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We had in class following question which I had no idea how to get to the correct answer:

The carrion crow and the hooded crow are fertile together, but their reproductive success is reduced by 50%. In a certain region exist two populations of both species of roughly the same size. Thus, mixed couples occur in about 10% of all cases. What is the percentage of hybrids in the F1-generation?

(Translated from German.)

I thought it would be something like 5/95 = 5,3% but apparently the answer is 1%. Why?

Tyto alba
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adjan
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    why do you think it would be 5/95? and how, according to you, can the answer be 1%? please show some research effort before asking question here. – another 'Homo sapien' Feb 06 '17 at 12:02
  • 1% is the answer according to the solution. 5/95 because 90% of the couples have 100% success, and 10% have 50% sucess. @another'Homosapien' – adjan Feb 06 '17 at 12:11
  • Homework questions are off-topic here unless you show some research effort. Tell us what research effort you have put on the question, lest this question is likely to be closed. – another 'Homo sapien' Feb 06 '17 at 12:17
  • Well I told you what I thought was the correct approach. I did not find any hints in our material. I did further research before posting here. I didnt find any source dealing with mathematical discussions of this topic. Further, to me my solution seems correct. @another'Homosapien' – adjan Feb 06 '17 at 12:34

1 Answers1

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In order to make your reasoning clearer, you should use more formal notations and explain your thinking step by step. Here's a proposition.

Let's use the following notations :

  • $C$: the total number of couples (mixed and not mixed)
  • $r$: the reproductive success
  • $F1_h$: the number of hybrids in the F1 generation
  • $F1_{nh}$: the number of non hybrids in the F1 generation

You are looking for the percentage of hydrids in the F1 generation, which is: $x = \frac{F1_h}{F1_h + F1_{nh}}$

You know that $10~\%$ of the couples are mixed couples and that their reproductive success is reduced by $50~\%$. This can be written: $\begin{cases} 0.9\cdot C \cdot r = F1_{nh} \\ 0.1\cdot C\cdot \frac{r}{2} = F1_h \end{cases}$

Thus: $\displaystyle x = \frac{F1_h}{F1_h + F1_{nh}} = \frac{0.05\cdot C\cdot r}{C\cdot r\cdot 0.95} = \frac{0.05}{0.95}$

This is indeed the result you suggested. So the correction you were given might not be correct. Or maybe there was some more information in your homework that you ignored...

Flo
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