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:
Social problem-solving is generally considered to apply to four different types of problems: Impersonal problems, for example, shortage of money; Personal problems, for example, emotional or health problems; Interpersonal problems, such as disagreements with other people.
Explanation:
When blood is in the lungs,
where oxygen is plentiful, oxygen easily binds to the first subunit and then quickly fills up the remaining ones. Then, as blood circulates through the body, the oxygen level drops while that of carbon dioxide increases. In this environment, hemoglobin releases its bound oxygen.
To show he presence of a trait you would need to follow it through multiple generations in a pedigree.