I believe the left chamber has a higher concentration of neon gas because it has more neon atoms in it, and the right chamber has a higher concentration of helium gas because it has more helium atoms in it.
The right answers are:
A-present in eukaryotic genomes ==> Both exons and introns
B-generally absent from bacterial genomes ==> Introns
C-part of the final mRNA strand ==> Exons
D-code for an amino acid sequence ==> Exons
E-removed from initial mRNA strand prior to translation ==> Introns
F-present in the DNA used as the template for transcription ==> Both exons and introns
In the genes of eukaryotic organisms, the exons are the segments of an RNA precursor that are conserved in the RNA after splicing and that are found in mature RNA in the cytoplasm. The segments of the RNA precursor that are removed during splicing are called in opposition to introns. Exons are mainly found in messenger RNAs (mRNAs) encoding proteins. Some mRNAs may sometimes undergo an alternative splicing process in which one or more exons may be excised or some introns preserved in rare cases.
Answer: c) amino acid
Explanation: A codon is an mRNA sequence which contains three nucleotides that codes for a particular amino acid. The codons on the mRNA are read by the ribosome during translation and the amino acid coded for by each codon is used to make a protein. There are 64 different codons in existence, each amino acid is coded for by at least one codon. Some amino acids have more than one codon. For example, the amino acid Leucine is coded for by six codons: UUA, UUG, CUU, CUC, CUA and CUG while the amino acid phenylalanine is coded for by two codons: UUU and UUC.
Answer:
50%
Explanation:
Glucose is a simple sugar with a total of 6 carbon atoms in its structure. Pyruvate has a total of three carbon atoms. Two molecules of pyruvate are obtained per glucose by glycolysis. None of the carbon of glucose is released in the form of CO2 during glycolysis. Therefore, the radio-labeled C-1 of glucose will be the component of the carbon skeleton of one of the total of two pyruvate molecules produced during glycolysis. So, 50% of the pyruvate will exhibit radioactivity.
