What do u mean <span>Cystic fibrosis is controlled by recessive alleles(cc). if two parents without cystic fibrosis have a child with cystic fibrosis, what is the probability ratio that their next child will have cystic fibrosis?</span>
The exons of the pre -mRNA contain the protein -coding regions.
Pre mRNA is the first made mRNA transcript and requires undergo many post transcriptional modifications for the formation of a mature mRNA.
The exons are the regions of the pre-mRNA that are found in the mature RNA, after the splicing of introns takes place.
RNA splicing is the process in which the non-coding segments of the RNA, which are known as introns, are removed by the help of small nuclear ribonucleoproteins. These SnRNPs make the spliceosome, which catalyzes the process of splicing.
After the introns are removed by the splicing process, the exons are covalently joined, which forms the mature mRNA.
Learn more about mRNA from here:
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Answer:
I think the awnser is D. Hope it helps
Hey there!
Nuclei with a high neutron-to-proton ratio usually undergo a beta emission.
Beta emissions, unlike some other types, is radioactive decay throughout these particles, and can have half lives like other types of exponential (usually) decay. Neutrino, along with <em>beta rays,</em> are coming from, or emitted from the nucleus - hence the name.
Hope this helps!
Answer:
(a) 0.16
(b) 1
Explanation:
Let Probability that ticks in the Midwest carried Lyme disease, P(L) = 0.16
Probability that ticks in the Midwest carried HGE disease, P(H) = 0.10
Probability that ticks in the Midwest carried either Lyme disease or HGE disease, P(
) = 0.10
(a) Probability that a tick carries both Lyme disease (L) and HE (H) is given by
P(L 
H);
As we know that P(A 
B) = P(A) + P(B) - P(A B)
So, in our question;
P(L H) = P(L) + P(H) - P(L H)
0.10 = 0.16 + 0.10 - P(L H)
P(L H) = 0.16 + 0.10 - 0.10 = 0.16
Therefore, the probability that a tick carries both Lyme disease (L) and HE (H) is 0.16 or 16% .
(b) <em>Conditional Probability P(A/B) is given by</em> = 
So, the conditional probability that a tick has HE given that it has Lyme disease is given by = P(H/L)
P(H/L) = 
= 
= 1 .
