The phenotype of crooked wings (cw) in Drosophila melanogaster is caused by a recessive mutant gene that independently assorts with a recessive gene for hairy (h) body. Assume that a cross is made between a fly with normal wings and a hairy body and a fly with crooked wings and normal body hair. All F1 flies from this cross were wild-type, and these flies were crossed among each other to produce 1024 F2 offspring. Which phenotypes would you expect among the offspring in the F2 generation, and how many of each phenotype would you expect

Answer and Explanation:

Available data:

As the whole F1 progeny were wild type, we can assume that their parents were homozygous for their respective traits. This is:

First cross:

Parental)  CWCW hh     x     cwcw HH

Gametes) CWh  CWh  CWh  CWh  

                 cwH  cwH   cwH  cwH

F1) 100%  CWcw Hh

Second cross:

Parental) CWcw Hh    x     CWcw Hh

Gametes) CWH  CWh  cwH  cwh

                CWH  CWh  cwH  cwh

Punnet square)     CWH             CWh             cwH             cwh

              CWH   CWCWHH    CWCWHh    CWcwHH      CWcwHh

              CWh    CWCWHh    CWCWhh    CWcwHh       CWcwhh

              cwH    CWcwHH     CWcwHh     cwcwHH       cwcwHh

              cwh     CWcwHh     CWcwhh     cwcwHh        cwcwhh

F2)  1024 Individuals

To calculate how many individuals of each phenotype would be in the F2, we can perform a three simple rule:

16 100% of the offspring 1024 individuals

9 56.25% of the offspring 576 individuals

3 18.75% of the offspring 192 individuals

3 18.75% of the offspring 192 individuals

1 6.25%  of the offspring 64 individuals.

Of the total 1024 individuals,

the match duh

explanation:

idk im sorry.