Respuesta :
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.
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Mendel was the first to explain the transmission of phenotypic characters and the independent assortment of the genes.
In the given data, F2 progeny of phenotypic traits are shown.
In the above observations, the gray body long wings and ebony body vestigial wing are parental combinations.
Also, the gray body and vestigial wings, and ebony body long wings are the recombinants.
The given ratio in the progeny indicates that gray bodies and long wings are expressed.
The genes for the two traits are independently assorted, which means that genes are unlinked present on the same chromosome.
Now,
For the F1 progeny:
- Gg Ll - G for Grey and L for Long
- The cross between F1 progeny and true gray vestigial will be:
- GGll x GgLl
- Gametes: GL Gl gL gl Gl
The cross between Gl and hybrid will result in a 50% chance of flies having the gray body and vestigial wings.
b) In the above given F2 progeny, the cross between true gray body and long wings with true ebony body and vestigial wings, will result in the independent assortment of the genes.
Given:
- Parental gametes - GGLL x ggll
- Gametes produced - GL and gl
For the F1 progeny, all the offspring will have genotype GgLl (Gray body and long wings but in heterozygous condition)
The above cross can be shown in the Punnett square, which is given in the attachment below.
To know more about Mendelian ratio, refer to the following link:
https://brainly.com/question/13567784
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