Answer: Water is drawn in through small openings on a sponge's body called OSTIA.
The supportive skeleton of a sponge contains SPICULES which provide protection from predators.
Explanation:
Sponges refer to any of the aquatic animals of the phyllum porifera which is normally characterized with porous structures and has horn-like calcareous internal framework.
Sponges usually have multiple pores called "OSTIA" on their bodies that allow water to enter them.
Most sponges are sessile and because of this, they can't run away from predators. They usually have sharp "SPICULES" which are part of their skeleton that provide them some certain degree of defense. Apart from this, they also produce toxins which poisons predators that attempt to consume them
K,I,m,r,n,s,p,q,o,t. That’s it in order
Mixed Acid Fermentation is a biological process by which a six carbon sugar is converted into complex mixture of acids. It is an example of anaerobic fermentation reaction which is commonly seen in bacteria.
The fermentation produces lactate, acetate, succinate, formate, ethanol, 
and 
and their formation depends upon the presence of certain key enzymes in the bacterium.
The Methyl Red test is used in detecting occurring of mixed acid fermentation pathway when glucose is provided to microbes by using ph indicator. If the fermentation pathway has taken place, the mixture of acids will make the solution very acidic and causes red color change. The Methyl Red test belongs to the group called as IMViC tests.
To learn more about fermentation here
brainly.com/question/13050729
#SPJ4
Answer: The GCF is 4
Explanation: The factors of 52 are 52, 26, 13, 4, 2, 1. The factors of 92 are 92, 46, 23, 4, 2, 1. So the GCF is 4.
Answer: mother: XX^aa, father: X^YAa, son: X^YAa, daughter: X^X^aa.
Explanation:
Color blindness is a genetic disorder that affects the ability to distinguish colors. It is hereditary and is transmitted by an X-linked recessive allele. If a male inherits an X chromosome with the altered allele he will be color blind. In contrast, females, who have two X chromosomes, will only be colorblind if both of their X chromosomes have the altered allele. This is because <u>males have one X chromosome and one Y chromosome, while females have two X chromosomes</u>.
If the woman has normal vision, that means she cannot have both chromosomes affected. She can only have one affected chromosome (be a carrier) or none at all. Also, if she has blue eyes, which is a recessive trait, then both alleles are recessive. But the eye color is not on the X chromosome. For example, her eye color genotype can only be aa, because if she had at least one dominant allele she would have brown eye color. As for the other trait, she can be XX^, with X^ being an affected (carrier) allele or XX, i.e. both normal. So in summary, her genotype can be XXaa or XX^aa
If she has a brown-eyed male child who is also colour blind, he has inherited the allele for colour blindness from his mother, since the father does not pass on an X chromosome to the male children, only the Y. With this we can now rule out the mother's XXaa genotype since she had to have passed on her affected X^ chromosome. Then the genotype of the mother is XX^aa. And since her mother can only pass on one allele to (recessive) because she does not have allele A, the dominant that determines her brown eye color can only come from the father. So the genotype of this son is X^YAa. The female daughter has color blindness and blue eyes. So she had to inherit the affected X^ chromosome from the mother (which we already know she has) and an affected X^ chromosome from the father, because the daughter needs to inherit both affected X^ chromosomes to develop the disease. And if she also has blue eyes, she had to have inherited a recessive allele from the mother and another from the father. So with this information we can say that the father's genotype can only be X^YAa. Because the father must have both A and A alleles of the same eye color, because he passed the dominant one to the son and the recessive one to the daughter. At last, the genotype of the daughter is X^X^aa.
