Question:
<em>The diagram shows the potential energy changes for a reaction pathway.</em>
Part 1: Does the diagram illustrate an endothermic or an exothermic reaction? Give reasons in support of your answer.
Part 2: Describe how you can determine the total change in enthalpy and activation energy from the diagram and if each is positive or negative.
Answer:
Part 1: The diagram illustrates an endothermic reaction as the products has a higher potential energy than the reactants do. There is a positive slope of the diagram and there is enough energy to meet the activation energy requirement.
Part 2: You can determine the total change in enthalpy and activation energy from the diagram by the potential energy of the reactants. If the reactants have a high potential energy, then the enthalpy is also high, and if the reactants have a low potential energy, then the enthalpy is low. You can determine if the diagram is positive or negative by knowing if its an endothermic or exothermic reaction. An endothermic reaction is positive because the products are higher than the reactants and a exothermic reaction is negative because the reactants are higher than the products.
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.