Answer:
Tongue-rolling is dominant, and both parents were heterozygous
Explanation:
Tongue rolling is a type of variation in human beings. When allelic genes are not identical as in Tt, the condition is referred to as heterozygous. An individual with such a genotypic condition is referred to as a heterozygote.
Both parents had a genotype of Tt and they could roll their tongues because T is dominant over the t gene.
T t
T TT Tt
t Tt tt
TT : Tt : tt
1 : 2 : 1
In a heterozygous parent Tt, the chance of producing T gamete from both parents is each 1/2 each. Similarly the chance of forming t gamete is 1/2. Therefore, the chance of gamete T and t fusing to get genotype Tt is 1/2*1/2=1/4.
The ratios of phenotypes are 3/4 will be able to roll the tongue and 1/4 will not be able to roll the tongue.
The mode of inheritance was monohybrid inheritance since one characteristic like tongue-rolling is controlled by a single pair of hereditary factors contributed by both parents.
Because it creates ATP, or active transport against the concentration gradient.
A because everything is fueled because of it. It does not consume or hunt anything.
Suppose that the proportion of the white crest alleles (r) is given by w and that of the Red crest allele (R) is given by p. We have that p+w=1. The probability that an individual has 2 r alleles is given by w*w since for each allele position the probability is w. Only these individuals have a White phenotype. Hence, we get that w^2=
; the right hand side is the proportion of white birds in the total population. Doing the calculations, this yields that w=0.37. From this, we calculate that p=0.63. The possible ways we have heterozygous individuals are the combinations Rr and rR. The probability for each of those is p*w. Thus, the total probability is 2pw. This is equal to 0.466=0.47. This is the fraction of the future population that is going to be heterozygous assuming the conditions of the Handy-Weinberg equilibrium like random reproductive matching etc.
