Answer:
Protiens
Explanation:
because the DNA or RNA is translatted to sequence.
Answer:
a. Stabilizing selection
Explanation:
Babies lower than 5.5lbs would suffer from various disorder because of low birth weight as their body would not be able to compete as well they would have weak immune system and malnourishment. Whereas high birth weight carry the equally above risk as their body wont be able to cope up and organs become insensitive to various hormones. for the survival the birth weight should be within a median range so that it would be favorable for themselves and surroundings.
Stabilizing selection occurs when a population stabilizes on a particular trait value and genetic diversity decreases. As in this example, an average baby weight has to favour against extreme variation for it's better survival.
therefore, this is an example of stabilizing selection.
If you were to draw a punnet square of two individuals with the genotypes Ss, there would be 1 box with ss, 2 boxes with Ss, and 1 box with SS.
The possible genotypes are SS, Ss, and ss.
The possible phenotypes are sufferers of sickle cell (ss) and non-sufferers (Ss and SS)
<span><u>Mitosis </u>
</span><span><span>
Mitosis </span>and meiosis are simply cell division processes that occurs differently, they're characteristically divergent from each other according to their function and structure. Mitosis is the cell division that happens in all cells in the human body except sperm and egg cells. They produce diploid cells. Meiosis on the other hand is responsible for the cell division of the gametes, spermatogenesis (sperm cells) and oogenesis (egg cells), such haploid cells. </span>
Answer:
CH₄ is the excess reagent and SO₄²⁻ is the limiting reagent.
0.040 g
Explanation:
Step 1: Write the balanced reaction for the methane anaerobic oxidation
CH₄ + SO₄²⁻ ⇒ S²⁻ + CO₂ + 2 H₂O
Step 2: Establish the theoretical mass ratio between the reactants
According to the balanced equation, the theoretical mass ratio of CH₄ to SO₄²⁻ is 16.04:96.06 = 0.1670:1
Step 3: Establish the experimental mass ratio between the reactants
50 mg (0.050 g) of CH₄ react with 60 mg (0.060 g) of SO₄²⁻. The experimental mass ratio of CH₄ to SO₄²⁻ is 0.050:0.060 = 0.83:1
Comparing both mass ratios, we can deduce that CH₄ is the excess reagent and SO₄²⁻ is the limiting reagent.
Step 4: Calculate the mass excess of CH₄
The mass of CH₄ that reacts with 0.060 g of SO₄²⁻ is:
0.060 g SO₄²⁻ 16.04 g CH₄/96.06 g SO₄²⁻ = 0.010 g CH₄
The mass excess of CH₄ is:
0.050 g - 0.010 g = 0.040 g
