a cross between a white horse and a chesnut horse produce all light brown horses

Answers

Answer 1

Answer:

Incomplete dominance

Explanation:

- Eijiro <3

Answer 2

Answer:

incorrect.

Explanation:

there is a possibility. but in reality color does not effect the foal.

My Friesians grandfather was an Akhal teke that was gold. He gave birth to all black offspring as you can see in the attachment.

A Cross Between A White Horse And A Chesnut Horse Produce All Light Brown Horses

Related Questions

Suppose a trait is controlled by a gene that has one dominant allele (G) and one recessive allele (g).

Which of the following crosses would be expected to produce the greatest variety of genotypes among the offspring?

Answers

Answer:

Gg x Gg

Explanation:

Gg x Gg crosses would be expected to produce the greatest variety of genotypes among the offspring.

What is Punnett Square?

A Punnett square is defined as a graphical representation of the possible genotypes of the offspring resulting from a particular cross or breeding event. Creating a Punnett square requires knowledge of the genetic makeup of the parents.

For a monohybrid cross, Punnett squares are 2X2 squares containing four boxes where each represents a fertilization event between the original gametes. While the second type is used to predict the outcome of breeding experiments where two traits are being observed and the Punnett square is large with sixteen boxes.

In the above example, if a Gg x Gg cross occurs, the genotype for the children will be Gg, which would be expected to produce the greatest diversity of genotypes among the offspring, i.e. 100%.

Thus, Gg x Gg crosses would be expected to produce the greatest variety of genotypes among the offspring.

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What experiment is left overnight what would be the observation fresh potato in sugar solution and water

Answers

The potato would expand or it’s mass will increase bc there is more solute in the potato. The potato is placed into a hypotonic solution, thus the water will move into the potato.

The chromosome are arranged to show
ll
13
O A. homologous pairs
O B. tetrads
OC. independent assortment
O D. nucleotides

Answers

Answer:

D. nucleotides

i'm not sure

The answer is D nucleotides

How does heat affect the thermal energy of an object that is colder than the air that surrounds it?

Answers

Answer:

Heat decrease the thermal energy of an object as energy moves to the air. Heat causes the thermal energy of the object to increase as energy moves from warmer air to the colder object

I dont know what is the answer sorry!

How can a chloroplast, a structure found in plant cells but not in animal cells, provide energy for BOTH plants and animals?

Answers

Answer:

Chloroplasts are found in plant cells only because chloroplasts contain chlorophyll which is essential for photosynthesis. Chlorophyll traps sunlight and uses it to prepare food for plants by the process of photosynthesis.

Photosynthesis takes place in the _______________ and releases __________________.
A. mitochondria, oxygen
B. chloroplast, oxygen
C. mitochondria, carbon dioxide
D. chloroplast, carbon dioxide

Answers

Answer:

B. chloroplast, oxygen

Explanation:

This is because plants take in the carbon, water, air,and it transforms the water into oxygen and carbon into glucose. So it absorbs light, and the chloroplasts are special structures that are responsible for that absorption of light, They are green, because of chlorophyll, which gives the green pigment to leafs

photosynthesis takes place inside them, so they can absorb light to start the process.

Which new animal adaptation may be favored if average

temperatures in a temperate forest decreases by 25

degrees C over 1500 years?

Answers

Answer:

If the average temperatures of temperate forests fell by about 25ºC during the period of 1500 years, the natural modification of the animals that would be favored would be that which allows an adaptation to the new climatic circumstances by the fauna of said forests, that is , to the notable drop in temperature. Thus, for example, animals would develop more voluminous furs, which would allow them to heat their bodies in a better way.

The following picture is what part cell division?

Answers

Answer:

ANNYEONG HASEYO JONEON BANGTANSONYEONDAN HOGEON MAKNAE JEON JUNGKOOK IMNIDA


The real length of one villus is 0.8 mm
Calculate the image length if the villus is viewed at a magnification of 20
Use the equation:
size of image
magnification =
size of real object
[3 marks]

Answers

Answer: 16 mm

Explanation:

If I understood the given equation correctly, it could be rewritten as:

size of image = size of real object * magnification

Plug in the given values, and we get

size of image = 0.8 * 20 = 16 (mm)

PLEASE ANSWERR

Which of the following is the independent variable for a controlled experiment
investigating how waste production affects a certain person's ecological
footprint?

A. Opinion the person has about environmental impact
B. Measurement of the amount of waste produced
C. Identification of the person
D. Calculation of resulting ecological footprint

Answers

Answer: Measurement of the amount of waste produced.

Explanation:

Measurement of the amount of waste produced is the independent variable for a controlled experiment in this case. Therefore, the correct statement is option B.

What is the independent variable?

The independent variable is the variable that can be changed in a controlled experiment. In this case, the researcher can manipulate the amount of waste produced followed by observation of its effect on the ecological footprint.

The researcher can manipulate the amount of waste by changing the factors such as the type of waste produced, recycling facilities available, etc. under controlled conditions.

Once the independent variable is manipulated, the researcher can observe its effect on the dependent variable. In this case, the ecological footprint of the individual is a dependent variable. The dependent variable is the variable that can be affected by the independent variable and is measurable.

Therefore, the measurement of the amount of waste produced is the independent variable for a controlled experiment.

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factors that affect blood flow through the bone marrow​

Answers

Insufficient amount of nutrients such as iron, vitamin B12, or folate, affects the bone marrow's ability to produce normal red blood cells; those produced may be small (microcytic), large (macrocytic), and/or may have decreased amounts of hemoglobin inside them (hypochromic).

how many molecules of glucose are produced in photosynthesis

Answers

Answer: One molecule of glucose (C6H12O6) and six molecules of oxygen (O2)

Explanation:

Six molecules of carbon dioxide (CO2) combine with six molecules of water (H2O) in the presence of light energy. This produces one molecule of glucose (C6H12O6) and six molecules of oxygen (O2)

How does a planet's distance from the Sun affect its surface temperature and the length of its orbit (year)?

PLS IS FOR TODAYY

HELP MEE PLSS

Answers

Answer:  In general, the surface temperatures decreases with increasing distance from the sun. Venus is an exception because its dense atmosphere acts as a greenhouse and heats the surface to above the melting point of lead, about 880 degrees Fahrenheit (471 degrees Celsius)

Explanation:

I Hope This Helps You! Have a Great day! ^v^

Select the correct answer from the choices provided in the underlined sections

A plant grown in low light will have (lighter / darker) flower color while a plant grown in bright light will have a (light / darker) flower color.

Answers

Answer:

darker and lighter

Explanation:

Hope it helps :)

A plant grown in low light will have darker flower color while a plant grown in bright light will have lighter flower color.

The plants which grow in high light intensities are called?

The plants which have the ability to survive in the high intensity of sunlight they are called heliophytes.  

Prosopis, Betula, Teak, etc are the examples of heliophytes.The plants which grow in low light intensities are called Sciophytes. Example includes Aegle, Nycanthes.

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How did the insertion of a nucleotide in a DNA sequence result in a phenotypic change in an organism

Answers

Answer:

if the nucleotide insertion occurs within the coding region of a gene (frameshift mutation), then it will change the open reading frame (ORF) of the resulting protein and thus also will alter the phenotype of the individual  

Explanation:

A mutation is any change in the DNA sequence of the genome of a cell/organism. A point mutation is a genetic mutation where a single nucleotide is inserted, deleted or changed from the original DNA sequence. The insertion of a single nucleotide in the coding region of a gene is called frameshift mutation since this nucleotide base shifts the way the DNA sequence is read. A frameshift mutation changes the meaning of the codons that follow the inserted nucleotide base and thereby protein function is generally lost, thereby also affecting the resulting phenotype associated with this protein.

A radioactive material has a half-life of 30 days. You begin with 4 grams of radioactive material.


a. How much of the original is left unchanged after 90 days?


b. After 120 days?


Give explanation. Don't go.ogle.

Answers

Answer:

a. 0.5 grams b. 0.25

Explanation:

Just use the half-life formula(I linked it below)

a. 4(1/2)^90/30=1/2

b. 4(1/2)^120/30=1/4

A. 0.5 grams is left unchanged after 90 days.

B. 0.25 grams is left unchanged after 120 days.

What is Half life?

The half-life of a radioactive isotope is defined as the time it takes for half of the radioactive isotope to decay. It is stable in specific radioactive isotopes. It is unaffected by circumstances. It is also independent of the initial amount of that isotope.

It can be find out by this formula:-

[tex]N=N_o (1/2)^n\\\\[/tex]

where, N=  quantity of the substance remaining

[tex]N_o[/tex]= initial quantity of the substance

n= number of half lives

For above given example,

Given that radioactive element X has a half-life 30 days

Hence the number (n) of half-lives in 90 days

n=90/30

n=3

The amount N of radioactive element X left after n=3 half-lives with initial amount [tex]N_o[/tex]=4 gm

[tex]N=N_o (1/2)^n\\\\\\N= 4 (1/2)^3\\N= 4(1/8)\\N= 0.5 gms\\\\\\[/tex]

For 120 days, n=120/30

so, n= 4

Then,

[tex]N=4(1/2)^4\\N=4(1/16)\\N=0.25 gms[/tex]

Thus, A. 0.5 grams is left unchanged after 90 days.

B. 0.25 grams is left unchanged after 120 days.

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