Answer:
prokaryotic cells do not have a nucleus or membrane bound organelles.They also have small rings of double-stranded extra-chromosomal DNA called plasmids.
Answer:
The next generation average time to flowering will be 98 days.
Explanation:
Before answering the question, we need to know a few concepts.
- Artificial selection is the selecting practice of a specific group of organisms in a population -that carry the traits of interest- to be the parents of the following generations.
- Parental individuals carrying phenotypic values of interest are selected from the whole population. These parents interbreed, and a new generation is produced.
- The selection differential, SD, is the difference between the mean value of the trait in the population (X₀) and the mean value of the parents, (Xs). So,
SD = Xs - X₀
- Heritability in the narrow-sense, h², is the genetic component measure to which additive genetic variance contributes. The heritability might be used to determine how the population will respond to the selection done, R.
h² = R/SD
- The response to selection (R) refers to the metric value gained or lost from the cross between the selected parents. R can be calculated by multiplying the heritability h², with the selection differential, SD.
R = h²SD
R also equals the difference between the new generation phenotypic value (X₁) and the original population phenotypic value (X₀),
R = X₀ - X₁
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Now that we know these concepts and how to calculate them, we can solve the proposed problem.
<u>Available data: </u>
- trying to decrease the maturation time in a population of sunflowers.
- the population mean time to flower is 100 days → X₀
- Chosen parental Plants mean flowering time is 90 days → Xs
- the narrow-sense heritability for flowering time is 0.2 → h²
According to what we sow previously, we need to find out the value of X₁, which reflects the next generation average time to flowering.
- We know that R = X₁ - X₀, so we need to clear this formula to calculate X₁
X₁ = R + X₀
We already know that X₀ = 100 days,
Now we need to calculate R.
We know that h² = 0.2,
Now we need to calculate SD
Xs = 90 days → Parentals media flowering time
X₀ = 100 → Population media flowering time
SD = Xs - X₀
SD = 90 - 100
SD = - 10 days
Knowing this, we can calculate R
o h² = 0.2
o SD = - 10
R = 0.2 x (-10)
R = - 2
- Finally, once we know the R-value we can calculate the X₁ value
X₁ = R + X₀
X₁ = - 2 + 100
X₁ = 98
Answer:
Red blood.
Explanation:
Red blood is for oxygenated blood, while blue blood is for deoxygenated blood.
The blood in Isabelle's left ventricle would be red, in other words, oxygenated. The left ventricle has oxygenated blood because this blood was first in the right ventricle. Then, it went to the pulmonary arteries, which led the blood to the capillaries close to the alveoli in the lungs. In this area, the deoxygenated blood dropped the CO₂ and took O₂ becoming oxygenated blood. Now, this blood is color-coded red and will go to the pulmonary venules, then to the pulmonary vein, and from there, it will go to the left atrium. Lastly, it will go to the left ventricle to start the systemic circulation, which is the one that distributes the oxygenated blood in the body.
Answer and Explanation:
Throughout the world, the most different species of organisms are insects. As compared to the other species, the number of insect species is greater. The exact number of insect species can be estimated from previous and current studies. Approximately nine hundred thousand species of insects are known. Insectivorous birds are those birds that feed on insects. For numerous growing insectivorous birds, insects are used as a source of protein. One of the most common kinds of insects is beetles. They are found everywhere. Beetles eat fruits, leaves, and other parts of the plant. The reproductive and fertility ability of insects is remarkable. In habitat 2, beetles and insectivorous birds were living. If all insectivorous birds and remaining beetles were removed from habitat, two and five hundred additional light tan beetles and 500 dark brown beetles were released in habitat 2. After one more week, one thousand dark brown beetles would be expected to be recaptured in the habitat 2.
