Answer:
linkage with approximately 33 map units between the two gene loci
Explanation:
If two genes are not linked, number of recombinants and parental offspring will be equal. Here it is clearly visible that recombinants are less than parental offspring hence the genes are linked. Given, the offspring are in following numbers:
AaBb = 106 = Parental
aabb = 94 = Parental
Aabb = 48 = Recombinant
aaBb = 52 = Recombinant
Recombination frequency = (Number of recombinants/ Total progeny) * 100 = (100/300) * 100 = 33.33 %
1% recombination frequency= 1 map unit of distance between the two gene loci. So here the distance between the two gene loci is approximately 33 map units.
Hence, these results are consistent with linkage with approximately 33 map units between the two gene loci
Answer:
at right angles or perpendicular to the direction they are moving
Each one is correct except d
B) Peyer’s patches. Located in the mucosa and extend to the submucosa of the small intestine
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