A substance that activates plasminogen might be useful to:
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Answer:
The phenotypic ratio for the neurofibromatosis gene would be 1:1.
This is because woman has one dominant allele for neurofibromatosis, so there would be 50% of female gametes with dominant allele for neurofibromatosis. On the other hand, man is normal, so there would be all normal male gametes. Therefore, half of their progeny would have neurofibromatosis from mother.
Further explanation:
A). Dominant autosomal genetic disease.
A dominant autosomal genetic disorder can be defined as disorder, in which one mutated copy of gene (allele) is sufficient to affect the person. It is known as autosomal disorder as genes for this disease are present on autosomes.
B). Red-Green Color blindness.
Red-green color blindness is a sex-linked recessive genetic disorder, which mutated gene is present on X chromosome (sex chromosome). It is know as recessive as it is expressed only if both the recessive copies of X chromosomes are present.
C). Phenotype.
Phenotype of an individual can be defined as a set of features or characteristics, which are observable or expressed physically. For example, height, color of eye or hair, or observable traits for a disease
D).Phenotypic ratio of for diseased gene in children.
People with autosomal dominant disorder are affected with disorder either in heterozygous or homozygous dominant conditions. In this case, female is heterozygous and affected.
Assume genotype of woman is Aa (where 'A' represents dominant allele and 'a' represents recessive allele for neurofibromatosis).
On the other hand, man is normal for neurofibromatosis as he has both recessive alleles for neurofibromatosis. Assume genotype of man is aa (both recessive alleles).
Even if the man is affected with color blindness, there is no role of color blindness (a sex-linked disorder) in inheritance of neurofibromatosis, which is an autosomal disorder.
So, the phenotypic ratio for neurofibromatosis can be determined as:
Parents: Woman X Man
(Aa) (aa)
Gametes: ↓ ↓
A a a
(50%) (50%) (100%)
Children: Aa (with diseased phenotype) = 50%
aa (normal normal phenotype) = 50%
(For more information please find the Punnet square attached with answer).
Thus, phenotypic ratio for the neurofibromatosis would be 1:1 as 50% of their children are affected with disease.
Learn more:
https://brainly.in/question/9964724
Grade: 5-9.
Subject: Biology
Subchapter: Genetics
Keywords: Autosomal dominant genetic disorder, color-blindness, X-linked recessive trait, phenotypic ratio.
