In a mating between IV-2 and IV-5, what is the chance that the child produced would have both nail-patella syndrome and alkapton
1 answer:
The chance that the child produced would have nail-patella syndrome from this mating is 1/8 (12.5%) and the chance that the child has alkaptonuria is zero (0%). It is a case of recessive phenotypes.
<h3>Pedigree and genetic inheritance</h3>The answer is incomplete, but I have now included the Figure to answer it. The vertical lines indicate nail-patella syndrome, while horizontal lines indicate alkaptonuria.
Pedigree is a diagram showing the genealogical relationships in different individuals, which is usually used to represent the inheritance pattern of a given phenotype.
In this case, the nail-patella syndrome and alkaptonuria are recessive phenotypes, which means that individuals may be heterozygous carriers but they need to have both mutated alleles to express each condition.
In the pedigree, it is possible to observe that the individual IV5 expresses both syndromes (nail-patella syndrome and alkaptonuria), whereas there is a 1/8 chance that the individual IV2 is a carrier of the nail-patella syndrome.
Learn more about pedigree here:
You might be interested in
Hardy Weinberg Equilibrium
Explanation:
- Hardy Weinberg Equilibrium formula for allelic frequency is:
- Hardy Weinberg Equilibrium formula for genotypic frequency is:
- Here,
=frequency of dominant allele
=frequency of recessive allele
=frequency of dominant genotype
=frequency of recessive genotype
=frequency of heterozygous genotype
- frequency of affected individuals=0.0004(given)
here,affected individuals means the one's having Cystic Fibrosis and since it is recessive disease so we can write frequency of recessive genotype as: =0.0004
- If
=0.0004
will be 0.02;frequency of recessive allele will be 0.02
- Using the formula
- hence,frequency of dominant allele will be 0.98
- Frequency of dominant genotype will be
- Frequency of heterozygous genotype will be