Describe how viruses and prions can alter cell functions

If two genes are found on different chromosomes, or if they are far enough apart on the same chromosome that the chance of a cro

nika2105 [10]

Complete question:

A wild-type tomato plant (Plant 1) is homozygous dominant for three traits: solid leaves (MM), normal height (DD), and smooth skin (PP).

Another tomato plant (Plant 2) is homozygous recessive for the same three traits: mottled leaves (mm), dwarf height(dd), and peach skin (pp).

In a cross between these two plants (MMDDPP x mmddpp), all offspring in the F1 generation are wild type and heterozygous for all three traits(MmDdPp).

Now suppose you perform a testcross on one of the F1plants (MmDdPp x mmddpp). The F2 generation can include plants with these eight possible phenotypes:

solid, normal, smooth

solid, normal, peach

solid, dwarf, smooth

solid, dwarf, peach

mottled, normal, smooth

mottled, normal, peach

mottled, dwarf, smooth

mottled, dwarf, peach

Assuming that the three genes undergo independent assortment, predict the phenotypic ratio of the offspring in the F2 generation.

Answer and Explanation:

Cross 1:

Parental) MMDDPP x mmddpp

F1) MmDdPp

Cross 2: Testcross

Parental) MmDdPp x mmddpp

Gametes) MDP MdP MDp Mdp mDP mdP mDp mdp

mdp mdp mdp mdp mdp mdp mdp mdp

Punnet Square)

mdp

MDP MmDdPp

MdP MmddPp

MDp MmDdpp

Mdp Mmddpp

mDP mmDdPp

mdP mmddPp

mDp mmDdpp

mdp mmddpp

F2) 8/64 = 1/8 MmDdPp, solid leaves, normal height, smooth skin

8/64 = 1/8 MmddPp, solid leaves, dwarf height, smooth skin

8/64 = 1/8 MmDdpp, solid leaves, normal height, peach skin

8/64 = 1/8 Mmddpp, solid leaves, dwarf height, peach skin

8/64 = 1/8 mmDdPp, mottled leaves, normal height, smooth skin

8/64 = 1/8 mmddPp, mottled leaves, dwarf height, smooth skin

8/64 = 1/8 mmDdpp, mottled leaves, normal height, peach skin

8/64 = 1/8 mmddpp, mottled leaves, dwarf height, peach skin

Another way of predicting the phenotypic ratio of the offspring in the F2 generation is by performing a testcross for each gene by separately and then multiplying their genotypic proportions. This is:

Leaves Gene

Parental) Mm x mm

Gamtes) M m m m

Punnet Square) M m

m Mm mm

F2) 1/2 Mm

1/2 mm

High Gene

Parental) Dd x dd

Gamtes) D d d d

Punnet Square) D d

d Dd dd

F2) 1/2 Dd

1/2 dd

Skin Gene

Parental) Pp x pp

Gamtes) P p p p

Punnet Square) P p

p Pp pp

F2) 1/2 Pp

1/2 pp

To get the phenotypic ratio of the offspring we multiply the genotypic proportion for each trait by the genotypic proportions of the other two traits, like this:

For plants with solid leaves, normal height, smooth skin

solid leaves ratio -> 1/2

normal height ratio -> 1/2

smooth skin ratio -> 1/2

MmDdPp ratio: 1/2 x 1/2 x 1/2 = 1/8

For plants with solid leaves, dwarf height, smooth skin

MmddPp ratio: 1/2 x 1/2 x 1/2 x 1/2 = 1/8

For plants with solid leaves, normal height, peach skin

MmDdpp ratio: 1/2 x 1/2 x 1/2 x 1/2 = 1/8

For plants with solid leaves, dwarf height, peach skin

Mmddpp ratio: 1/2 x 1/2 x 1/2 x 1/2 = 1/8

For plants with mottled leaves, normal height, smooth skin

mmDdPp ratio: 1/2 x 1/2 x 1/2 x 1/2 = 1/8

For plants with mottled leaves, dwarf height, smooth skin

mmddPp ratio: 1/2 x 1/2 x 1/2 x 1/2 = 1/8

For plants with mottled leaves, normal height, peach skin

mmDdpp ratio: 1/2 x 1/2 x 1/2 x 1/2 = 1/8

For plants with mottled leaves, dwarf height, peach skin

mmddpp ratio: 1/2 x 1/2 x 1/2 x 1/2 = 1/8

3 0

3 years ago

Does glycolysis occur inside or outside the mitochondria

icang [17]

Answer: Glycolysis occurs outside the mitochondria in the cytoplasm of the cell.

Glycolysis is a metabolic process in which glucose is broken down into two molecules of pyruvate ( 3 carbon containing molecule) and produces energy molecules as 2 ATP ( adenosine triphosphate) and 2 NADH molecules. Glucose is one of the most important energy fuel that is utilized by living organisms.

Glycolysis occurs in the cytoplasm of all living cells. Thus, it occurs outside the mitochondria.

7 0

3 years ago

: Which of the following is not a sign of sepsis? Which of the following is not a sign of sepsis? increased white blood cell cou

natka813 [3]

Answer: The correct answer to the question is DILATED PUPILS.

DILATED PUPILS is not a sign of sepsis.

Explanation: Sepsis can be defined as the way our body system respond to infection or invasion of a foreign body into the body system.(usually bacterial infection)

When Sepsis becomes severe,it is injurious to organs of the body(affects their function in a negative way).

Sepsis occurs when the body's immune system tends to fight infection in the body by the release of certain chemicals(cytokines and prostaglandins),the body instead of reacting in a coordinated manner to these chemicals rather reacts out of proportion to these chemicals released to fight infection and this will lead damage to organs of the body system.

Sepsis is characterised by Fever(above 101°F (38.3°C) or below 96.8°F (36°C), increased in respiratory rate (more than 20 breaths per minute), Increased heart rate (more than 90 beats per minute),increased white blood cell count,low blood pressure and mental confusion.

Treatment of Sepsis includes the administration of antibiotics and intravenous fluid, administration of drugs to support blood pressure and administration of steroids.

5 0

4 years ago

4. When plates slide against each other as seen on the San Andreas fault,many__________ are known to occur.

Zina [86]

Plate boundary is the type of boundary which occurs when plates slide against each other as seen on the San Andreas fault in which many earthquakes are known to occur.

<h3>What is an Earthquake?</h3>

This is characterized by the ground shaking as a result of movement in the Earth crust.

Plate boundary is responsible for Earthquakes on San Andreas fault thereby making it the most appropriate choice.

Read more about Plate boundary here brainly.com/question/21668651

4 0

2 years ago

Consider the atps that can be generated via substrate-level phosphorylation. Will glycolysis be useful for generating any atps d

iris [78.8K]

Indeed, glycolysis would be useful for creating ATP as it produces a net increase of ATPs. One glucose particle produces two ATP in a cycle. On the off chance that the ETPUM experiences the procedure of aging, at that point the natural atoms would then be able to be used as the essential electron transporter. Two ATPs would be delivered by means of TCA cycle and yes it will be sufficient to help development.

Further Explanation:

Phosphorylation:

Phosphorylation is the synthetic expansion of a phosphoryl gathering (PO3-) to a natural atom. The expulsion of a phosphoryl gathering is called dephosphorylation. Both phosphorylation and dephosphorylation are done by catalysts (e.g., kinases, phosphotransferases).

ATP:

Adenosine triphosphate is an unpredictable natural synthetic that gives vitality to drive numerous procedures in living cells, for example muscle withdrawal, nerve motivation proliferation, and substance amalgamation. Found in all types of life, ATP is regularly alluded to as the "atomic unit of money" of intracellular vitality move.

ATP utilized for:

ATP Molecule. The Adenosine triphosphate (ATP) particle is the nucleotide referred to in organic chemistry as the "atomic cash" of intracellular vitality move; that is,ATP can store and move substance vitality inside cells.

TCA:

Tricyclic antidepressants (TCAs) are a class of drugs that are utilized basically as antidepressants. TCAs were found in the mid 1950s and were promoted later in the decade. They are named after their concoction structure, which contains three rings of particles.

Subject: biology

Level: college

Keywords: Phosphorylation, ATP,ATP utilized for, TCA.

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3 0

3 years ago

Read 2 more answers