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FIgure 10-15 Illustration of a 1000-base-pair DNA with two segments of 150 contiguously unwound bases each
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such segments If we keep the left-most unwound segment at the left end of the molecule, the other segment (of 150 unwound base pairs) can be shifted one base pair at a time along the molecule For this 150-base-pair segment, the situation is identical to the case illustrated in Fig 10-14 There are 701 ways to arrange this 150base-pair segment Four of these 701 ways are shown as a, b, c, and d in Fig 10-15 Now consider moving the left-most unwound segment just one base pair to the right This gives us a single helical base pair at the left end of the molecule (see drawings e and f in Fig 10-15) For this arrangement of the left-most segment, there are now 700 ways to arrange the other unwound segment If we then move the left-most segment again by one base pair, there are now only 699 ways to arrange the other (the right-most) unwound segment We can continue this logic to find that the total number of ways to arrange two 150-base-pair unwound segments along a 1000-base-pair molecule is 701 1 700 1 699 1 1 1, or 246,051 This is an amazing result! By simply dividing the 300-base-pair segment into two, we ve increased the number of ways to arrange the unwound base pairs from 701 to 246,051 That s a factor of over 350 times! This tells us that based only on the number of ways to arrange melted base pairs, it is 350 times more likely that the 300 based pairs will melt in two segments of 150 base pairs each than it is that they will melt as a single segment of 300 base pairs Now consider three segments of 100 base pairs each If we keep the leftmost 100-base-pair segment fixed at the left end of the molecule, the logic for the next two segments (the middle and the right-most segment) is the same as
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for the previous case of two segments, which gave us 246,051 ways to arrange just those two segments Now if we move the left-most segment just one base pair to the right and repeat the process for the other two 100-base-pair segments, we get 700 1 699 1 1 1, or 245,350 ways to arrange the two rightmost segments Move the left-most segment again, and we get 699 1 698 1 1 1, or 244,650 If we continue this logic, we find that there are 57,657,951 ways to arrange three contiguous unwound segments of 100 base pairs each In going from just one segment to three segments, we have increased the number of ways to arrange the unwound bases from less than a thousand to more than 57 million! You can just imagine what increasing the number of segments to 4 or 8 or 20 would do And we have not even yet considered having different numbers of base pairs in each of the segments For example, take the case of two segments We considered 150 base pairs in each segment What about the case of 149 base pairs in one segment, and 151 base pairs in the other And so on This means that the case of two segments does not really give us 246,051 ways to arrange the segments, but 299 times 246,051 ways, or 73,569,249 ways Similarly the case of three segments was undercounted If we include various lengths for the three segments (eg, instead of 100, 100, 100, consider 99, 100, 101, and 98, 100, 102, and 98, 99, 103, etc) the result is more than 25 3 1012 ways
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