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algol [13]
3 years ago
9

A client has a mean arterial blood pressure (map) of 97 mmhg and an intracranial pressure (icp) of 12 mmhg. what is the cerebral

perfusion pressure (cpp) for this client? record your answer using a whole number
Biology
1 answer:
ololo11 [35]3 years ago
8 0
Cerebral perfusion pressure or CPP can be calculated by subtracting the mean arterial blood pressure with intracranial pressure(ICP). Mean arterial blood pressure(MAP) itself calculated by adding (2*systole + diastole)/3 
The <span>cerebral perfusion pressure for this client would be:
CPP= MAP-ICP
CPP= 97mmHg- 12mmHg= 85mmHg</span>
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Ludmilka [50]

Answer:

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6 0
2 years ago
(c) Based on the data in Table 1, describe why the dominant alleles for body color and wing shape are the alleles that produce a
Brrunno [24]

Answer:

Check the explanation

Explanation:

Consider the below table of F2 data of phenotypic flies: we can reconstruct the table as seen in the second attached image below:

From the above observations it is clear the Gray body long wings and Ebony body vestigial wings are the parental combinations and that the Gray body vestigial wings and Ebony body long wings are the recombinants.

Also the ratio indicates that Gray body is dominant over ebony body and that long wings is dominant over vestigial wings.

here the genes for the two characters have shown independent assortment which means that the genes are unlinked if located on the same chromosome or are located  on different chromosomes.

Now F1 hybrid= GgLl (G for Grey and L for Long)

Cross between F1 hybrid and true breeding Gray vestigial (GGll)

GgLl x GG ll

Gametes-----------> GL Gl gL gl Gl

        GL                          Gl                      gL                  gl

Gl    GGLl                      GGll                  GgLl               Ggll

   (Gray long)      (Gray vestigial)     (gray Long)     (Gray vestigial)

Therefore the probability of getting the flies with gray body vestigial wings= 2/4= 50%

b) The reason why the students obtained the above given F2 results involving a cross between true breeding Gray body and long wings with true breeding Ebony body and vestigial wings is that the genes for the two charcters asssort independently in F2 generation and that the genes are not linked as:

Parents------------------> GGLL x ggll

Gametes -----------------> GL gl

F1---------------------> GgLl (Gray long but in heterozous condition)

Now GgLl x GgLl

Gametes GL Gl gL gl   GL Gl gL gl

Here gametes assorted independently and hence in F2 generation we got the above results (U can show the results in the form of punnett square.

4 0
3 years ago
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allsm [11]
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3 0
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Georgia [21]

Answer:

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