Color Blindness is x-linked recessive. Therefore, it could not be heterozygous; the daughter would not be colorblind, but rather have normal vision.
The probability that the mother produces a gamete with the allele for dimples is 50%. This is because each parent only passes on one of their two alleles for a given trait to their offspring, and in this case, the mother has a 50% chance of passing on the allele for dimples.
If both parents carry the trait then there is a only a 25 percent chance the pregnancy will abort but there is a 75 percent chance the child will carry the trait and that can cause abnormalities during the pregnancy and afterwards.
The egg is the gamete produced by the female.
The probability of a child being color blind depends on the parents' genetic makeup. If the mother is a carrier of the color blindness gene located on the X chromosome and the father does not carry the gene, the chances are 0% for a daughter and 50% for a son to inherit color blindness. If the father is color blind and the mother is a carrier, the chances are 50% for a daughter and 50% for a son to inherit color blindness.
We'llsay F is dominant for freckles and f is recessive for non-freckled. The father is ff The mother is Ff The child is ff. Probability of this cross producing a homozygous recessive child is 50%. There isn't a precise term for this cross.
50%
50% (apex)
The probability that the mother produces a gamete with the allele for dimples is 50%. This is because each parent only passes on one of their two alleles for a given trait to their offspring, and in this case, the mother has a 50% chance of passing on the allele for dimples.
If the mother has one LL allele for color lameness, it means she can only pass on the L allele to her offspring. Since she has one allele for normal vision, that allele does not affect the color trait directly. Therefore, the probability that her child will inherit the LL genotype for color lameness is 100%, as the mother can only provide the L allele.
A male with red-green color blindness has the genotype X^cY, where X^c represents the X chromosome carrying the recessive allele for color blindness. He inherited the recessive allele from his mother, who must have at least one X chromosome with the color blindness allele (either X^cX or X^cY). Since males inherit their X chromosome from their mothers and their Y chromosome from their fathers, the mother's genotype determines whether the son will express the condition.
Color blindness is a sex-linked trait that is carried on the X chromosome. Since males inherit their single X chromosome from their mother, if the mother carries the allele for color blindness, her son will inherit it and be colorblind. Females need two copies of the allele to be colorblind, so they can be carriers without exhibiting the trait.
This depends entirely on the genotype of the parents. The probability of getting a specific genotype is the probability of getting the correct allele from mother (1/2) multiplied by the probability of getting the correct allele from father (1/2) multiplied by the number of ways this can occur. The probability of getting a phenotype, if the phenotype is dominant, is the sum of the probability of getting two dominant alleles, and the probability of getting one dominant allele. If the phenotype is recessive, the probability is equal to the probability of getting two recessive alleles.
(Apex Learning) She has at least one recessive color blindness allele.
In sexual reproduction each parent contributes only one allele to the offspring. This is why meiosis takes diploid cells and makes them haploid. The process meiosis separates the homologous pairs, separating the alleles from each other. Thus, each gamete produced has only one allele for each trait. When the male gamete (sperm) fuses with the female gamete (egg) and fertilization takes place, the resulting zygote has two alleles; one from the father and one from the mother.
In this scenario, the mother is heterozygous for normal skin pigmentation (Aa) and the father is homozygous recessive for albinism (aa). The possible genotypes for their child are Aa (normal pigmentation) and aa (albino). Using a Punnett square, there is a 50% probability that the child will be albino (aa).
If the couple has a color-blind son, it would indicate that color blindness is caused by an X-linked recessive allele. This is because sons inherit their single X chromosome from their mother, who carries the recessive allele for color blindness but does not express it due to her second X chromosome providing the normal color vision gene.
The probability would be 0.5 or 50%. A heterozygous woman will pass on the X chromosome with the recessive allele to 50% of her sons, and since the disorder is recessive, the son would only have the disorder if the X chromosome with the recessive allele is inherited from the mother.