Question

The mitochondrial DNA sequence that is shared by two species has a steady mutation rate. Scientists determine the sequences of these two species to be as follows:

Species A: CAGGCCATTATG
Species B: CCAGCCTATAGG

This DNA sequence has a known mutation rate, which the scientists used to calculate that the species diverged from a common ancestor 80 million years ago. Using this information, how much more time do you predict will pass before these species differ by a total of eight base pairs?

160 million years
32 million years
48 million years
240 million years

Answer 1

I keep coming up with 128.

Because 80 million years / 5 differences = 16.

16 * 8 = 128

Answer 2

@ganeshie8

Answer 3

@Jhannybean

Answer 4

Lol

Answer 5

Oh this is like permutation kind of problem. No?

Answer 6

I guess so.

Answer 7

base pairs are like G-C, A-T, etc.

And clearly they're not transcribed into U or anything.

Answer 8

So I would look for the base pairs first, and perhaps because they repeat at every interval or so.... (only a guess), that to predict 80 years you need to find where the pattern starts repeating once again. Again, this is just an opinion, lol.

Answer 9

Um, okay...?

Answer 10

So how do you find the Base Pairs?

Answer 11

So not all the letters in the sequence are base pairs?

Answer 12

Any ideas?

Answer 13

Oh I kind of get it.

Answer 14

So, I was on the right track sort of.

First you need to count the number of base changes that happen in the sequence.

Answer 15

Let me point them out, one sec.

Answer 16

Yes, that's 5.

Answer 17

\(Species~A \! \!: C\color{red}A\color{red}GGCCA\color{red}TTA\color{red}TG\)
\(Species~B \!\!: C\color{red}C\color{red}AGCCT\color{red}ATA\color{red}GG\)

Answer 18

Oh..that's four..lol.

Answer 19

No, wait that is 5..xd

Answer 20

OK yeah I was trying to type that out in \(LaTeX\) but I couldnt.

Answer 21

\(Species~A \! \!: C\color{red}A\color{red}GGCC\color{red}A\color{red}TTA\color{red}TG\)
\(Species~B \!\!: C\color{red}C\color{red}AGCC\color{red}T\color{red}ATA\color{red}GG\)

Answer 22

Im so bad at it <.< but anyways..

Answer 23

Lol

Answer 24

All 5 differences aren't base pairs, are they?

Answer 25

So you already have 5 changes, and you can predict the initial (or current) change which is \(\frac{5}{80} = \frac{ 1\ \ mut}{16 \ \ years}\)

Answer 26

Yes, I got that far.

Answer 27

1 mutation happens every 16 years, and you need 3 more.

Answer 28

What do you do?

Answer 29

\(\dfrac{5}{80} = \dfrac{ 1\ \ mut}{16 \ \ years}\)

I did 5 differences / 80 million years = 1 difference 16 million years

So 8 different base pairs * 16 million years = 128 million years.

But 128 million years isn't an option.

Answer 30

no but see we're already have done 5 different base pairs

8-5 = 3 pairs left.

3 * 16 million years = 48 million years.

Answer 31

That's how much time is left for all 8 to occur within the time frame of 80 years, (our approx)

Answer 32

Oh..

Answer 33

Thanks a bunch.

Answer 34

No problem :)

Answer 35

oh yeah this totally interests me, remind me to never click any link you send me on skype again xD

Answer 36

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