Pentahybrid cross

1 ANSWERS

1. amomipais82
   

   Hi There,
   I am not sure what exactly your assignment is, so I will approach it in two ways:
   
   If your assignment was to find the chances of producing a child with a specific genotype (let's say RRiiYYPpAa), then all you would have to do is look at the individual probabilities for each trait (thanks to the law of independent assortment). So, the chances of RR (from Rr X Rr) is 1/4, the chances of ii (from Ii X Ii) is also 1/4, the chances of YY (from Yy X Yy) is also 1/4, the chances of Pp (from Pp X Pp is 1/2), and the chances of Aa (from Aa X Aa) is also 1/2. Therefore, the probability of having an offspring with the above genotype is:
   1/4 X 1/4 X 1/4 X 1/2 X 1/2 = 1/256
   You can apply this method to any assigned genotype.
   
   If your assignment was simply to show the cross with a Punnet square, then it is no different than a dihybrid cross (just this time with a lot more work). Apply the same process, by writing out all of the possible gametes from each parent, set it up in a table, and write out each combination. Of course this method could be used for the first case in which you were assigned a genotype, but it would take much more time.
   
   Hope this helps, and hopefully you do not have to draw the whole Punnet square to earn those 100 pts!
   
   --in case you need a demonstration of the general method for Punnet square drawing, you can look at this site:
   http://web.science.oregonstate.edu/bio/c

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