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Ravi's program joins the gametes with a + b but forgets to sort them.
Contributed by StemPy LabTry it on paper first. A hint, the key idea and the worked solution are here when you need them.
Ravi's program joins the gametes with a + b but forgets to sort them. For Tt with Tt it builds ['TT', 'Tt', 'tT', 'tt']. How many different keys does Counter show without the sort, and how many with it?
Need a reminder? The key ideaShow
A gamete is one letter of a parent: list("Tt") gives ['T', 't']. A child is "".join(sorted(a + b)), so the dominant capital letter comes first. Two loops over the parents' letters fill the Punnett square, and Counter counts the genotypes or the phenotypes. Tt with Tt gives 1 TT : 2 Tt : 1 tt, which is 3 tall to 1 short.
The worked solution
Have you tried it?
Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Step 1. Without the sort. TT, Tt, tT and tt are four different strings, so Counter shows keys, one each.
Step 2. With the sort. sorted("tT") gives ['T', 't'], so tT becomes Tt. The keys are TT, Tt and tt: keys.
Step 3. Why it matters. Only the sorted count shows the 1:2:1 ratio of genotypes.
Answer: keys without sorting, with it.
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