Complete question:
Part C: Using the Punnett square to make predictions
Use the completed Punnett square in Part B to answer the questions below about the F2 generation.
Note: You will find the Punnet square in the attached files.
Note that questions 3 and 4 require a more advanced evaluation of probabilities than do questions 1 and 2 because you have to consider all possible groupings and orders of three F2 seeds. If you need help with these calculations, use Hint 3.
1) What is the probability that an F2 seed chosen at random will be yellow?
2) What is the probability that an F2 seed chosen at random from along the yellow seeds will breed true when selfed?
3) What is the probability that three F2 seeds chosen at random will include at least one yellow seed?
4) What is the probability that three F2 seeds chosen at random will include one green seed and two yellow seeds?
Answer:
1) 3/4
2) 1/3
3) 63/64
4) 27/64
Explanation:
Y is the dominant allele and expresses yellow color y is the recessive allele and expresses green colorSeeds with YY and Yy genotype are yellowSeeds with yy genotype are greenParental) Yy x Yy
Gametes) Y y Y y
Punnet Square)
(1/2) Y (1/2) y
(1/2) Y (1/4) YY (1/4) Yy
(1/2) y (1/4) yY (1/4) yy
F2) 1/4 YY
2/4 = 1/2 Yy
1/4 yy
1) There will be 3/4 of probabilities of getting a yellow F2 seed
1/4 YY + 2/4 Yy = 3/4 Y-
The symbol "-" means that in its position there might be either a Y or y allele.
2) As there are only three yellow possible genotypes for yellow seeds, the fourth genotype for green seeds is not considered. Our pool now is only yellow seeds. So the probability of getting a pure breeding yellow seed among all yellow seeds is 1/3.
1/3 YY
1/3 Yy
1/3 yY
In the pool of yellow seeds, there are three genotypes, and only one of them corresponds to pure breeding yellow seed.
3) To calculate the probability of getting at least one yellow seed among the three seeds randomly chosen, we need to multiply and then perform addition. This is, when we take three seeds from the pool, we have different possibilities of getting seeds with different genotypes. So:
the three chosen seeds could be yellow. In this case, we multiply their probabilities. The probability of getting one yellow seed is 3/4, so, the probabilities of getting three yellow seeds are: 3/4 x 3/4 x 3/4 = 27/64There might also be one green sees and two yellow seed, so we multiply their probabilities in the order in which we might get them. This is:
Two yellow seeds and one green seed: 3/4 x 3/4 x 1/4 = 9/64One yellow seed, one green seed, and one yellow seed: 3/4 x 1/4 x 3/4 = 9/64one green seed, and two yellow seeds: 1/4 x 3/4 x 3/4 = 9/64And we could also get two green seeds and only one yellow seed:
Two green and one yellow seed: 1/4 x 1/4 x 3/4 = 3/64one green, one yellow and one green: 1/4 x 3/4x 1/4 = 3/64one yellow seed and two green seeds: 3/4 x 1/4 x 1/4 = 3/64Now, having all the possibilities, we just need to add all the possibilities:
27/64 + 9/64 + 9/64 + 9/64 + 3/64 + 3/64 + 3/64= 63/64.
Note that we did not include the possibility of getting three green seeds (1/4 x 1/4 x 1/4). This is because we were asked to calculate the probabilities of getting AT LEAST one yellow seed.
4) To calculate the probability of getting two yellow seeds and one green seed among the three seeds randomly chosen, we should multiply probabilities and then sum them up.
The orders in which we could get the seeds are:
Two yellows and one green: 3/4 x 3/4 x 1/4 = 9/64One yellow, one green, and one yellow: 3/4 x 1/4 x 3/4 = 9/64one green seed and two yellow seeds: 1/4 x 3/4 x 3/4 = 9/64Now, having all the possibilities, we just need to add all the possibilities:
6/64 + 9/64 + 9/64 = 27/64.
Answer:
1) 3/4
2) 1/3
3) 63/64
4) 27/64
Explanation:
1) yellow seed could either be GG, Gg or Gg from Punnett square, therefore, there is a 3/4 probability
2) true-breeding = either GG or gg. Therefore, out of the 3 yellow seed options, only 1 is true-breeding, therefore, it's a 1/3 probability
3) "there are 7 possible groupings that have at least 1 yellow seed; the sum of their individual probabilities yields the final answer of 63/64. Note that you could have also calculated this answer by subtracting the probability of the one category that doesn’t fit the criterion (Green, Green, Green) from the total probability (1 – 1/64 = 63/64)"
4) "there are 3 possible groupings that have 1 green and 2 yellow seeds; the sum of their individual probabilities yields the final answer of 27/64"
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Which statement describes one aspect of an earthquake's magnitude?
It is measured by the Mercalli scale.
It is based on the size of seismic waves.
O It is not usually affected by the amount of fault movement.
It is used to determine damage caused by earthquakes.
Answer:
b. It is based on the size of seismic waves
Explanation:
just did the test
The statement that describes one aspect of an earthquake's magnitude is: "It is based on the size of seismic waves."
What is an earthquake's magnitude?An earthquake's magnitude is a measure of the amount of energy released by the earthquake, and it is determined by measuring the amplitude (height) of the seismic waves generated by the earthquake. The size of the seismic waves is directly related to the amount of energy released by the earthquake, and this is used to calculate the earthquake's magnitude.
The most commonly used scale for measuring earthquake magnitude is the Richter scale, which is based on the amplitude of the largest seismic wave recorded on a seismogram. Other scales, such as the moment magnitude scale, are based on the seismic moment, which takes into account the size of the fault rupture and the amount of slip on the fault.
Learn more about earthquake, here:
https://brainly.com/question/29500066
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150 ml of 54% CaCI2 solution contains how many grams of CaCI2?
Answer:
mass of CaCl₂ = 67.4 g of CaCl₂
Explanation:
A 45% CaCl₂ Solution contains 45 g of solute per 100 mL of solution.
mass concentration = mass of solute (g)/ volume of solution (L)
100 mL of solution = 100 * 1L /1000 = 0.1 L
Mass concentration of solution = 45 g/0.1 L = 450 g/L
Molar concentration of solution = mass concentration / molar mass
molar mass of CaCl₂ = 111 g/mol
molar concentration = 450 g/L / 111 g/mol = 4. 05 mol/L
Number of moles of CaCl₂ present in 150 ml of 4.05 mol/L solution = molar concentration * volume (L)
number of moles of CaCl₂ = 4.05 * 150 * 1 L/1000 = 0.6075 moles
mass of CaCl₂ present in 0.6075 moles = number of moles * molar mass
mass of CaCl₂ = 0.6075 * 111 = 67.4 g of CaCl₂