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Two parents with genotype AaBbcc and aaBbCc are crossed. If the genes assort independent to one another, the probability that progeny of the genotype aabbcc in the F1 would be ______(Round off to two decimal places).
Correct answer is between '0.05,0.07'. Can you explain this answer?
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Two parents with genotype AaBbcc and aaBbCc are crossed. If the genes ...
If the genes assorts independently, each of the genetic cross for the three genes can be assumed as independent event. So, probability of aa is ½ from the Aa X aa; Probability of bb is 1/4 from the BbX Bb; Probability of cc is ½ from the cc X Cc. Therefore applying the product rule, the Probability of aabbccoffpring = ½ X ¼ X ½ = 1/16 or 0.0625.
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Two parents with genotype AaBbcc and aaBbCc are crossed. If the genes ...
To determine the probability of obtaining progeny of the genotype aabbcc in the F1 generation, we need to consider the principles of Mendelian genetics and the laws of independent assortment.
The genotype of the first parent is AaBbcc, which means it has heterozygous alleles for the A and B genes and homozygous recessive alleles for the C gene. The genotype of the second parent is aaBbCc, which means it has homozygous recessive alleles for the A and C genes and heterozygous alleles for the B gene.
According to the law of independent assortment, the alleles for different genes segregate independently of one another during gamete formation. This means that the alleles for the A gene will segregate independently of the alleles for the B and C genes.
To determine the possible genotypes of the offspring, we can create a Punnett square:
AaBbCc AaBbCc
aaBbCc aaBbCc aaBbCc
aaBbCc aaBbCc aaBbCc
In the Punnett square, each box represents a possible combination of alleles from the two parents. Since the genes assort independently, the probability of each box is equal.
There are a total of 8 boxes in the Punnett square, and we are interested in the genotype aabbcc. From the Punnett square, we can see that only one box has the genotype aabbcc. Therefore, the probability of obtaining progeny of the genotype aabbcc is 1 out of 8.
To calculate the probability as a decimal, we divide 1 by 8:
Probability = 1/8 ≈ 0.125
Rounded off to two decimal places, the probability is approximately 0.13.
Since the correct answer is between 0.05 and 0.07, the given answer of 0.13 is incorrect.
Therefore, the correct answer cannot be determined based on the given information.
The genotype of the first parent is AaBbcc, which means it has heterozygous alleles for the A and B genes and homozygous recessive alleles for the C gene. The genotype of the second parent is aaBbCc, which means it has homozygous recessive alleles for the A and C genes and heterozygous alleles for the B gene.
According to the law of independent assortment, the alleles for different genes segregate independently of one another during gamete formation. This means that the alleles for the A gene will segregate independently of the alleles for the B and C genes.
To determine the possible genotypes of the offspring, we can create a Punnett square:
AaBbCc AaBbCc
aaBbCc aaBbCc aaBbCc
aaBbCc aaBbCc aaBbCc
In the Punnett square, each box represents a possible combination of alleles from the two parents. Since the genes assort independently, the probability of each box is equal.
There are a total of 8 boxes in the Punnett square, and we are interested in the genotype aabbcc. From the Punnett square, we can see that only one box has the genotype aabbcc. Therefore, the probability of obtaining progeny of the genotype aabbcc is 1 out of 8.
To calculate the probability as a decimal, we divide 1 by 8:
Probability = 1/8 ≈ 0.125
Rounded off to two decimal places, the probability is approximately 0.13.
Since the correct answer is between 0.05 and 0.07, the given answer of 0.13 is incorrect.
Therefore, the correct answer cannot be determined based on the given information.
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Two parents with genotype AaBbcc and aaBbCc are crossed. If the genes assort independent to one another, the probability that progeny of the genotype aabbcc in the F1 would be ______(Round off to two decimal places).Correct answer is between '0.05,0.07'. Can you explain this answer?
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Two parents with genotype AaBbcc and aaBbCc are crossed. If the genes assort independent to one another, the probability that progeny of the genotype aabbcc in the F1 would be ______(Round off to two decimal places).Correct answer is between '0.05,0.07'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about Two parents with genotype AaBbcc and aaBbCc are crossed. If the genes assort independent to one another, the probability that progeny of the genotype aabbcc in the F1 would be ______(Round off to two decimal places).Correct answer is between '0.05,0.07'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two parents with genotype AaBbcc and aaBbCc are crossed. If the genes assort independent to one another, the probability that progeny of the genotype aabbcc in the F1 would be ______(Round off to two decimal places).Correct answer is between '0.05,0.07'. Can you explain this answer?.
Two parents with genotype AaBbcc and aaBbCc are crossed. If the genes assort independent to one another, the probability that progeny of the genotype aabbcc in the F1 would be ______(Round off to two decimal places).Correct answer is between '0.05,0.07'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about Two parents with genotype AaBbcc and aaBbCc are crossed. If the genes assort independent to one another, the probability that progeny of the genotype aabbcc in the F1 would be ______(Round off to two decimal places).Correct answer is between '0.05,0.07'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two parents with genotype AaBbcc and aaBbCc are crossed. If the genes assort independent to one another, the probability that progeny of the genotype aabbcc in the F1 would be ______(Round off to two decimal places).Correct answer is between '0.05,0.07'. Can you explain this answer?.
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Two parents with genotype AaBbcc and aaBbCc are crossed. If the genes assort independent to one another, the probability that progeny of the genotype aabbcc in the F1 would be ______(Round off to two decimal places).Correct answer is between '0.05,0.07'. Can you explain this answer?, a detailed solution for Two parents with genotype AaBbcc and aaBbCc are crossed. If the genes assort independent to one another, the probability that progeny of the genotype aabbcc in the F1 would be ______(Round off to two decimal places).Correct answer is between '0.05,0.07'. Can you explain this answer? has been provided alongside types of Two parents with genotype AaBbcc and aaBbCc are crossed. If the genes assort independent to one another, the probability that progeny of the genotype aabbcc in the F1 would be ______(Round off to two decimal places).Correct answer is between '0.05,0.07'. Can you explain this answer? theory, EduRev gives you an
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