Answer:
The genotype of the F1 was wy+/w+y.
Explanation:
One of the given options has a typo: the red eye-brown body offspring count should be 56 instead of 561.
<u>We have two genes with two alleles each:</u>
Red eyes (w+) is dominant over white eyes (w).
Brown body (y+) is dominant over yellow body (y).
The phenotypes of the F2 tesulting from a test cross (F1 x wy/wy) are:
- wy+/ey (white-eye, brown body): 670
- w+y/wy (red-eye, yellow body): 650
- wy/wy (white-eye, yellow body): 38
- w+y+/wy (red-eye, brown body 56
If the genes w and y are linked, two phenotypes in the F2 will be much more abundant than the other two. Recombination during meiosis is a rare event, so the most abundant phenotypes are the parentals (the ones present in the F1 parent).
Every individual in the offpsring has a <em>wy</em> chromosome, as this was the gamete inherited from the test cross individual.
In this case, the most abundant gametes are wy+ and w+y, so the genotype of the F1 was wy+/w+y.
Notice how when recombination occurs in the F1 parent, the recombinant gametes appear: wy and w+y+, which are the less abundant in the F2 progeny.
Answer: C
Explanation: Because the larger animals might starve if they don’t any smaller fishes to feed on unless they also leave the area or change there diet.
Answer:
The direction of an electric current is by convention the direction in which a positive charge would move. Thus, the current in the external circuit is directed away from the positive terminal and toward the negative terminal of the battery. Electrons would actually move through the wires in the opposite direction.
