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3 years ago

what type of mountains are commonly formed at ocean-ocean and ocean-continental convergent plate boundaries?

2 answers:

4 0

The Cascade Mountain Range is a line of volcanoes above the melting oceanic plate. The Andes Mountain Range of western South America is another example of a convergent boundary between an oceanic and continental plate. Here the Nazca Plate is subducting beneath the South American plate.

mixas84 [53]3 years ago

4 0

Hello! I hope this helps :

Most divergent plate boundaries are underwater.. While the process is volcanic, volcanoes and earthquakes along oceanic spreading ridges are not as violent as they are at convergent plate boundaries.

your answer should be the divergent plates.

You might be interested in

During which phase of mitosis do the nuclear membrane, nucleolus, and nucleus dissolve? metaphase prophase anaphase telophase

Mekhanik [1.2K]

Prophase
During the prophase stage of mitosis, the nucleus seems to disband itself. The nucleolus suspends itself and finally, the nuclear membrane disassembles itself. This is because of the undergoing process of copying and dividing.
Mitosis is a functional cell dividing process that helps the cells repair or rejuvenate itself because of apoptosis and other factors.

4 0

3 years ago

Read 2 more answers

A new organism begins to form when egg and sperm cells join in the process called fertilization

Irina18 [472]

Yes, when the egg and a sperm cell meet fertilization has taken place

6 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 color. 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

Suppose you have one plant with the genotype TTPpYy, and another plant that is recessive for all three traits. If you were to cr

musickatia [10]

Answer:

16/64 = 1/4

Explanation:

This is a typical trihybrid cross involving three genes T, P and Y. A plant with genotype TTPpYy is crossed with a plant recessive for all traits (ttppyy).

According to Mendel's law of independent assortment, each allele for each gene will get sorted into the following 8 gametes with only 4 different: TPY, TPy, TpY, Tpy, TPY, TPy, TpY and Tpy.

The recessive parent, ttppyy will produce tpy, tpy, tpy, tpy, tpy, tpy, tpy and tpy.

Hence, using a punnet square, 64 offsprings will be produced with only 16 of them heterozygous dominant for the three traits with genotype (TtPpYy). Hence, proportion is 16/64 equivalent to 1/4.

6 0

3 years ago

How can science be used to address "big" issues such as "global warming"?

Katarina [22]

global warming is caused by different gases ,, hence they are affecting the environment by increasing temperature ,, only science can solve such a big problem because during global warming all the reactions that take place are scientific..

5 0

3 years ago