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padilas [110]
3 years ago
5

The blood sample consists of dead cells that are breaking apart into their basic

Biology
2 answers:
satela [25.4K]3 years ago
5 0
I think the answer is dna
arsen [322]3 years ago
4 0

Answer:

I think the answer is b. DNA

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Round seeds and yellow seed color are dominant to wrinkled seeds and green seed color. Which Punnett square represents the corre
svetoff [14.1K]
A punnet square is a diagram that is use to predict an outcome of a particular cross or breed experiments. So base on the following data you have given, the possible answer would be that the punnet square data in horizontal are is RY, RY,rY,rY. In the vertical are is RY Ry RY Ry. I hope you are satisfied with my answer and feel free to ask for more 
 
4 0
3 years ago
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A inhibitor of carbonic anhydrase would:______
Naily [24]

Answer:

c. increase blood pH.

d. interfere with oxygen binding to hemoglobin.

Explanation:

A inhibitor of carbonic anhydrase increases the pH of the blood due to prevent the production of carbonic acid from the carbon dioxide formed during cellular respiration. This carbonic acid decreases the pH of the blood so inhibitor of carbonic anhydrase is responsible for the increase in the pH. It also interfere with oxygen binding to hemoglobin because oxygen bind with hemoglobin in low pH but the pH is higher due to inhibitor of carbonic anhydrase.

8 0
3 years ago
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Assume that two genes are 80 map units apart on chromosome II of Drosophila and that a cross is made between a doubly heterozygo
STatiana [176]

Answer:

The recombination frequency between two genes exhibits a positive correlation with the distance between them, that is, farther they are, and more will be the chance of recombination. Thus, recombination frequency is used to signify distance among the two genes, for example, 1 percent recombination frequency demonstrates distance of 1 map unit.  

Let us consider that the heterozygous female of genotype AaBb can generate four kinds of gametes, that is, AB, Ab, aB and ab. Of these, the two gametes are the outcomes of recombination, or it can be said that 50 percent are recombinants. Thus, it can be concluded that in case of two linked genes, the maximum probable recombination frequency is 50 percent.  

This shows that any genes, which are distant than 50 map units will function as unlinked and will function as if they were on distinct chromosomes, and the frequency of recombinant frequency will be 50 percent.  

In the given question, it is given that the map distance between the two genes is 80 map units, that is, more than 50 map units. The maximum probable recombinant offspring will be 50 percent of the entire offspring.  

6 0
3 years ago
please help me I need much help as needed thanks.) The dependent variable in this investigation is the (1) size of the test tube
ryzh [129]

(4) amount of protein digested

8 0
3 years ago
-3X+3(2X-6)=6 answer
Dahasolnce [82]
-3x + 3(2x - 6) = 6

-3x +6x - 18 = 6

3x - 18 = 6

3x = 6 + 18

3x = 24

x = 8

hope this helped, blessing! :)
5 0
3 years ago
Read 2 more answers
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