Hey there!
The answer would be ultrasound.
The ultrasound is considered safe because it uses sound waves for the detection of tissue, organ, baby etc. These waves are considered safe and transfer only a small amount of heat to the tissue in which the ultrasound is done. A little amount of heat is not at all harmful to the body in any means. Early defects of baby can be detected through this procedure, the merits of the process is much higher than its demerits.
Hope this helps !
Answer:enzymes
Explanation:during digestion, food is broken down by chewing in the mouth.enzymes also acts on foods to reduce them into simpler constituents.enzymes acts on food in the mouth, stomach, intestine etc.
Enzymes that acts on proteins helps to break the peptides bonds present in proteins.they break up the polypeptide chains into amino acids .An example is trypsin .the conditions necessary for these enzymes to acts may be specific.some require acidic environment while others require basic environment.pepsin for example requires stomach hydrochloric acid to be converted from it's inactive form, pepsinogen.
The resultant Amino acids are then absorbed in the small intestine
The correct answer here is certainly B. They allow scientists to perform new investigations and test new ideas they may not have been able to do in the past because scientists have been able to do the other 3 options without modern technology.
Answer:
C. It is a male with atleast one dominant allele
Explanation:
In the given pedigree, the two normal parents of the generation I have one daughter with the attached earlobe. Since the trait is recessive, the daughter should be homozygous recessive to express the trait. The genotype of the daughter (shaded circle in generation II) is "aa". To have a daughter with "aa" genotype, both the parents should have one copy of "a" allele. So, the genotype of both parents is "Aa".
In generation II, individual A is non-shaded square. Squares represent males in a pedigree. Since its not shaded, it does not have attached earlobe. Both the parents are heterozygous dominant for attached earlobes (Aa x Aa = 1/4 AA : 1/2 Aa : 1/4 aa). The genotype of this individual may be AA or Aa.
