5'-AGA CTT AGC GCT AAA CGT GGT-3'
3'-TCT GAA TCG CGA TTT GCA CCA-5'
RNA-5'-AGA CUU AGC GCU AAA CGU GGU-3'
AA seq-Arg-Leu-Ser-Ala-Lys-Arg-Gly
Inversion: make sure to maintain correct 5'-3' and vis versa order
so..
**If any wants to know why you need to invert the chains its to maintain polarity
DNA-5'-AGA ACG TTT AGC GCT AAG GGT-3'
3'-TCT TGC AAA TCG CGA TTC CCA-5'
RNA 5'-AGA ACG UUU AGC GCU AAG GGU-3'
AA seq-Arg-Thr-Phe-Ser-Ala-Lys-Gly
25. a. Val-Pro-Arg-Ser-Ser-His
b. Val-Ser-Cys-Leu-Ser-Tyr
c. 2x2x2x2…so 16 polypeptide sequences
28. Both strands can be transcribed (tsc)
DNA(top) 5'-CTAGGTGACCTAGCTTAA-3'
RNA(tsc) 3'-GAUCCACUGGAUCGAAUU-5'-no start codon, 3 stop codons (underlined)
DNA(bot) 3'-GATCCACTGGATCGAATT-5'
RNA(tsc) 5'-CUAGGUGACCUAGCUUAA-3'-no start codon, 4 stop codons (underlined)
for each mRNA strand, there are a possible 3 reading frames-to find reading frame, look for either the start codon or stop codon (always go from 5'-3' in reading of stop codon on mRNA)
-The mRNA transcribed from top part of DNA has 2 reading frames b/c stop codon is part of 2/3 reading frames
-The mRNA transcribed from bottom DNA has 3 reading frames b/c stop codon is part of 3/3 reading
frames
Possible reading frame…5'-CUAGGUGACCUAGCUUAA-3' note for this reading frame, the very first C is the start of it. and the underlined stop codon is obviously the end of it.
Challenge Prob.
DNA-5'-CCGTATATATTATGATCAATATGCATGCTCTCGGGGGTCACACT-3'
3'-GGCATATATAATACTAGTTATACGTACGAGAGCCCCCAGTGTGA-5'
a. 5'-GAUCAAUAUGCAUGCUCUCGGGGG-3'
b. Asp-Gln-Tyr-Ala-Cys-Ser-Arg-Gly
