Question

How many 5-element DNA sequences

a) end with A?

b) start with T and end with G?

c) contain only A and T?

d) do not contain C?

Hint: Recall that a DNA sequence is a sequence of letters, each of which is one of A, C, G, or T. Thus by the product

rule there are 4^5 = 1024 DNA sequences of length five if we impose no restrictions.

discrete math

discrete structures

Answer 1

DEFINITIONS

Product rule If one event can occur in m ways AND a second event can occur in n ways, the number of ways the two events can occur in sequence is then m • n.

Sum rule If an event can occur either in m ways OR in n ways (non-overlapping), the number of ways the event can occur is then m + n.

Subtraction rule If an event can occur either in m ways OR in n ways (overlapping), the number of ways the event can occur is then m + n decreased by the number of ways that the event can occur commonly to the two different ways.

SOLUTION

There are 4 possible DNA bases: A, C, G and T

(a)

First base: 4 ways

Second base: 4 ways

Third base: 4 ways

Fourth base: 4 ways

Fifth base: 1 way (since it has to be A)

We need to use the product rule, because the first event is picking the first base, the second event is picking the second base, ...., the 5th event is picking the 5th base.

4.4.4.4.1 = 256

(b)

First base: 1 way (since it has to be T)

Second base: 4 ways

Third base: 4 ways

Fourth base: 4 ways

Fifth base: 1 way (since it has to be G)

We need to use the product rule, because the first event is picking the first base, the second event is picking the second base, ...., the 5th event is picking the 5th base.

1.4.4.4.1 = 64  

(c)

First base: 2 ways (since it has to be A or T)

Second base: 2 ways (since it has to be A or T)

Third base: 2 ways (since it has to be A or T)

Fourth base: 2 ways (since it has to be A or T)

Fifth base: 2 ways (since it has to be A or T)

We need to use the product rule, because the first event is picking the first base, the second event is picking the second base, ...., the 5th event is picking the 5th base.

2.2.2.2.2 = 32

(d)

First bane: 3 ways (since it cannot be C)

Second base: 3 ways (since it cannot be C)

Third base: 3 ways (since it cannot be C)

Fouth base: 3 ways (since it cannot be C)

Fifth base: 3 ways (since it cannot be C)

We need to use the product rule, because the first event is picking the first bane, the second event is picking the second base, ...., the 5th event is picking the 5th base.

3.3.3.3.3 = 243

RESULT

(a) 256 DNA sequences

(d) 64 DNA sequences

(c) 32 DNA sequences

(d) 243 DNA sequences