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V125BC [204]
3 years ago
13

Which amino acid is least likely to produce a hydrogen bond with water or polar molecules?

Chemistry
1 answer:
KonstantinChe [14]3 years ago
3 0
9 amino acids (alanine, cysteine, glycine, isoleucine, leucine, methionine, phenylalanine, proline, valine) have no hydrogen donor or acceptor atoms in their side chains.
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2C2H2 + 5O2 → 4CO2 + 2H2O

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What is the entropy of this collection of training examples with respect to the positive class B. What are the information gains
Alenkinab [10]

The data set is missing in the question. The data set is given in the attachment.

Solution :

a). In the table, there are four positive examples and give number of negative examples.

Therefore,

$P(+) = \frac{4}{9}$   and

$P(-) = \frac{5}{9}$

The entropy of the training examples is given by :

$ -\frac{4}{9}\log_2\left(\frac{4}{9}\right)-\frac{5}{9}\log_2\left(\frac{5}{9}\right)$

= 0.9911

b). For the attribute all the associating increments and the probability are :

  $a_1$   +   -

  T   3    1

  F    1    4

Th entropy for   $a_1$  is given by :

$\frac{4}{9}[ -\frac{3}{4}\log\left(\frac{3}{4}\right)-\frac{1}{4}\log\left(\frac{1}{4}\right)]+\frac{5}{9}[ -\frac{1}{5}\log\left(\frac{1}{5}\right)-\frac{4}{5}\log\left(\frac{4}{5}\right)]$

= 0.7616

Therefore, the information gain for $a_1$  is

  0.9911 - 0.7616 = 0.2294

Similarly for the attribute $a_2$  the associating counts and the probabilities are :

  $a_2$  +   -

  T   2    3

  F   2    2

Th entropy for   $a_2$ is given by :

$\frac{5}{9}[ -\frac{2}{5}\log\left(\frac{2}{5}\right)-\frac{3}{5}\log\left(\frac{3}{5}\right)]+\frac{4}{9}[ -\frac{2}{4}\log\left(\frac{2}{4}\right)-\frac{2}{4}\log\left(\frac{2}{4}\right)]$

= 0.9839

Therefore, the information gain for $a_2$ is

  0.9911 - 0.9839 = 0.0072

$a_3$     Class label      split point       entropy        Info gain

1.0         +                        2.0            0.8484        0.1427

3.0        -                         3.5            0.9885        0.0026

4.0        +                        4.5            0.9183         0.0728

5.0        -

5.0        -                        5.5            0.9839        0.0072

6.0        +                       6.5             0.9728       0.0183

7.0        +

7.0        -                        7.5             0.8889       0.1022

The best split for $a_3$  observed at split point which is equal to 2.

c). From the table mention in part (b) of the information gain, we can say that $a_1$ produces the best split.

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