paracrine signaling
endocrine signaling
direct signaling across gap junctions
Answer:
The correct answer is the third option - endocrine signaling.
Explanation:
The endocrine signaling takes place when endocrine cell organs or glands release hormones that flow through the bloodstream and that act on distantly located target cells in the body. The endocrine system includes several glands that release or secretes different hormones.
The pituitary gland transfers signals to other endocrine glands to release or inhibit their hormone production according to the brain signals. The adrenocorticotropic hormone (ACTH) is released by the anterior pituitary gland that stimulates cortisol release in the adrenal glands in stressed conditions.
Thus, the correct answer is option third - endocrine signaling.
The pituitary gland secreting hormones is an example of endocrine signaling, where hormones are secreted directly into the bloodstream and carried to target cells.
The secretion of hormones by the pituitary gland is an example of endocrine signaling. Endocrine signaling refers to the process where glands secrete hormones directly into the bloodstream, which are then carried to the target cells elsewhere in the body. In contrast, autocrine signaling involves cells responding to their own signals, paracrine signaling refers to cells communicating with nearby cells, and direct signaling across gap junctions involves communication between adjacent cells through specific channels.
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I cant draw it for you but just look up images of what it looks like an just copy it from that.
Enzymes are affected by temperature and pH in their ability to catalyze chemical reactions. Extreme temperatures and pH values can denature enzymes, causing them to lose structure and function. The active site of an enzyme is sensitive to changes in the local environment, affecting its ability to bind substrates.
Enzymes are subject to influences by the local environment, including temperature and pH. Increasing or decreasing the temperature outside of an optimal range can affect the chemical bonds within the enzyme's active site, making them less suitable for binding substrates.
Extreme temperatures can cause enzymes to denature, losing their structure and function. Similarly, extreme pH values can also denature enzymes and affect their ability to bind substrates.
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answer:
a balance and a graduated cylinder
Explanation
Density is mass per unit and it is property characteristic of a substance. The arrangement of the mass of atoms and their size is what determines the the density of a substance.
For a student to measure the density of seawater he/she is required to use a balance and a graduated cylinder where a balance will be used to measure the mass of the mineral sample and then a graduated cylinder to determine the volume.
Answer:
2 out of every 5 or 2/5
Explanation:
there are 200 black cats because there are 300 calico there are 500 total so it would be 200/500 or 2/5
b. p=0.725.q = 0.275; P=0.06. H=0.56, Q=0.51
c. p=0.4.q = 0.6: P=0.12. H=0.56, Q=0.32
d. p=0.725.q = 0.275: P=0.34. H=0.57. Q=0.09
Answer:
a. p=0.725, q = 0.275: P=0.51, H=0.43, Q=0.06
Explanation:
Let state our given parameters from the question:
Frequencies of coat color genes at the C locus for population 1 are .85 for C
This implies that the Allelic frequency C for population p1 =0.85
Frequencies of coat color genes at the c locus for population 1 are .15 for c
This implies that the Allelic frequency c for population q1 = 0.15
Frequencies for Care .6 i.e p2= 0.6
Frequencies for care .4 i.e, let that be q2= 0.4
The table below shows a diagrammatic representation of the above expression:
Alllelic Frequency C c
Population 1 (p1) 0.85 (q1) 0.15
Population 2 (p2) 0.6 (q2) 0.4
Now, from above: let think of the table as a punnet square and then cross it together;
(p1) = 0.85 (q1) = 0.15
p2 = 0.6 p1p2 p2q1
= 0.6 × 0.85 = 0.15 × 0.6
= 0.51 (P) = 0.09 (H)
q2 = 0.4 p1q2 q1p2
= 0.85 × 0.4 = 0.4 × 0.15
=0.34 (H) = 0.06 (Q)
From the above table, the heterozygous are represented by (H)
∴ Frequency of heterozygous can be calculated as:
= 0.09 + 0.34
= 0.43
Thus, we can conclude that the progenyF1 genotypic frequencies are:
P= 0.51
H= 0.43
Q= 0.06
Now, let us calculate the allelic frequencies, p and q in F1
p = P + 1/2 × (H)
= 0.51 + (1/2 × 0.43)
= 0.51 + 0.215
= 0.725
q =Q + 1/2× (H)
= 0.06 + (1/2 × 0.43)
= 0.06 × 0.215
= 0.275
Hence, p=0.725, q = 0.275: P=0.51, H=0.43, Q=0.06 , This makes option a the correct answer.