A)Exons 2 and 3 must have the same number of nucleotides. B)Exons 2 and 3 must contain an integral number of codons (that is, the number of nucleotides divided by 3 must be an integer). C)Exons 2 and 3 must contain a number of nucleotides that when divided by 3, leaves the same remainder (that is, 0, 1, or 2). D)Exons 2 and 3 must have different numbers of nucleotides. Multiple Choice

Statement C is the only one that is necessarily true for exons 2 and 3. It is also true for exons 7 and 8. While statements A and B could be true, they don’thave to be. Because the protein sequence is the same in segments of the mRNA that correspond to exons 1 and 10, neither choice of alternative exons (2 versus 3, or 7 versus 8) can alter the reading frame. To maintain the normal reading frame—whatever that is—the alternative exons must have a number of nucleotides that when divided by 3 (the number of nucleotides in a codon) give the same remainder. Since the sequence of the a-tropomyosin gene is known, it is possible to check to see the actual state of affairs. Exons 2 and 3 both contain the same number of nucleotides, 126, which is divisible by 3 with no remainder.

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the solute can move "downhill," from regions of higher to lower concentration, relying on the specificity of the protein carrier to pass through the membrane. this process is called passive transport or facilitated diffusion, and does not require energy.