It is the rib cage (thoracic cage)
Answer: The probability is 100%.
Explanation: The chromosome X is related to gender. A man carries two different chromosomes (XY) and a woman carries two similar chromosomes (XX). The offspring of a couple will inherited:
- If it is a boy, he inherites the chromosome Y from his father and X from his mother;
- If it a girl, she inherites X from her father and X from her mother;
In the question, the woman is color-blind, which means she carries the recessive alleles: 
The man, on the other hand, is normal, so his X-linked genotype will be: 
.
Now, since they have a daughter, she inherited her double X from each of her parents, which means she inherited one 
from her mother and one 
from her father.
So, the daughter's genotype can only be heterozygotic in other words, the probability is 100%.
There is not any nucleus and nuclear bound organelles in prokaryotes....
Answer:
1/8 (12.5%)
Explanation:
An autosomal recessive disease is an inherited disease in which an individual need to receive both defective alleles at the same gene <em>locus</em> to be expressed in the phenotype. In this case, both parents are carriers of the recessive mutant allele associated with the sickle cell anaemia trait, thereby both parents are heterozygous, ie., each parent has one copy of the normal allele 'H' and one copy of the defective mutant allele 'h' associated with this condition. In consequence, their first child has a 1/4 (25%) chance of having sickle-cell anaemia. Moreover, the chance of having a girl is 1/2 and the chance of having a boy is 1/2, thereby the final chance of having a girl sickle cell anaemia individual is 1/4 x 1/2 = 1/8 (12.5%).
- Parental cross for sickle cell anaemia trait = Hh x Hh >>
- F1 = 1/4 HH (normal); 1/2 Hh (normal); 1/4 hh (sickle cell anaemia) >>
- Sex proportion of sickle cell anaemia individuals = 1/8 female sickle cell anaemia individuals + 1/8 male sickle cell anaemia individuals (1/8 + 1/8 = 1/4)
I believe the answer is B. Hope I help! Sorry if not. ):)
