Remind the parents to observe the child for :
1. excessive bleeding.
2. infection at the surgical incision site.
If any of these are observed, the child should get immediate medical attention.
A is true
hope i helped :D
Answer:
The correct answer would be option D.
Explanation:
True breeding is a breeding procedure in which the parents would always produce their progeny with the same phenotype characters as the parents carry. This means for every trait these parents are homozygous so their offspring must show a similar phenotype.
In this case, all cattle are true-breeding which means they homozygous alleles show the progeny will also show similar traits which are short stature with brown hides and short horns.
Thus, the correct answer is option D.
I believe the answer must be B when considering the options.
I hope this helps, have a great day, and God Bless!
Brainliest is always appreciated :)
Answer:
- The lac operon can be activated by the binding of allolactose to the repressor protein, releasing it from DNA and thereby allowing for transcription to occur.
- In response to low glucose levels, cAMP is upregulated; the binding of cAMP to the cAMP receptor protein triggers the activation of the operon.
Explanation:
Lactose operon or lac operon (includes lacZ, lacY and lacA genes) is found in some bacteria and the products of its genes are involved in lactose metabolism. So, this operon is active (genes are transcribed) when lactose is present and glucose is absent (or at low level). The operon is regulated by the lac repressor which acts as a lactose sensor and catabolite activator protein (CAP) which acts as a glucose sensor.
When there is lactose (in the form of allolactose) lac repressor detects it and stops being repressor. This enables transcription.
CAP detects glucose (via cAMP) and activates transcription when glucose levels are low.
