The amount of guanine in a organism always equals the amount of

Igneous rock is formed from the cooling and crystallization of magma or partially melted rock. Igneous rock can be either intrus

11111nata11111 [884]

Answer:

A

Explanation:

Intrusive igneous rocks cool from magma slowly because they are buried beneath the surface, so they have large crystals. Extrusive igneous rocks cool from lava rapidly because they form at the surface, so they have small crystals.

4 0

2 years ago

A cell biologist carefully measured the quantity of DNA in grasshopper cells growing in cell culture. Cells examined during the

xz_007 [3.2K]

Answer:

100 units

Explanation:

The interphase of the cell cycle includes three stages: G1, S, and G2. During the S phase of interphase, the cell enters the process of DNA replication. DNA replication duplicates the existing DNA molecule and doubles the DNA content of the cell. The cell accommodates the newly formed DNA in two sister chromatids of chromosomes. If a cell of grasshopper contained 200 units of DNA in the G2 phase, the cell would have 100 units of DNA in the G1 phase. The S phase would have doubled the DNA content and the G2 phase had 200 units.

7 0

2 years ago

Thyroxine, an important hormone that controls the rate of metabolism in the body, can be isolated from the thyroid gland. If 0.4

MA_775_DIABLO [31]

Answer:

The molar mass of thyroxine is 18,2 g/m

Explanation:

To solve this we must apply the colligative property of cryoscopic descent whose formula is:

T° sv pure - T° sl = Kf x molality

So let's replace our values

5,444°C - 5,144°C = 5,12 °C/m x molality

0,3°C = 5,12 °C/m x molality

0,3°C / 5,12 m/°C = molality

2,5 m

Now you should know that molality is moles from solute in 1kg of solvent but you dont have 1kg you have just 10 g. Let's convert for the rule of three

10 g = 0'010 kg

If 1 kg has 2,5 moles of thyrosine, how many for 0,010 kg

moles = (0,010kg * 2,5 moles)/1kg

moles = 0,025

Now we have the moles of thyrosine in our solution and we were informed that we dissolved 0,455 g so let's get the molar mass (mass of compound for one mole)

moles = mass/ molar mass

molar mass = mass/moles

molar mass = 0,455g/0,025 moles

molar mass = 18,2 g/m

3 0

2 years ago

ipn [44]

Answer:

Plant Reproduction and Life Cycle. The life cycle of a plant is very different from the life cycle of an animal. Humans are made entirely of diploid cells (cells with two sets of chromosomes, referred to as ''2n''). ... Plants, however, can live when they are at the stage of having haploid cells or diploid cells.

Explanation:

In an animal life cycle, male and female parents each create sex cells (sperm and eggs) that unite to form a fertilized egg and develop into an offspring organism. Plants, likewise, have sperm and eggs in their life cycles, but these are produced by an intermediate stage between the adult and the offspring.

7 0

2 years ago

Read 2 more answers

Genetic linkage mapping for a large number of families identifies 4% recombination between the genes for Rh blood type and ellip

Cloud [144]

Answer:

0.2404

Explanation:

The genes R/r and E/e are linked and there is 4% recombination between them.

The possible genotypes and phenotypes are:

RR or Rr: Rh+ blood type
rr: Rh- blood type
EE or Ee: elliptocytosis
ee: normal red blood cells

Tom and Terri each have elliptocytosis (they are E_), and each is Rh+ (they are R_).

Tom's mother has elliptocytosis (E_) and is Rh- (rr), so she has the genotype Er/_r. His father is healthy (ee) and has Rh+ (R_), so he has the genotype eR/e_. Tom must have inherited his E allele from his mother and his R allele from his father, so he has the genotype eR/Er.

Terri's father is Rh+ (R_) and has elliptocytosis (E_), while Terri's mother is Rh- (rr) and is healthy (ee) with the genotype er/er. Terry could only receive the chromosome er from her mother, and because she is heterozygous for both genes the dominant alleles were both received from her father. Terri's genotype is ER/er.

The frequency of recombination is 4%, so 4% of the produced gametes will be recombinant. There are two possible recombinant gametes, so each will appear 2% of the times (a frequency of 0.02).

Tom will produce the following gametes:

eR, parental (0.48)
Er, parental (0.48)
er, recombinant (0.02)
ER (recombinant (0.02)

Terri will produce the following gametes:

ER, parental (0.48)
er, parental (0.48)
Er, recombinant (0.02)
eR, recombinant (0.02)

A child Rh- with elliptocytosis has the genotype rrE_. This can happen from the independent combination of the following gametes from Tom and Terri respectively:

Er (0.48) × er (0.48) = 0.2304 Er/er
Er (0.48) × Er (0.02) = 0.0096 Er/Er
er (0.02) × Er (0.02) = 0.0004 er/Er

And the total probability of having a rrE_ child will be 0.2304 + 0.0096 + 0.0004 = 0.2404

6 0

2 years ago