Answer:
Explanation: 1) Individual who is bald but neither parents has a widow's peak: As having a peak is dominant and the individual is bald with parents that doesn't have the trait, the genotype is recessive, in the case ww;
2) Individual has a widow's peak: The individual's genotype can be WW, homozygous for the trait or Ww, heterozygous for the trait, depending on the genotype of the parents;
3) Individual can't roll their tongue: Being able to roll the tongue is a dominant characteristics, so if the individual can't roll their tongue means the genotype is recessive. But, there is little evidence that this trait is dominant or inheritable;
4) Individual has a widow's peak but their mother does not: In this case, the genotype of the individual is heterozygous (Ww), because the mother is recessive (ww) and the offspring has the trait, so the father must have had it;
5) If an individual does not have a widow's peak (ww), which genotype would be impossible? If this individual crosses over with another individual with the same trait, their children will be all recessive for widow's peak (ww), so there is no possibility of an offspring with the trait;
If the initial individual crosses over with an individual with widow's peak, two possiblities can occur:
- If the crossover is with an individual homozygous for the trait (WW): all their children will have widow's peak with genotype Ww. So, the possibility of not having the peak is 0;
- If the crossover is with an individual heterozygous for the trait (Ww), there will be a probability of 25% for the children to have genotype WW, a probability of 50% for the offspring to be heterozygous (Ww) and has the trait and probability of 25% to be recessive (ww) and therefore doesn't have the trait;
Neuroglial Cells is the answer
Answer:
Frequency of allele A1- 0.41
Explanation:
In Hardy weinberg equilibrium,
P refers to the dominant allele
q refers to the recessive allele
The allele frequency will be p+q=1
The genotypic frequency is- P²+q²+2pq=1
P²= genotype of dominant trait ( A1A1)- 77
2pq= genotype of heterozygotes (2pq)- 65
q²= genotype of recessive trait (A2A2)- 123
Total number of offsprings= 77+ 65+ 123
= 265
Now to calculate allele frequency of A1=
= 77/265 + 1/2( 65//265)
= 0.290+ 0.122
= 0.413
Thus, 0.41 is correct.
Answer;
cells are very different but have similar properties
Explanation;
-Even though there are many different types of cells, they all share similar characteristics. All cells have a cell membrane, organelles organelles, cytoplasm, and DNA. All cells are surrounded by a cell membrane, all cells contain organelles, all cells contain cytoplasm and all cells contain DNA.
-Cells may differ in their number such that some organisms are made of only a single cell while other organisms are made of billions of cells
Answer:
Prokaryotic DNA can be found in the cytoplasm whereas eukaryotic DNA is found in the nucleus, enclosed by the nuclear membrane.
Explanation:
