Answer:
d is zero (0)
Explanation:
ψ(x)=ce^κx+de^−κx
Let say x = - 1000
We have something like
ce^κ* -1000 = approximately zero
Where we assume k to be 1
C becomes zero as it tend to negative infinity
But
If k is 1
de^−1* 1000 =d * 2 * 10^434
To have a zero function d = 0
So 0 * 2 * 10^434 = 0
When you aren't showering as regularly, your skin can become oily and salty leading to blemishes and breakouts. Although your skin can get all sweaty during the day anyway, not bathing means these bacteria are not being thoroughly removed or cleaned.
Answer:
a. p53 activates transcription of WAF1.
Explanation:
<u>WAF1 transcription occurs independent of p53 during oxidative stress so p53 does not play any role in cell cycle arrest in the signaling pathway which involves WAF1 . </u>
In rest of the mentioned options, p53 plays a role directly or indirectly. During double stranded lesion in DNA in G1 phase, a sensor protein known as ATM binds the DNA lesion site. ATM is a serine/threonine kinase which phosphorylates another kinase known as chk2. After phosphorylation,<u> chk2 stabilizes transcription factor p53.</u> p53 further acts as a transcription factor for the synthesis of a protein known as p21 which inhibits G1 phase specific CDK and ultimately cell is arrested in G1 phase. The cell remains in arrested state until the DNA lesion is fully repaired. <u>Hence, p53 indirectly blocks G1 to S transition with the help of p21. </u>
As such <u>WAF1 transcription factor involving pathway</u><u> </u><u>also requires p21 protein for causing cell cycle arrest but in this pathway p21 is not synthesized with the help of p53. </u>
Answer:
1/3
Explanation:
<em>A standard monohybrid cross is a cross that follows the dominance/recessive pattern from Mendel's experiment.</em>
It means the purple flower color is dominant over the white flower color.
Assuming the allele for purple flower color is P and that of the white flower color is p, a standard monohybrid cross will involve a true breeding PP and pp.
PP x pp: Pp, Pp, Pp and Pp. All the F1 offspring will have purple flowers with Pp genotype.
At F2: Pp x Pp = PP, Pp, Pp, and pp.
3/4 or 75% of the F2 offspring have purple flower color out of which 1 is true breeding for the trait.
Hence the fraction of the purple flowered peas in the F2 that is expected to be true-breeding (PP) is 1 out of 3.
Answer: The nurse needs to perform the following check on the child.
1 check for tachycardia(irregular beating of the heart.)
2.Capillary refill than 3 seconds
3.Sunken eyes .and fontanel(soft membranous gaps of infant skull).
These are all signs that the body electrolyte balance is deleted. And urgent need to adjust the is by the nurse.
Explanation: