The organism that would have the most variation in the DNA of its offspring is the cat (Option C). Meiosis is a type of cell division that generates more genetic variability than asexual types of reproduction.
Meiosis is a type of reductional cell division by which a parental cell produces 4 daughter cells (gametes), each containing half of the genetic material.
Animals (e.g., cats) generate gametes by meiosis which fuse during fertilization to produce new offspring.
Both amoeba and bacteria reproduce by a type of asexual reproduction called binary fission. Moreover, yeasts also reproduce asexually by a process called budding and fission.
Both asexual and sexual types of reproduction generate genetic variability by the emergence of new mutations in daughter cells.
Meiosis generates much more genetic variability than asexual types of reproduction due to two different processes:
- Random assortment of chromosomes, which produces new allele combinations.
- Recombination, i.e., by the exchange of genetic material (DNA) between non-sister chromatids during Prophase I.
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brainly.com/question/7002092
The particles in a solid are tightly packed and locked in place. Although we cannot see it or feel it, the particles are vibrating in place.
As these molecules heat up, they will vibrate more vigorously, and will eventually turn to water, then gas.
Answer:
C
Explanation:
The concept behind, is mole ratio of Al:FeO
<h3>
Answer:</h3>
1.25 moles (R.T.P.) or 1.34 moles (S.T.P.)
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Explanation:</h3>
- 1 mole of a gas occupies a volume of 24 liters at room temperature and pressure (R.T.P.)
- On the other hand, 1 mole of a gas will occupy 22.4 Liters at standard temperature and pressure (S.T.P.)
Therefore, at R.T.P.
30.0 Liters will be equivalent to;
= 30.0 L ÷ 24 L
= 1.25 moles
At S.T.P
30.0 Liters will be equivalent to;
= 30.0 L ÷ 22.4 L
= 1.34 moles
Thus, 30.0 L of helium gas are equivalent to 1.25 moles of He at R.T.P. and 1.34 moles at S.T.P.
Answer:
P₂ = 299.11 KPa
Explanation:
Given data:
Initial volume = 600 mL
Initial pressure = 70.00 KPa
Initial temperature = 20 °C (20 +273 = 293 K)
Final temperature = 40°C (40+273 = 313 K)
Final volume = 150.0 mL
Final pressure = ?
Formula:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
Solution:
P₂ = P₁V₁ T₂/ T₁ V₂
P₂ = 70 KPa × 600 mL × 313 K / 293K ×150 mL
P₂ = 13146000 KPa .mL. K /43950 K.mL
P₂ = 299.11 KPa
