There is this question which was asked in my exam.

A diploid cell contain 6 chromosomes. How many possible random arrangement of homologous chromosomes could occur during metaphase-I? (consider that allele for only one character present on chromosome and they are in heterozygous condition.)

Options are :





Answer given is : ii)8

But my answer is : i)4

My solution :

Aa | Aa | Aa | Aa

Bb | Bb | bB | bB

Cc | cC | Cc | cC

Here, 1 column represents 1 possiblity.

These are the possible combinations according to me in metaphase-I.

Am i missing something?

Please tell me where am i wrong.

PS : i would have loaded the picture of my solution but it ain't getting loaded. Sorry for any inconvenience.

Thank you. =)


2 Answers 2


The correct answer is 8 as given by the formula 2n , where n is the number of chromosome in haploid cell.


During prophase 1, synapsis take place as a result of which bivalents are formed. Now each bivalent can arrange itself in two possible manner ( known as independent orientation). Since in this question there are 6 chromosomes which means 3 bivalents, the possible number of arrangement is 2 × 2 × 2 = 8

The photo here explain independent orientation in a cell having two chromosome pair. You can work this out for 3 chromosomal pair as well.

Your doubt:

As far as I can understand your doubt is that arrangement one is actually same as arrangement third ( with reference to the photo). Well, it might seem like this but if you look carefully the two non homologous chromosomes are arranged in different manner in them .. relative to each other and the homologous partner. And hence the whole arrangement is considered as different arrangements and not the same.

Thus, for 3 chromosomal pair, there are 8 possible arrangement of homologous chromosomes which could occur during metaphase-I.

Independent orientation during metaphase 1


The answer would be 8. The best way to explain this would be by the formula for number of possible arrangements i.e. 2^n.

As mentioned in the question, the cell is diploid with 6 chromosomes, which means that 2n = 6 implies n = 3.

So number of possible arrangements = 2^3 = 2*2*2 = 8.

If the two homologs are A and B, mom's being A1, B1, C1 and dad's being A2, B2, C2 -

The eight possible combinations are A1B1C1, A1B1C2, A1B2C1, A1B2C2, A2B1C1, A2B2C1, A2B1C2, A2B2C2.

Hope this helps.

  • $\begingroup$ Yes, i get that there are 8 arrangements possible in total. But what I'm getting confused with is it says - 'Metaphase-I '. Now, in metaphase -I : the homologous chromosomes are still together and haven't separated yet. So, if my one pair is let's say A1B1C1 on one side then automatically on the other side of the metaphasic plate, A2B2C2 will be present. Same method for other combinations. So, overall only 4 combinations are possible( Am i thinking wrong?). But, i guess, everyone's doing the final calculation bcz i have asked 4 teachers and they all are applying the same method. $\endgroup$ Apr 16, 2018 at 2:11
  • $\begingroup$ Please point out any mistakes in my thinking. Thanks! $\endgroup$ Apr 16, 2018 at 2:12

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