Which of the following represent the reactants of the net equation of the urea cycle?
A. NH4+, urea, aspartate, and HCO3-
B. NH4+, ATP, fumarate, and HCO3-
C. NH4+, AMP, aspartate, and urea
D. NH4+, ATP, aspartate, and HCO3-
Nirenberg and colleagues determined a portion of the genetic code using filter binding assays that included ribosomes, trinucleotides and tRNA molecules charged with radioactively-labeled amino acids. Choose the answer below that correctly answers the following three questions in order:
1) If the trinucleotide sequence is CAG, which amino acid will be bound to the filter paper?
2) If the wobble position of the trinucleotide sequence is changed to a U, which amino acid will be bound to the filter paper?
3) True or False: Two distinct tRNA molecules with different anticodon sequences MUST be used to recognize the codons AGA and AGG.
1) Gln, 2) none, 3) True
2) Gln, 2) Tyr, 3) True
1) Gln, 2) Tyr, 3) False
1) Asp, 2) His, 3) True
1) Asp, 2) none, 3) False
1) Gln, 2) none, 3) False
1) Gln, 2) His, 3) True
1) Asp, 2) none, 3) True
1) Val, 2) His, 3) True
1) Val, 2) Tyr, 3) False
1) Val, 2) His, 3) False
1) Val, 2) Tyr, 3) False
1) Gln, 2) His, 3) False
1) Asp, 2) His, 3) False
Why is the S value for the eukaryotic ribosome (80S) smaller than the sum of the S values for the small (40S) and large (60S) ribosomal subunits? Choose the ONE best answer.
S values correspond to Swedish units, which are a measure of mass in an electric field, but ribosomes have no charge, so their mass needs to be reduced by 20% (100-20=80).
It is not known why they do not add up, probably because sedimentation is not an exact science and evolution is not perfect, so the ribosomes are not as large as they should be (100S versus 80S).
S units refer to shape AND sedimentation, so shape and sedimentation need to both be taken into account and then multiply by 0.80 to get the actual value for the ribosome (80S).
None of these answers is correct.
S values correspond to Svedberg units, which are a measure of sedimentation in a centrifuge, however S values are not additive because shape affects sedimentation rates.
Ribosomes form a sombero shape made up of two smaller subunits, the brim of the sombero and the crown of the sombero, which together have a smaller mass than a top hat.
Eukaryotes are generally larger than prokaryotes, so it makes sense that the eukaryotic ribosome is 80S and bacterial ribosome is 70S.
S values correspond to Svedberg units, which are a measure of sedimentation in a centrifuge, S values are additive as long as you subtract 20% to account for density (100-20=80).
a) Method 1: It takes 40 minutes for bi-directional replication, so it would take ~80 minutes for unidirectional.
Method 2: If using the genome size and rate of polymerization, it would be 4.6 x 106 divided by 103 nucleotides/sec, which is 4.6 x 103 seconds or 76 minutes.
b) The genome size is 3.3 x 109 and the replication rate is 50 nucleotides/sec, so with one origin, it would take 6.6 x 107 seconds, but there are 104 origins, so would take only 6.6 x 103, now divide by 60 seconds in a minute and total time with assumptions is 110 minutes for haploid and 220 minutes for diploid, which is 3 hours and 40 minutes. Since there ARE chromosomes and telomeres, it actually takes ~8 hours.
Assume that you have identified all the cis-acting sites in a prokaryotic genome for a particular
A. 2, 5, 7 B. 3, 5, 7 C. 4, 6, 7 D. 2, 3, 4 E. 1, 3, 5
trans-acting transcription factor. What is the function of this transcription factor when binding to these sites?