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Delicious77 [7]
2 years ago
15

Drag each tile to the correct box. Explain the process in which hormones secreted by the pancreas function with respect to incre

ased glucose levels in the blood. Synthesis of hormones by beta cells release of hormones from the receptors intake of glucose molecules from the blood by specific transporters high amount of glucose in the blood, sending signals toward the pancreas binding of hormones with receptors on the liver
Biology
2 answers:
olga55 [171]2 years ago
4 0

Answer:

The correct order is:

  1. High amount of glucose in the blood, sending signals toward the pancreas.  
  2. Synthesis of hormones by beta cells.
  3. Binding of hormones with receptors on the liver.
  4. Intake of glucose molecules from the blood by specific transporters.
  5. Release of hormones from the receptors  

Explanation:

  • The pancreas produces two hormones important for the regulation of blood glucose levels.
  • The two hormones, glucagon and insulin, increase and decrease the blood sugar respectively.
  • Insulin lowers the blood glucose by stimulating the body cells to increase the intake of glucose.
  • The brain senses high/low blood glucose, then, sends signals to the pancreas to initiate the synthesis of insulin or glucagon by the beta cells.
  • These hormones bind to receptors on body cells; on hepatic cells, in case of insulin.
  • The receptors initiate cellular signaling that facilitates intake or exit of glucose in or out of body cells.
Leviafan [203]2 years ago
4 0

Answer:

high amount of glucose in

the blood, sending signals

toward the pancreas

↓

 synthesis of hormones

by beta cells

↓

 binding of hormones with

receptors on the liver

↓

 intake of glucose molecules

from the blood by specific

transporters

↓

release of hormones

from the receptors

Explanation: everyone is welcome:)

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Explanation:

We have three genes in the same chromosome and are trying to determine their order and relative distance to each other.

<u>The genes and possible alleles are:</u>

  • claret (c/c+)
  • spineless (s/s+)
  • hairless (h/h+)

All mutations are recessive: two copies of the mutant allele are needed for the fly to show that trait.

<u>Parental cross</u>

  • claret hairless male:\frac{s^+\  c\ h}{s^+\  c \ h}
  • spineless female: \frac{s\  c^+\ h^+}{s\  c^+ \ h^+}

Each parent can produce 1 type of gamete only, so the F1 will be homogeneous:

<u>F1</u>

<u>\frac{\ s\  \ \  c^+\ \ h^+}{s^+\   \ c \ \ \ h}</u>

During meiosis, the F1 females can produce 8 types of gametes: 2 parentals and 6 recombinants (two of them, the result of a double crossing over).

If they are test crossed to homozygous recessive males (which can only produce a \frac{s  \ c\   h}{} gamete), the following phenotypes are obtained (I just write the alleles they inherited from the female fly, as the ones that came from the male are the same for all of them):

  • 321 spineless (s c+ h+) ----> Parental
  • 309 claret, hairless (s+ c h) ----> Parental
  • 130 claret, spineless (s c h+) ----> Recombinant
  • 140 hairless  (s+ c+ h) ----> Recombinant
  • 32 hairless, claret, spineless  (s c h) ----> Recombinant
  • 38 WT  (s+ c+ h+) ----> Recombinant
  • 18 claret (s+ c h+) ----> Double Recombinant
  • 12 hairless, spineless (s c+ h) ----> Double Recombinant

The phenotypes observed in the highest frequency are always the parentals, and the ones in the lowest frequency are always the double recombinants.

<u>To determine the order of the genes:</u>

  1. we have to write down the genotype of the F1 female three times, changing the order of the genes each time.
  2. Then, we hypothesize what the double recombinant gametes would look like.
  3. When the theoretical double recombinants we obtain are the same as the ones observed in the F2, we know that <em>that </em>is the correct order of the genes.

In this problem, only if the middle gene is h+/h the double crossing over gives us the observed double recombinant gametes, therefore <u>hairless</u> is the middle gene.

\frac{s\ h^+\ c^+}{s^+\  h\ c}

<u> Double recombinants:</u>

  • s h c+ ----> spineless hairless
  • s+  h+ c  ----> claret

<u>To determine the distance between the genes:</u>

Genetic distance (m.u.) = Recombination Frequency x 100

  • Distance between the spineless and hairless genes:

Distance \ [s-h]= \frac{number\ of\  recombinants \ [s-h]}{Total number of individuals}  * 100\\\\\\Distance \ [s-h]= \frac{32+38+12+18}{1000}  * 100\\\\Distance \ [s-h]= 10\  map\ units

  • Distance between the hairless and claret genes:

Distance \ [h-c]= \frac{130+140+12+18}{1000}  * 100\\\\Distance \ [h-c]= 30\  map\ units

<h3><u>The gene map for these genes is:</u></h3>

spineless -----------------hairless ---------------------------claret

                   10 m.u.                            30 m.u.

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