Consider two genes, A and B$,$ with alleles A$,$ a and B$,$ b respectively. These genes are incompletely linked on the same chromosome. Consider an individual heterozygous for both these genes such that the alleles $$A$$ and $$B$$ are on one homolog and the alleles $$a$$ and $$b$$ are on the other homolog. Draw two individual tetrads that would form during meiosis in such an individual. Assume crossing over in exactly one of the two tetrads involving one pair of non$-$sister chromatids between the two genes. The remaining two non$-$sister chromatids do not undergo crossing over. $($Assume that the above individual is a human male, so every daughter cell at the end of meiosis forms a male gamete$).$ What is the number of gametes of each type that would be produced following meiosis involving both of these tetrads? $($AB$=$ noncross over gamete, ab$=$ noncrossover gamete, Ab$=$ crossover gamete, aB$=$ crossover gamete$)$

Group of answer choices

$3$ AB$,3$ ab$,1$ Ab$,1$ aB

$1$ AB$,1$ ab$,3$ Ab$,3$ aB

$1$ AB$,3$ ab$,1$ Ab$,3$ aB

$1$AB$,3$ ab$,3$ Ab$,1$ aB

$3$ AB$,1$ ab$,3$ Ab$,1$ aB