Answer:
Capillary refill time
Explanation:
Capillary refill time refers to the time that it takes for color to return to an external capillary bed after pressure has been applied (which causes blanching). Usually, on healthy individuals, capillary refill time takes less than 2 seconds. If the time is much longer, this could indicate problems such as shock, dehydration or peripheral artery disease.
Specific dye bind to specific nucleotides.
Answer:
In molecular biology, a hybridization probe is a fragment of DNA or RNA of variable length which can be radioactively or fluorescently labeled. It can then be used in DNA or RNA samples to detect the presence of nucleotide substances that are complementary to the sequence in the probe.DNA probes are stretches of single-stranded DNA used to detect the presence of complementary nucleic acid sequences (target sequences) by hybridization. DNA probes are usually labelled, for example with radioisotopes, epitopes, biotin or fluorophores to enable their detection.
spanish
En biología molecular, una sonda de hibridación es un fragmento de ADN o ARN de longitud variable que puede marcarse de forma radiactiva o fluorescente. Luego puede usarse en muestras de ADN o ARN para detectar la presencia de sustancias nucleotídicas que son complementarias a la secuencia en la sonda. Las sondas de ADN son tramos de ADN monocatenario utilizados para detectar la presencia de secuencias complementarias de ácido nucleico (secuencias diana) por hibridación Las sondas de ADN generalmente están marcadas, por ejemplo, con radioisótopos, epítopos, biotina o fluoróforos para permitir su detección.
Explanation:
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 think d is the correct answer or not
