Explanation:
- Dull fruit (D) is dominant over glossy (d) fruit.
- Orange friut (R) is dominant over cream fruit (r).
- Bitter cotyledon (B) is dominant over non-bitter cotyledons (b).
The three genes are independent.
<h3>a)</h3>
- A plant homozygous for dull, orange fruit and bitter cotyledons has the genotype <em>DDRRBB</em><em>.</em>
- A plant homozygous for glossy, cream fruit and non-bitter cotyledons has the genotype <em>ddrrbb</em>.
100% of the F1 will have dull, orange fruit and bitter cotyledons and will be heterozygous for all genes: <em>DdRrBb</em>.
<u>The F2 will have 8 possible phenotypes:</u>
- 27 D_R_B dull, orange fruit and bitter cotyledons
- 9 D_R_bb dull, orange fruit and non-bitter cotyledons
- 9 D_rrB_ dull, cream fruit and bitter cotyledons
- 3 D_rrbb dull, cream fruit and non-bitter cotyledons
- 9 ddR_B_ glossy, orange fruit and bitter cotyledons
- 3 ddR_bb glossy, orange fruit and non-bitter cotyledons
- 3 ddrrB_ glossy, cream fruit and bitter cotyledons
- 1 ddrrbb glossy, cream fruit and non-bitter cotyledons
<h3>b)</h3>
An F1 plant is crossed with a plant that has glossy, cream fruit and non-bitter cotyledons.
<h3>
DdRrBb x ddrrbb </h3>
The F1 individual can produce 8 types of gametes: DRB, DRb, DrB, Drb, dRB, dRb, drB and drb. The cream fruit and non-bitter cotyledons individual can only produce drb gametes. This cross also produces a progeny with the following genotypes and phenotypic proportions:
- 8 DdRrBb dull, orange fruit and bitter cotyledons
- 8 DdRrbb dull, orange fruit and non-bitter cotyledons
- 8 DdrrBb dull, cream fruit and bitter cotyledons
- 8 Ddrrbb dull, cream fruit and non-bitter cotyledons
- 8 ddRrBb glossy, orange fruit and bitter cotyledons
- 8 ddRrbb glossy, orange fruit and non-bitter cotyledons
- 8 ddrrBb glossy, cream fruit and bitter cotyledons
- 8 ddrrbb glossy, cream fruit and non-bitter cotyledons
Answer:
B
Explanation:
It would not be D because D is the location of a common ancestor, AKA an animal that does not haver jaws or bones. It would not be C because the question states that the fish has jaws, and C does not have jaws since the arrow does not point to that. It would not be A because the question states the fish has "NO TRUE BONES", meaning that there are no bones present. And in A, there is the bones arrow pointing to it, meaning that A has bones since that trait is on the same horizontal line as A. Only leaving us with option B left, since it came from an ancestor with jaws but not from an ancestor with bones. Sorry, I'm not the best with explaining but hopefully that makes sense why it is B, it's like a family tree, you just have to follow it carefully :) I hope that helps
Answer:
B) energy is captured in the bonds linking amino acids that compose lipase.
Explanation:
Anabolic reactions are chemically defined as synthesis reactions in opposition to Catabolic reactions, in which a degradation of chemical compounds occurs. Thus, in anabolic transformations simpler reagents are used to produce more complex substances. In the particular case of the synthesis of a lipase, a number of amino acids are chemically bonded to form a polymerized structure, clearly chemically more complex, known as polypeptide.
An increase in chemical complexity always requires an investment of energy, considering that building a larger molecule implies bonding more reagents or precursor into the final product. Thus, the increase in the energy content of the product (in this case the lipase) when comparing to each precursor (individuals amino acids) will be reflected in the linking (covalent) bonds between amino acids. Of course, there is always a certain amount of energy that is lost as heat during any chemical transformation (not all the energy invested in the synthesis will be captured in the product).
The overall of all anabolic reaction which take place inside the cell is called anabolism, while the whole set degradation chemical reaction occurring within the cell is known as catabolism. Being said that, it is clear that anabolism and catabolism, when being considered jointly, conform the metabolism as a whole and coordinated cellular process.
Answer:
If mother 1 is a carrier of hemophilia, it is likely that her son would in fact have hemophilia while the son of mother 2 does not. Mother 1 would have the genotype XᴴXʰ, meaning she has one recessive allele for hemophilia, while the father would have the genotype XᴴY, and would neither be afflicted with hemophilia nor carry it. If you do a punnett square, it shows that the son of mother 1 would have a 50% chance of having hemophilia, since he wouldn't have a second X chromosome with a dominant allele to mask the recessive hemophilia allele. Also, if mother 2 has the homozygous dominant genotype XᴴXᴴ despite the father having hemophilia and the genotype XʰY, a punnett square for this couple proves that their son would have a 0% chance of having hemophilia. Therefore, it is very likely that their sons were not switched at birth and the correct answer is D: The fact that the father in couple 2 has hemophilia would not predispose his son to hemophilia. The first couple has no valid claim. Hope this helped!
