The muscle responsible for action in a single direction is called

Anika [276]

I believe photosynthesis

5 0

4 years ago

Diabetic ketoacidosis (dka) in a client with type 1 diabetes occurs when the lack of insulin leads to the release of which physi

masha68 [24]

The answer to the above question is Fatty acids.

<h3>What are Fatty acids?</h3>

A fatty acid is a carboxylic acid with an aliphatic chain that is either saturated or unsaturated. This definition applies to chemistry, particularly biochemistry. The majority of naturally occurring fatty acids have an unbranched chain with 4 to 28 carbon atoms, which is an even number. In some species, like microalgae, fatty acids make up a significant portion of the lipids (up to 70% by weight), but in other organisms, they are present as one of the three main types of esters: triglycerides, phospholipids, or cholesteryl esters rather than in their solo form. Fatty acids are crucial nutritional sources of energy for animals in any of these forms, as well as crucial cellular building blocks.

To learn more about Fatty acids with the help of given link:

brainly.com/question/26353151

#SPJ4

6 0

1 year ago

Chemical energy is stored in molecules in is released through which process?

zaharov [31]

Photosynthesis I know that I took the test

7 0

3 years ago

What are brownfields

Daniel [21]

The lands which are lef unuse after it is affected by environmental contamination.

<u>Explanation:</u>

Brownfields are abandoned or unused fields where reuse is hindered by real or observed environmental contamination. The places may include localities destroyed by hazardous materials, petroleum or mine-scarred land.
Brownfields are found in a city or town's manufacturing section, on areas with discontinued factories or industrial buildings, or other earlier contaminating operations like steel millhouses, refineries or landfills. It is mainly found in urban cities where there is normally a stronger potential to redevelop these localities.
Some of the most frequent contaminants known at Brownfield sites are from fuels such as grease, propellant, diesel and kerosene from subterrene storage tanks, floor drains, External storage of barrels and machine, and cleaning solvents.

3 0

3 years ago

Three linked autosomal loci were studied in smurfs.

cupoosta [38]

Answer:

height -------- color --------- mood

(13.2cM) (14.5cM)

C=0.421

I = 0.579

Explanation:

We have the number of descendants of each phenotype product of the tri-hybrid cross.

Phenotype Number

pink, tall, happy 580

blue, dwarf, gloomy 601

pink, tall, gloomy 113

blue, dwarf, happy 107

blue, tall, happy 8

pink, dwarf, gloomy 6

blue, tall, gloomy 98

pink, dwarf, happy 101

Total number of individuals = 1614 = N

Knowing that the genes are linked, we can calculate genetic distances between them. First, we need to know their order in the chromosome, and to do so, we need to compare the phenotypes of the parental with the ones of the double recombinants. We can recognize the parental in the descendants because their phenotypes are the most frequent, while the double recombinants are the less frequent. So:

Parental)

Pink, tall, happy 580 individuals

Blue, dwarf, gloomy 601 individuals

Simple recombinant)

Pink Tall Gloomy 113 individuals

Blue, Dwarf, Happy 107 individuals

Blue Tall Gloomy 98 individuals

Pink Dwarf Happy 101 individuals

Double Recombinant)

Blue Tall Happy 8 individuals

Pink Dwarf Gloomy 6 individuals

Comparing them we realize that parental and double recombinant individuals differ in the position of the gene codifying for <u>color</u><u>.</u> They only change in the position of Blue and Pink. This suggests that the position of the color gene is in the middle of the other two genes, height and mood, because in a double recombinant only the central gene changes position in the chromatid.

So, the alphabetic order of the genes is:

---- height ---- color ----- mood ----

Now we will call Region I to the area between Height and Color, and Region II to the area between Color and Mood.

Once established the order of the genes we can calculate distances between them, and we will do it from the central gene to the genes on each side. First We will calculate the recombination frequencies, and we will do it by region. We will call P1 to the recombination frequency between Height and color genes, and P2 to the recombination frequency between color and mood.

P1 = (R + DR) / N

P2 = (R + DR)/ N

Where: R is the number of recombinants in each region (the ones that have an intermediate phenotypic frequency), DR is the number of double recombinants in each region, and N is the total number of individuals. So:

Region I

Tall------ Pink--------happy (Parental) 580 individuals

Dwarf ---Pink------- Happy (Simple Recombinant) 101 individuals

Dwarf--- Pink-------Gloomy (Double Recombinant) 6 individuals

Dwarf----Blue-------Gloomy (Parental) 601 individuals

Tall ------Blue------- Gloomy (Simple Recombinant) 98 individuals

Tall ----- Blue------- Happy (Double Recombinant) 8 individuals

Region II

Tall------ Pink--------happy (Parental) 580 individuals

Tall-------Pink------- Gloomy (Simple Recombinant) 113 individuals

Dwarf----Pink------- Gloomy (Double Recombinant) 6 individuals

Dwarf----Blue-------Gloomy (Parental) 601 individuals

Dwarf ----Blue-------Happy (Simple Recombinant) 107 individuals

Tall ----- Blue------- Happy (Double Recombinant) 8 individuals

In each region, the highlighted traits are the ones that suffered recombination.

P1 = (R + DR) / N

P1 = (101+6+98+8)/1614

P1 = 213/1614

P1 = 0.132

P2= = (R + DR) / N

P2 = (113+6+107+8)/1614

P1 = 234/1614

P1 = 0.145

Now, to calculate the recombination frequency between the two extreme genes, height and mood, we can just perform addition or a sum:

P1 + P2= Pt

0.132 + 0.145 = Pt

0.277=Pt

The genetic distance will result from multiplying that frequency by 100 and expressing it in map units (MU). One centiMorgan (cM) equals one map unit (MU).

The map unit is the distance between the pair of genes for which every 100 meiotic products, one results in a recombinant product.

Now we must multiply each recombination frequency by 100 to get the genetic distance in map units:

GD1= P1 x 100 = 0.132 x 100 = 13.2 MU = 13.2 cM

GD2= P2 x 100 = 0.145 x 100 = 14.5 MU = 14.5 cM

GD3=Pt x 100 = 0.277 x 100 = 27.7 MU = 27.7 cM

To calculate the coefficient of coincidence, CC, we must use the next formula:

CC= observed double recombinant frequency/expected double recombinant frequency

Note:

-observed double recombinant frequency=total number of observed double recombinant individuals/total number of individuals

-expected double recombinant frequency: recombination frequency in region I x recombination frequency in region II.

CC= ((6 + 8)/1614)/0.132x0.145

CC=0.008/0.019

CC=0.421

The coefficient of interference, I, is complementary with CC.

I = 1 - CC

I = 1 - 0.421

I = 0.579

8 0

3 years ago