Answer:
Explanation:
a. Since both the parents are carriers, they have one allele that is the gene for the normal trait and one gene that is for the sickle cell trait. So the genotypes of the parents are Hh
b. If we use a punnett square:
H h
H HH Hh
h Hh hh
Genotypic ratio: 1 HH: 2Hh: 1hh
Phenotypic ratio: 1 normal blood cell: 2 sickle cell carriers: 1 sickle cell disease
Answer:
I think it's A
must lie btn a clearly defined interval
Such changes would occur mostly likely near or in the active binding site of the enzyme.
Because the drugs used are competitive inhibitors of the <span>HIV reverse transcriptase enzyme, it means that they connect directly to the active binding site of this enzyme not allowing it to preform its function. If the mutations impede this drugs to work, it is probably because they alter the active binding site of the enzyme, not allowing the drug to bind and have its competitive behaviour permitting the enzyme to work normally. </span><span /><span>
</span>
Answer:
endoplasmic reticulum
Explanation:
ER is full with proteins than help to synthesize phospholipids in plant cells
Yes, it is possible.
In this case both of the parental plants were heterozygotes and they manifested dominant allele in their phenotype, which is round seed.
P: Aa x Aa
F5: AA, Aa, aA, aa - possible genotypes in fifth generations.
A- dominant allele (round seeds); a- recessive allele (wrinkled seeds)
Wrinkled phenotype is manifested only if there are two recessive alleles present.
