The probable genotype of the IV-3 with the perspective of autosomal inheritance will be A because he does not have condition A. Therefore both parents must be in condition A.
The probable genotype of IV-3 with the perspective of set=linked inheritance will be BbY this is because he has B condition. Therefore the mothers' genotype must be XBXb and the father's genotype must be XBY.
The probability of IV-4 having condition A is 1/36
75% of the kernels born will be Purple, the dominant gene. And, 25% will be the other color. The phenotype might be 3:1 i think, well need anything tag me :)
Answer:
B. No
Explanation:
First, let's watch what it looks like when a population is not evolving. If a population is in a state called Hardy-Weinberg equilibrium, the frequencies of alleles, or gene versions, and genotypes, or sets of alleles, in that population will stay the same over generations (and will also satisfy the Hardy-Weinberg equation). Formally, evolution is a change in allele frequencies in a population over a very long period of time, so a population in Hardy-Weinberg equilibrium is not evolving.
Answer:
B) You measure the quantity of the appropriate pre-mRNA in various cell types and find they are all the same.
Explanation:
In the above scenario, we are using the control gene and we should know about the control gene is that its expression is either at transcription level or at translation level. For this situation we should examine the pre-mRNA stage of control gene in various cells because it is control gene so it should be present in various cells.
So, the best practice is to quantify the pre-mRNA level of gene then we can effectively solve the experimental problem and find the gene expression either it is at transcription or translation level.
Answer:
to get the heat of the planet
Explanation:
