In snakes, being rude (R) is dominant to being respectful (r), and being sneaky (S) is dominant to being sincere (s).
A female snake that is homozygous recessive for both the r and s trait is mated with a male snake that is homozygous dominant for both traits (R and S). Both the R and S traits assort independently from one another.
What is the probability that they will have an offspring that is respectful and sincere?
First, let's determine the possible gametes for each parent based on their genotypes:
The female snake is homozygous recessive for the r and s traits, so her gametes will all carry the recessive alleles: rs.
The male snake is homozygous dominant for the R and S traits, so his gametes will all carry the dominant alleles: RS.
Now, let's determine the possible genotypes of the offspring when these gametes combine:
Since the two traits assort independently, we can use the product rule to calculate the probability. The probability of obtaining an offspring that is respectful (r) and sincere (s) would be the product of the probabilities of each trait.
For the respectful trait:
Probability of inheriting r from the female = 1 (since she is homozygous recessive for r)
Probability of inheriting R from the male = 1/2 (as he could pass on either R or S)
So, the probability of inheriting r = 1 * 1/2 = 1/2.
For the sincere trait:
Probability of inheriting s from the female = 1 (since she is homozygous recessive for s)
Probability of inheriting S from the male = 1/2 (as he could pass on either R or S)
So, the probability of inheriting s = 1 * 1/2 = 1/2.
Now, we can calculate the probability of having an offspring that is respectful and sincere:
Probability of being respectful and sincere = Probability of being r * Probability of being s = 1/2 * 1/2 = 1/4.
Therefore, the probability that they will have an offspring that is respectful and sincere is 1/4.
(Choice A)
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(Choice B)
3
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start fraction, 3, divided by, 16, end fraction
B
3
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start fraction, 3, divided by, 16, end fraction
(Choice C)
16
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C
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(Choice D)
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D
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The correct answer is:
(Choice B)
3
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start fraction, 3, divided by, 16, end fraction
In rabbits, brown fur (B) is dominant to white fur (b) and short fur (H) is dominant to long fur (h).
A brown, long-furred rabbit (Bbhh) is crossed with a white, short-furred rabbit (bbHh). Both the B and H traits assort independently from one another.
What percentage of the offspring will be brown with long fur?
Choose 1 answer:
Choose 1 answer:
(Choice A)
100
%
100%100, percent
A
100
%
100%100, percent
(Choice B)
25
%
25%25, percent
B
25
%
25%25, percent
(Choice C)
50
%
50%50, percent
C
50
%
50%50, percent
(Choice D)
75
%
75%75, percent
D
75
%
75%75, percent
To solve this problem, we need to determine the possible gametes for each parent based on their genotypes:
The brown, long-furred rabbit (Bbhh) can produce the following gametes: Bh and bh.
The white, short-furred rabbit (bbHh) can produce the following gametes: bH and bh.
Now, let's determine the possible genotypes of the offspring when these gametes combine:
For the brown fur trait:
Probability of inheriting B from the first parent = 1/2 (as Bb can produce either B or b)
Probability of inheriting b from the second parent = 1 (since the second parent is homozygous recessive bb)
So, the probability of inheriting B = 1/2 * 1 = 1/2.
For the long fur trait:
Probability of inheriting h from the first parent = 1 (since the first parent is homozygous recessive hh)
Probability of inheriting h from the second parent = 1/2 (as Hh can produce either H or h)
So, the probability of inheriting h = 1 * 1/2 = 1/2.
Now, we can calculate the probability of having an offspring that is brown with long fur:
Probability of being brown with long fur = Probability of B * Probability of h = 1/2 * 1/2 = 1/4.
To convert this to a percentage, we multiply by 100:
Percentage = (1/4) * 100 = 25%.
Therefore, the percentage of the offspring that will be brown with long fur is 25%.
The correct answer is:
(Choice B)
25
%
25%25, percent
In a pea plant, tall (T) is dominant to short (t), and green peas (G) are dominant to yellow peas (g).
A tall plant with yellow peas (Ttgg) is crossed with a short plant with green peas (ttGg). Both the T and G traits assort independently from one another.
What is the chance that the offspring will be short with yellow peas?
Choose 1 answer:
Choose 1 answer:
(Choice A)
1
2
2
1
start fraction, 1, divided by, 2, end fraction
A
1
2
2
1
start fraction, 1, divided by, 2, end fraction
(Choice B)
3
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B
3
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start fraction, 3, divided by, 16, end fraction
(Choice C)
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C
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(Choice D)
3
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start fraction, 3, divided by, 4, end fraction
D
3
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start fraction, 3, divided by, 4, end fraction
To solve this problem, we need to determine the possible gametes for each parent based on their genotypes:
The tall plant with yellow peas (Ttgg) can produce the following gametes: Tg and tg.
The short plant with green peas (ttGg) can produce the following gametes: tG and tg.
Now, let's determine the possible genotypes of the offspring when these gametes combine:
For the short trait:
Probability of inheriting t from the first parent = 1 (since the first parent is homozygous recessive tt)
Probability of inheriting t from the second parent = 1/2 (as ttGg can produce either tG or tg)
So, the probability of inheriting t = 1 * 1/2 = 1/2.
For the yellow pea trait:
Probability of inheriting g from the first parent = 1 (since the first parent is homozygous recessive ttgg)
Probability of inheriting g from the second parent = 1 (since the second parent is homozygous dominant GG)
So, the probability of inheriting g = 1 * 1 = 1.
Now, we can calculate the chance that the offspring will be short with yellow peas:
Chance = Probability of t * Probability of g = 1/2 * 1 = 1/2.
Therefore, the chance that the offspring will be short with yellow peas is 1/2.
The correct answer is:
(Choice A)
1
2
2
1
start fraction, 1, divided by, 2, end fraction