Answer:
25% colorblind daughter: 25% colorblind son: 25% carrier daughters with normal vision: 25% normal son.
Explanation:
The genotype of a color-blind man is X^cY and the genotype of the heterozygous carrier female is X^cX. A cross between X^cY and X^cX would produce a progeny in following ratio=
25% colorblind daughter: 25% colorblind son: 25% carrier daughters with normal vision: 25% normal son.
Therefore, the couple is likely to have 50% normal son and 50% affected son. Likewise, the couple is likely to have 50% normal daughters and 50% colorblind daughters.
1. DNA strands separate
2. RNA polymerase attached to the gene
3.ribonucletides are assembled
4. mRNA searches from the gene
5. mRNA leaves the nucleus
6. mRNA attaches to ribosome
Answer:
only females
Explanation:
In humans, sex chromosomes in males and females are different. The sex chromosomes found in humans are X and Y chromosomes. X-linked trait is a trait which is inherited on the X- chromosome. According to the question, the trait is passed on a X-linked dominant condition, which means the condition is inherited on the abnormal dominant X-chromosome that will express itself even when in an heterozygous state with a normal X-chromosome.
Hence, a father affected by the condition will have a genotype; XY while a mother that does not have the condition will have a genotype: xx (two normal x chromosomes). Since the Father can only pass his X chromosome to his daughters and never his sons, all his daughters will inherit the condition (see the punnet square in attached image).
N.B: None of the sons will inherit the condition since the mother will pass normal X-chromosomes (x) to her sons.
By examining the tree bark and recording it’s properties. I would note if the cells move and duplicate, I would also note any living organisms which lay on the surface
