Answer:
The phenotype of the brown parent would be heterogeneous.
Explanation:
As albino is a recessive trait, both the alleles should be recessive for this trait to occur.
A punnet square is a diagram which is made to illustrate the possible phenotype of the offspring.
If the brown parent will be homologous, then a cross with albino parent will produce offspring with brown color having heterogeneous phenotype.
For production of albinos the brown parent should be heterogeneous for the trait.
Answer:
The frequency of recessive allele should be 0.30.
Explanation:
According to Hardy-Weinberg equilibrium, sum of both dominant allele frequency (p) and recessive allele frequency (q) should be equal to 1 (p+q = 1).
Brown hair = dominant (p)
white hair = recessive (q)
49% mice are brown hair
so dominant genotype frequency = 0.49
According to Hardy-Weinberg principle, square root of the genotype (homozygous) is equal to allele frequency.
√p =√0.49 = 0.70
The dominant allele frequency is 0.7
Now, by this value we can find the recessive allele frequency by
p + q = 1
1 - 0.7 = 0.3
So, the recessive allele (white hair mice) frequency is 0.30.
The start codon is almost always AUG
2 producers tell me if right
