b. The nose
c. The pharynx
Answer:
The correct answer is A. The larynx moves up against epiglottis when food is swallowed to prevent passage of food into it.
Explanation:
The epiglottis is a moist, cartilaginous structure that is part of the cartilaginous skeleton of the larynx. It also marks the boundary between the oropharynx and the laryngopharynx. The epiglottis obstructs the passage of the bolus at the time of swallowing preventing it from going to the respiratory system.
Larynx closure occurs when the vestibular and vocal folds approach the midline during swallowing. Occasionally, when you eat very fast, solid foods or liquids can enter the larynx.
Answer: tooth comb
Explanation: Strepsirhines are any member of the clade (a group of animals or other organisms derived from a common ancestor species) Strepsirrhini, one of the two suborders (a taxonomic category below order and above infraorder) of primates. They have a special lower incisor called a tooth comb which consists of long, flat teeth with microscopic grooves, and used for grooming the fur.
Strepsirhines, a suborder of primates that includes lemurs and lorises, have a distinct structure called a toothcomb, made up of closely packed lower incisors that stick out from the mouth. Apart from other unique features like a grooming claw, this toothcomb sets strepsirhines apart from other primates.
Strepsirhines, a suborder of primates that includes lemurs and lorises, have unique anatomical features that set them apart from other primates. They have incisors in their lower jaw that are packed closely together and stick out from the mouth in a structure called a toothcomb. This toothcomb, in combination with a clawlike second toe known as a grooming claw, is a hallmark of strepsirhines. The toothcomb is used for grooming - removing debris and parasites from their fur - and is also thought to assist in feeding.
Unlike the majority of mammals, which are diphyodonts and have two sets of teeth in their lifetime, strepsirhines' teeth do not necessarily get replaced. Strepsirhines are found primarily in the 'Old World' - parts of Africa, Asia, and Europe - though a number of species are indigenous to islands such as Madagascar.
#SPJ6
Answer:
Explanation:
The question is not complete. Remaining part of the question is as follows - Minimal growth medium for bacteria such as E. coli includes various salts with characteristic concentrations in the mM range and a carbon source. The carbon source is typically glucose and it is used at 0.5% (a concentration of 0.5 g/100 mL). For nitrogen, minimal medium contains ammonium chloride (NH4Cl) with a concentration of 0.1g/100 mL
How many cells can be grown in a 5 mL culture using minimal medium before the medium exhausts the carbon?
Solution -
We will first find the mass concentration of 0.5 g/100 mL of solution.
gram per ml of glucose
The chemical formula of glucose is
The molecular weight of glucose molecule is grams per mole
Now, we will find the number of moles of glucose in a 5 ml medium -
mole
The number of carbon atom in each glucose molecule is equal to six, thus, number of minimal carbon mole is equal to
mole
Number of carbon atoms is equal to
Carbons
One bacteria has carbon molecule.Thus, ml medium will have bacteria
Answer:
The genotype of a fly can be determined gray bodied is dominant over black body flies. The possible genotype of a fly is either YY or Yy.
Explanation:
The genotype of a gray bodied fly can be determined with the help of punnett square. The gray bodied fly is crossed with black bodied fly, the original genotype of the fly can be determined. This type of cross is known as test cross.
If the cross results in the formation of all gray bodied progeny, the fly is homozygous dominant with the genotype YY. The cross results in the mixture of progeny with gray and black body , then the genotype of a fly is heterozygous dominant Yy.
Thus, the genotype of a fly can be determined by the test cross. The fly may have genotype either YY or Yy.
Answer:
Vermicast.
Explanation:
Vermicast are worm castings, worm manure, or worm faeces.
Answer:
Vermicast.
Explanation:
lol I copied his answer to get points LOL
b. What is the probability that:
(i) The first child that is born will not have moles.
(ii) All of the children will have moles.
(iii) The first two children will have no moles and the last five will have moles.
(iv) Of the 7 children, 4 will have no moles and 3 will have moles
c. Assume this couple now have two children, one with moles and one without moles. What is the probability that the child born without moles is a carrier of the a-allele (ie heterozygous)?
Answer:
a. =
b. i. 0.75
ii. 0.000061
iii. 0.012
iv. 0.17
c. 0.67
Explanation:
a. The expansion of the binomial (p + q)7 would be such that:
=
b. Both couples are heterozygous:
Aa x Aa
AA Aa Aa aa
Since A is dominant over a,
probability of having mole (aa) = 1/4
probability of not having moles = 3/4
Therefore, the probability of the first child not having moles = 3/4 or 0.75
ii. Let the probability of not having mole = p and the probability of having mole = q. From the binomial expansion:
=
Probability that all of the children will have moles =
since p = 3/4 and q = 1/4
= = 0.000061
iii.Probability that the first two children will have no moles and the last five will have moles =
=
= 0.012
iv. Probability that 4 will have no moles and 3 will have moles out of the 7 children =
=
= 0.17
c. Probability that the child born without moles is a carrier of the a-allele = probability of heterozygous.
From the cross in (b), the genotypes of those born without moles are AA and 2Aa. Therefore, the probability of not having moles and be Aa is:
= 2/3 or 0.67