In humans, the ABO blood type is under the control of autosomal multiple alleles. Red-green color blindness is a recessive X-linked trait. If two parents who are both type A and have normal vision produce a son who is color-blind and type O, what is the probability that their next child will be a female who has normal vision and is type O?
Hi can someone please help me. Thank you
To find the probability of having a female child who has normal vision and is type O, we need to consider the inheritance of two traits: blood type and color blindness.
1. Blood type:
Both parents are type A, which means they carry two copies of the A allele (A/A). However, since they have a son who is type O, we can determine that they both must also carry a copy of the O allele. Therefore, their genotypes can be represented as A/O.
2. Color blindness:
Since the son is color-blind, we can conclude that the mother must be a carrier of the recessive color-blindness allele (Xc) and the father must have a normal vision (X). Therefore, the mother's genotype can be represented as Xc/X, and the father's genotype as X/Y.
To determine the probability of having a female child who has normal vision and is type O, we need to consider the inheritance of these traits separately:
1. Blood type:
If the parents are both A/O, there is a 50% chance of passing on the O allele to their child.
2. Color blindness:
Since the father does not carry the color-blindness allele, there is no chance that the child will be color-blind.
Considering the two traits are independent:
1. The probability of passing on the O allele for blood type is 50%.
2. The probability of having a female child who does not inherit the color-blindness allele is 100%.
To obtain the probability of both events occurring together, we multiply the probabilities:
50% (blood type) x 100% (normal vision)
= 0.5 x 1
= 0.5 or 50%
Therefore, the probability that the parents' next child will be a female who has normal vision and is type O is 50%.
Certainly! I'll be glad to help you with that.
To determine the probability of having a female child who has normal vision and is blood type O, we need to consider two separate events: the inheritance of the ABO blood type and the inheritance of color blindness.
Let's break down the problem step by step:
1. ABO blood type: The parents are both type A, which means they each have two A alleles. Since A and O are codominant, their child might have inherited one A allele from each parent or an A allele from one parent and an O allele from the other.
2. Color blindness: The son being color-blind tells us that he inherited the recessive allele for color blindness from his mother, as only females can carry the recessive allele while males express it. This means the mother must be a carrier for color blindness and has a single recessive X allele.
Now, let's consider the possible genotypes of the parents and calculate the probability of having a female child with normal vision and blood type O:
Father's genotype: AA (blood type A, normal vision)
Mother's genotype: XAXa (blood type A, carrier for color blindness)
Probability of passing on alleles:
- The father can only pass on an A allele (100% chance).
- The mother can pass on either an A or O allele for blood type (50/50 chance), and either an XA or Xa allele for color vision (50/50 chance).
Using a Punnett square or by multiplying the probabilities together, we can calculate the probability of each possible genotype for the female child:
Genotype AAXA:
- Probability of inheriting A allele for blood type from the mother: 1 (100%)
- Probability of inheriting A allele for blood type from the father: 1 (100%)
- Probability of inheriting XA allele for color vision from the mother: 1 (100%)
- Probability of being a female: 0.5 (50% chance)
Genotype AAXa:
- Probability of inheriting A allele for blood type from the mother: 1 (100%)
- Probability of inheriting A allele for blood type from the father: 1 (100%)
- Probability of inheriting Xa allele for color vision from the mother: 0.5 (50% chance)
- Probability of being a female: 0.5 (50% chance)
The probability of having a female child with normal vision and blood type O is the sum of the probabilities of these two genotypes:
Probability = (0.5 * 1 * 1 * 1) + (0.5 * 1 * 0.5 * 1) = 0.5 + 0.25 = 0.75
Therefore, the probability that their next child will be a female who has normal vision and is type O is 0.75, or 75%.
If they both have A type blood with an O son, their genotypes for blood are both AO. The probability of getting an O child is 25%. Check with a Punnett square.
If the parent both have normal vision, the mother must be a carrier for the color-blindness. The probability of getting a normal vision female is also 25%. Must get normal X from both mom and pop.
The probability of 2 or more events occurring is obtained by multiplying the probability of the separate events. You can do this.
I hope this helps. Thanks for asking.