Part A - Independent assortment of three genes
A wild-type tomato plant (Plant 1) is homozygous dominant for three traits: solid leaves (MM), normal height (DD), and smooth skin (PP).
Another tomato plant (Plant 2) is homozygous recessive for the same three traits: mottled leaves (mm), dwarf height (dd), and peach skin (pp).
In a cross between these two plants (MMDDPP x mmddpp), all offspring in the F1 generation are wild type and heterozygous for all three traits (MmDdPp).
Now suppose you perform a testcross on one of the F1 plants (MmDdPp x mmddpp). The F2 generation can include plants with these eight possible phenotypes:
•solid, normal, smooth
•solid, normal, peach
•solid, dwarf, smooth
•solid, dwarf, peach
•mottled, normal, smooth
•mottled, normal, peach
•mottled, dwarf, smooth
•mottled, dwarf, peach
Assuming that the three genes undergo independent assortment, predict the phenotypic ratio of the offspring in the F2 generation.
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You know the gametes formed by the homozygous recessive parent will always be the same, "mdp", so you need to find the possible gametes the other parent can form. You already know the genotype of the parent MmDdPp, so begin with all uppercase letters, then starting from the right, change the last letter every row, the second letter every second row, the third letter every fourth row (if more than three letters, just double the number of rows from the previous letter before the case of the letter changes). So for your heterozygous parent...
MDP
MDp
MdP
Mdp
mDP
mDp
mdP
mdp
Use these in a Punnett square showing just the possible gametes by each paret:
........|.mdp.|
MDP.|
MDp.|
MdP.|
Mdp..|
mDP.|
mDp.|
mdP.|
mdp.|
You'll note that the "letters" for the gametes that the heterozygous parent can produce follows the same pattern as the list of phenotypes that are possible for your cross. Just count how many you have of each phenotype, and use a ratio symbol (":") between the numbers.
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