For linked genes A, B, and C, the map distance A–B is 5 map units and the map distance B–C is 25 map units. If there are 10 doub
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Answer:
Check the explanation
Explanation:
Consider the below table of F2 data of phenotypic flies: we can reconstruct the table as seen in the second attached image below:
From the above observations it is clear the Gray body long wings and Ebony body vestigial wings are the parental combinations and that the Gray body vestigial wings and Ebony body long wings are the recombinants.
Also the ratio indicates that Gray body is dominant over ebony body and that long wings is dominant over vestigial wings.
here the genes for the two characters have shown independent assortment which means that the genes are unlinked if located on the same chromosome or are located on different chromosomes.
Now F1 hybrid= GgLl (G for Grey and L for Long)
Cross between F1 hybrid and true breeding Gray vestigial (GGll)
GgLl x GG ll
Gametes-----------> GL Gl gL gl Gl
GL Gl gL gl
Gl GGLl GGll GgLl Ggll
(Gray long) (Gray vestigial) (gray Long) (Gray vestigial)
Therefore the probability of getting the flies with gray body vestigial wings= 2/4= 50%
b) The reason why the students obtained the above given F2 results involving a cross between true breeding Gray body and long wings with true breeding Ebony body and vestigial wings is that the genes for the two charcters asssort independently in F2 generation and that the genes are not linked as:
Parents------------------> GGLL x ggll
Gametes -----------------> GL gl
F1---------------------> GgLl (Gray long but in heterozous condition)
Now GgLl x GgLl
Gametes GL Gl gL gl GL Gl gL gl
Here gametes assorted independently and hence in F2 generation we got the above results (U can show the results in the form of punnett square.
