the complementary base would be AACGTGC
the reason why this would be the complementary strand is because, you switch out the letter for the one its paired with so for reference
Thymine=Adenine
Guanine=Cytosine
Cytosine=Guanine
Adenine=Thymine
Answer:
During the gap between DNA synthesis and mitosis, the cell will continue to grow and produce new proteins. At the end of this gap is another control checkpoint (G2 Checkpoint) to determine if the cell can now proceed to enter M (mitosis) and divide.
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
The correct answer to the question about why can a Benedict solution be used to distinguish between a glucose and a sucrose is letter A. Benedict's Solution is used to test the presence of reducing sugar. Glucose is a reducing sugar while Sucrose is a non-reducing sugar since it has a glycosidic bonds between their structure that hinders the formation of open chain. In the end, Sucrose stuck at its cyclic form.
The answer is b
Because it makes sense