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2 years ago

I need help with this question. I’m really confused at the moment

Biology

1 answer:

arlik [135]2 years ago

8 0

I believe the answer should be mitosis.

Mitosis is a kind of cell division that produces 2 daughter cells that are genetically identical to both the other daughter cell and the parent cell. It is a process of exact duplication of the genetic material (chromosomes) in the parent cell, then dividing, each with the same original number and sequence of chromosome.

It is used for growth and development because they're the process of where the number of cells increases, and how? By mitosis. These new cells must be exactly identical to the other originally existing cells, as for example you wouldn't suddenly grow some skin cells as a replacement of nerve cells. Mitosis of nerve cells produces the same type of nerve cells, mitosis of skin cells produces skin cells.

Therefore, your answer should be mitosis.

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Who realized that fingerprints were unique to individuals?

Whitepunk [10]

1. I believe the answer is William Herschel.
Francis Galton is often credited with the discovery that everyone's fingerprints are unique and that they could therefore be used for identification. However his works is extensively based on the work of Dr. Henry Faulds, who appears to have a better claim to the discovery, along with British magistrate William Herschel.

2. I believe that the ratio of height to width indicate size and consistency in hand writing analysis. The ratio of height, width and size of letters is among the characteristics of handwriting analysis that entails whether the letters are consistent in height, width and size.

3. I believe the statement that describes the line quality in handwriting analysis is whether the letters are shaky or flow continuously. Line quality is an indication of speed and fluency; the better the quality, the more skilled, or at least fast, the movement of the pen.

4. The statement that best indicates pen pressure in hand writing analysis is the down strokes and upstrokes are an even width.
Pen pressure is the pressure equal or unequal when applied to upward and down ward strokes. The pen pressure of a writer indicates his physical and emotional energy.

5. I think what is measured in the ridge count is spacing between ridge lines.
Ridge counting is the the counting of the ridges that touch or cross the line of count drawn between the delta and core of loop patterns. A white space must always intervene between the delta and the first ridge to be counted.

8 0

3 years ago

How do transposable elements and short tandem repeats (STRs) differ? a. STRs occur within exons; transposable elements occur wit

Allisa [31]

Transposable elements and short tandem repeats are different because short tandem repeats make up only a small percentage of a given genome while transposable elements often make up larger parts of a genome. So the correct option is C.

<h3>What are transposable elements? </h3>

Transposable elements or transposons are a class of genetic elements. They can “jump” from one location to another, in a genome.

Even though they are called “jumping genes,” they have an integrated site in a genome and are maintained there. Also, most of them are eventually inactivated and can not move any longer.

Whereas Short tandem repeats or STRs are short repeats of DNA. They are also known as microsatellites or simple sequence repeats. They have a repetitive unit of 1-6 bp and form a series of 100 nucleotide lengths.

Therefore the correct option C.

Read more about transposons, here

brainly.com/question/12294587

#SPJ4

5 0

1 year ago

Which is an application of DNA technology in medicine?

dusya [7]

Answer:

determining hereditary diseases in individuals and families

Explanation:

DNA technology is used to join together DNA segments. DNA technology includes DNA sequencing and DNA cloning.

DNA technology is important to the pharmaceutical industry. By genetic modifications such as high nutrition value, it helps in improving the food production quality.

DNA technology helps in determining hereditary diseases in individuals and families.

6 0

2 years ago

Which layer of the atmosphere is closest to the surface of Earth?

bixtya [17]

Answer:

thermosphere

Explanation:

i think that one

3 0

3 years ago

Read 2 more answers

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