Answer:
I think it's the second one or first one
Answer:
Answered below.
Explanation:
The nerve fibres of the autonomic nervous system that connects the ganglia to the the central nervous system are called preganglionic fibres. They are divided into sympathetic preganglionic fibres and parasympathetic preganglionic fibres and both have acetylcholine as their neurotransmitter.
The sympathetic preganglionic is shorter compared to the parasympathetic preganglionic fibres. They originate from the hypothalamus and brainstem and project to the ganglia non the spinal cord. A ganglion is a cluster of nerves outside the central nervous system.
The preganglionic sympathetic fibres originating from T1-T2 innervate the cervical ganglion which innervates the muscles of the pupils (dilators).
The postganglionic fibres originating from the preganglionic fibres of T11-L3 passes through splenic, celiac and mesenteric ganglia to innervate the kidneys and renal vessels.
Answer:
The correct answer is - roundworms.
Explanation:
The answer is already mention in the question, however, the detailed answer is as follows:
The characteristics that are given in the question are true tissues, bilateral symmetry, and a pseudocoelom. Worms or helminths are known as primitive form of organization of the Bilaterians. All three group of worms or helmints have a basic bilateral symmetry.
These organisms inaugurated various characteristic that are found and carried by other animals such as true tissues, bilateral symmetry, and a pseudocoelom.
Thus, the correct answer is - roundworms.
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)
