What is not contained in our blood?A. Blood Plasma
B. Leukoctyes
C. Erythrocytes
D. Lymph

Answers

Answer 1

Answer: i think it might be lymph, but its not one of the first two

Answer 2

Answer:

Our blood contains blood plasma, leukocytes, and erythrocytes but does not contain lymph. Lymph is a separate body fluid, transported by the lymphatic system.

Blood consists of several components: blood plasma, leukocytes (white blood cells), and erythrocytes (red blood cells). Blood plasma is a yellowish liquid component of blood that holds the blood cells of whole blood in suspension. It makes up about 55% of the body's total blood volume. Leukocytes are the cells of the immune system that are involved in protecting the body against both infectious disease and foreign invaders. Erythrocytes, or red blood cells, are cells present in blood in order to transport oxygen. On the other hand, lymph, although a body fluid, is not a part of the blood. It is mainly composed of white blood cells and is transported by the lymphatic system, separate from the circulatory system that transports blood.

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Tail length in mammals is a heritable trait. A pig with a 6 cm tail was mated to a pig with a 30 cm tail. All F1 offspring had a tail of length 18 cm. An F2 generation was made by crossing F1 individuals. This resulted in many piglets whose tails ranged in 4-cm intervals from 6 to 30 cm (6, 10, 14, 18, 22, 26, 30). The majority of pigs had 18 cm tails, 1/64 had 6 cm tails, and 1/64 had 30 cm tails.These results are consistent with what genetic model?

(a) Two genes, each with two alleles that show dominance
(b) Two genes, each with two alleles that act additively
(c) Three genes, each with two alleles that show dominance
(d) Three genes, each with two alleles that act additively

An F2 piglet with an 18 cm tail is heterozygous at each of the tail-length loci, and is mated to an F2 piglet with a 6 cm tail. Write down the predicted offspring genotypes and calculate the predicted tail lengths. What is the expected frequency of piglets with a 14 cm tail length?

Answers

Answer:

The answer is (d) Three genes, each with two alleles that act additively.

Explanation:

1/ The genotype of P is AABBCC and aabbcc, for example. Then, F1 should be AaBbCc.

The genotype of individuals with n heterozygous pair is 2^n. Thus, in this case, the number of genotype in F2 should be 2^3 * 2^3 = 8 * 8 = 64. We can get this conclusion by analyzing the number of 6 cm tails in F2: 1/64 with genotype aabbcc, and the number of 30 cm tails: 1/64 with genotype AABBCC. These two genotypes is as same as the ancestor in this experiment.

The genotype of an F2 piglet with an 18 cm tail is AaBbCc. If these genes show dominance, the tail lenght of F2 will be 30 cm. And there are 7 possible phenotypes. Thus, we can conclude that the genes act additively.

2/ 18 cm tail F2: AaBbCc, and 6 cm tail: aabbcc.

The offspring genotypes are:

18 cm AaBbCc, 14 cm AaBbcc, 14 cm AabbCc, 10 cm Aabbcc
14 cm aaBbCc, 10 cm aaBbcc, 10 cm aabbCc, 6 cm aabbcc

The frequency of piglets with 14 cm tail length should be 3/8 = 37.5%.

Answer and Explanation:

1) These results are consistent with option (d)Three genes, each with two alleles that act additively .

Quantitative heritability: Refers to the transmission of a phenotypic trait in which expression depends on the additive effect of a series of genes.

Polygenic heritability occurs when a trait is due to the action of more than one gene that can also have more than two alleles. This can cause many different combinations that are the reason for genotypic graduation.

Quantitative traits are those that can be measure, such as longitude, weight, eggs laid per female, among others. These characters do not group individuals by precise and clear categories. Instead, they group individuals in many different categories that depend on how the genes were intercrossed and distributed during meiosis. The result depends on how each allele of each gene contributed to the final phenotype and genotype.

In the exposed example, there are 7 different phenotypes (6, 10, 14, 18, 22, 26, 30) and the majority of F2 individuals have 18-lengthed tails. This might be the heterozygotic phenotype for all the genes.The rest of the phenotypes are possible combinations of genes and their alleles. There are six possible combinations (apart from the 18 lenght form). This leads to 3 genes and two alleles in each gene.

So, until now we have:

A pig with a 6 cm tail was mated to a pig with a 30 cm tail. All F1 offspring had a tail of length 18 cm. This is:

Parental) 6cm x 30cm

F1) 18 cm

An F2 generation was made by crossing F1 individuals. This resulted in many piglets whose tails ranged in 4-cm intervals from 6 to 30 cm (6, 10, 14, 18, 22, 26, 30). This is:

Parental) 18 cm x 18 cm

F2) 6, 10, 14, 18, 22, 26, 30

The majority of pigs had 18 cm tails, this means that 18 cm phenotype is a heterozygote for each gene

1/64 had 6 cm tails, and 1/64 had 30 cm tails, this means that these two phenotypes are the extreme traits, that is the recessive homozygote and the dominant homozygote.

There are 7 phenotypes, one of them is the recessive form, the other is the dominant form and the majority is the heterozygotic form for every intervening gene. There are three genes with two alleles each:

Gene 1: allele A and a

Gene 2: allele B and b

Gene 3: Allele C and c

Phenotypes:

aabbcc: homozygotic recessive form

AABBCC: homozygotic dominant form

AaBbCc: heterozygotic form for every intervening gene

If the recessive form is 6cm length, and each dominant allele contributes in 4 cm to each phenotype, we get:

6 cm length = aabbcc (1/64)

10cm length = Aabbcc (A contributes 4cm)

14cm length = AaBbcc (A and B contribute 4 cm each)

18cm length = AaBbCc (The majority) (A, B and C contribute 4cm each)

22cm length = AABbCc

26cm length = AABBCc

30cm length = AABBCC (1/64)

2) An F2 piglet with an 18 cm tail is heterozygous at each of the tail-length loci, and is mated to an F2 piglet with a 6 cm tail.

Parental) AaBbCc x aabbcc

Gametes) ABC ABc AbC Abc aBC aBc abC abc

abc abc abc abc abc abc abc abc

Punnet square)

ABC ABc AbC Abc aBC aBc abC abc

abc AaBbCc AaBbcc AabbCc Aabbcc aaBbCc aaBbcc aabbCc aabbcc

F3 genotype and phenotype)

8/64 AaBbCc = 18 cm

8/64 AaBbcc = 14cm

8/64 AabbCc = 14cm

8/64 Aabbcc = 10cm

8/64 aaBbCc = 14cm

8/64 aaBbcc = 10cm

8/64 aabbCc = 10cm

8/64 aabbcc = 6cm

Each dominant allele contributes 4cm to the recessive homozygote form for each phenotype.

How are a carrot, an amoeba, and a mandrill. alike

Answers

A carrot, an amoeba (a cell/organism that has the ability to change shape), and a mandrill (a primate) are all alike in that they are all eukaryotes. This means that their cells share the same features of having a nucleus and other organelles within an enclosed membrane.

A student was conducting an experiment on the growth of tomato plants. The plantsreceived the same amount of sunlight and water, but the amount fertilizer each plant
received was different. Which of the following is the dependent variable in this
experiment?
a. Amount of sunlight
b. Amount of water
c. Amount of fertilizer
d. Growth of the plants

Answers

Final answer:

The growth of the plants is the dependent variable in this experiment on tomato plants' growth. The amount of sunlight, water, and fertilizer are independent variables that were controlled and varied to observe their impact on plant growth.

Explanation:

The dependent variable in this experiment is the growth of the plants. The independent variables are the amount of sunlight, amount of water, and amount of fertilizer. By keeping the amount of sunlight and water constant, and varying the amount of fertilizer, the student can determine the effect of fertilizer on plant growth.

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Archaea differ from bacteria in that archaea

Answers

Archaea have more complex RNA polymerases than Bacteria, similar to Eucarya. Unlike bacteria, archaea cell walls do not contain peptidoglycan. Archaea have different membrane lipid bonding from bacteria and eukarya. There are genetic differences.

he __________ system works with the nervous system to protect important organs such as the brain and spinal cord. A) skeletal B) lymphatic C) endocrine D) cardiovascular

Answers

The skeletal system since the brain is apart of the skeletal system.

Answer:

The correct answer would be skeleton

Explanation:

I had the same question

Rapid cell growth in preparation for cell divisionoccurs during
the G1 phase
the S phase.
O the G2 phase.
mitosis.

Answers

Final answer:

The G2 phase of the cell cycle is when rapid cell growth in preparation for cell division occurs. This is after DNA replication and before mitosis.

Explanation:

Rapid cell growth in preparation for cell division occurs during the G2 phase of the cell cycle. The cell cycle is a four-stage process consisting of the G1 phase (gap 1), the S phase (synthesis), the G2 phase (gap 2), and mitosis. During the G1 phase, cells grow and prepare for DNA replication which happens in the S phase. In the G2 phase, the cell experiences rapid growth in preparation for cell division, cell size increases dramatically, and proteins necessary for cell division are synthesised.