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
<span>If factors are only
rounded up, then the estimate is an overestimate. If factors are only rounded down, then the
estimate is an underestimate. When some factors are rounded up and some are rounded
down, it is harder to tell whether the estimate is an overestimate or an underestimate.</span>
Answer:
B
Explanation:
The finches needed different beak adaptions MOST to survive so those were most prominent. Hope this helped and have a blessed day. :)))
<span>not sure but i guess is sugar molecule</span>
