What happens in g1 phase.

1 answer:

5 0

G1 phase. G1 is an intermediate phase occupying the time between the end of cell division in mitosis and the beginning of DNA replication during S phase. During this time, the cell grows in preparation for DNA replication, and certain intracellular components, such as the centrosomes undergo replication.

You might be interested in

Step 1: drag the lac promoter to the stretch of dna. do not drag the lacz gene to the dna. what happens? why is this?

Marat540 [252]

RNA polymerase binds to the lac promoter and begins transcribing the stretch of DNA to messenger RNA. Therefore, in other words, the stretch of DNA is expressed. The DNA sequence of the lac promoter 'signals' where the RNA polymerase will bind and where transcription will begin. Promoters are found upstream of a gene to be transcribed.

4 0

3 years ago

scalloped (sd) is an X-linked recessive and ebony (e) is an autosomal recessive mutation. What proportion of scalloped, ebony fe

alisha [4.7K]

Answer:

1/16

Explanation:

scalloped (Xsd) is an X-linked recessive trait to Xsd+ (wild type)
ebony (e) is an autosomal mutation recessive to e+ (wild type).

The two genes are independent because they are located on different chromosomes.

<h3>Parental generation:</h3>

True breeding scalloped female wild type for ebony (Xsd Xsd e+e+) mates with a true breeding male mutant only for ebony (Xsd+ Y ee).

The female only produces Xsd e+ gametes. The male produces Xsd+ e or Y e gametes. Therefore, the F1 females will have the genotype Xsd Xsd+ e e+ and the F1 males will have the genotype Xsd Y e e+.

If you complete a Punnett Square with the gametes the two F1 individuals can produce, you will get all the F2 proportions. The scalloped, ebony females have a Xsd Xsd e e genotype and appear in a 1/16 proportion.