That is true, enzymes are specific.
Answer:
a: metaphase
Explanation:
the chromosomes are aligned at the metaphase plate
If a lot of tryptophan is present, the operon will be repressed.
Under intermediate amounts of tryptophan, the change to stop codons would cause the ribosome to stall and therefore would mimic as if there were no tryptophan present.
If no tryptophan present, the operon would be maximally expressed.
a - True. The ribosome would always be stalled in the leader sequence - so this wouldn't matter. However, the operon would still be transcribed because the repressor would be active (remember Trp operon is controlled both by attenuation and repression). True as long as there's not much tryptophan to activate the repressor!
b- True - With no tryptophan the repressor isn't engaged and the ribosome is stalled in the leader sequence.
c- Matters about how much tryptophan is available. So True if there's lots of tryptophan available.
d- I would choose this one as all of the above could be true depending on the levels of tryptophan.
Answer:
His process is wrong.
Explanation:
Bob must take the eye out of the fish. Then he must cut the eye. After he cuts the eye he needs to take a thin layer of the eye and put it on the glass. A microscope needs light to shine through the specimen. All Bob would see when he put the fish on there would be black. The thick amount of mass is blocking the light.
Answer:
Genotype: 100% or 4/4 of the progeny will be heterozygous for the trait, Yy.
Explanation:
<u>Available data:</u>
- Two plants differ in the seed color they produce.
- One plant produces green seeds, the other produces yellow seeds.
- Yellow is the dominant phenotype, over green which is the recessive phenotype
- The parental plants are true-breeding
Let us say that the allele Y expresses yellow color and is dominant over the allele y which expresses the green color and in the recessive one.
Cross: a green-seeded plant with a yellow-seeded plant
Parental) YY x yy
Phenotype) Yellow seeds Green seeds
Gametes) Y Y y y
Punnet square) Y Y
y Yy Yy
y Yy Yy
F1) Phenotype: 100% of the progeny will be yellow-seeded
Genotype: 100% or 4/4 of the progeny will be heterozygous for the trait
