All categories

3 years ago

The hershey chase experiment on bacteriophages showed the genetic information is carried in ____________.

Biology

1 answer:

Maksim231197 [3]3 years ago

4 0

Answer:

The hershey chase experiment on bacteriophages showed the genetic information is carried in DNA.

Explanation:

In 1952 Alfred Hershey and Martha Chase provided evidence that DNA is the genetic material. These scientists carried out an experimental work with the T2 virus, a bacteriophage that infects the Escherichia coli bacteria, which reproduces by attaching itself to the outer wall of the bacteria, injecting its DNA into it where it replicates and directs, the synthesis of phage proteins. Phage DNA is encapsulated within proteins and produces phages, which lyse or break the cell and releasing phage from the progeny. They used radioactive forms (isotopes) of phosphorus and sulfur. A radioactive isotope can be used as a marker to identify the location of a specific molecule, because any molecule that contains the isotope is radioactive and therefore easy to detect. The DNA contains phosphorus, but doesn't contain sulfur, therefore phosphorus-32 was used to mark the DNA, instead the protein has sulfur, but not phosphorus, so sulfur-35 was used. Hershey and Chase found that sulfur-35 is outside the cell while phosphorus-32 was inside, indicating that DNA was the physical support of the inheritance. This work allowed the scientists to conclude that it's the DNA and not the protein that enters the bacteria during phage reproduction and that only the DNA is transmitted to the phage of the progeny.

You might be interested in

1.)A car travels 10 miles east in 30 minutes.

Nitella [24]

4 because the distance is 40 and it hit the ground in 5 seconds

6 0

3 years ago

22. A plant geneticist is investigating the inheritance of genes for bitter taste (Su) and explosive rind (e) in watermelon (Cit

mylen [45]

Answer:

See the answer below

Explanation:

Using the formula for calculating Chi square ( $X^2$ ):

$X^2$ = $\frac{(O - E)^2}{E}$ where O = observed frequency and E = expected frequency.

The observed frequencies for the four phenotypes are 88, 62, 62, and 81 respectively.

For the expected frequency, the phenotypes are supposed to assort in 9:3:3:1 according to Mendel ratio.

Hence, expected frequencies are calculated as:

phenotype (1) = 9/16 x 293 = 164.81

phenotype (2) = 3/16 x 293 = 54.94

phenotype (3) = 3/16 x 293 = 54.94

phenotype (4) = 1/16 x 293 = 18.31

The $X^2$ is calculated thus:

Phenotype O E $X^2$

1 88 164.81 $\frac{(88 - 164.81)^2}{164.81}$ = 35.80

2 62 54.94 $\frac{(62 - 54.94)^2}{54.94}$ = 0.91

3 62 54.94 $\frac{(62 - 54.94)^2}{54.94}$ = 0.91

4 81 18.31 $\frac{(81 - 18.31)^2}{18.31}$ = 214.64

Total $X^2$ = 35.80 + 0.91 + 0.91 + 214.64 = 252.26

To the nearest tenth = 252

Degree of freedom = n - 1 = 4- 1 = 3

$X^2$ tabulated = 7.815

The calculated  $X^2$  exceeds the critical value, hence, the null hypothesis is rejected. The two genes did not assort independently. 

4 0

2 years ago

Which indicates the authenticity of a primary source document? Which best compares competition, cooperation, and predation betwe

arsen [322]

Answer: To know if a document is a primary source, check to see when the document is published, if it is an autobiography of a person, a diary, a letter, or the documents happen at the time of the event.

The competition between two organisms can be seen if there is only 1 source of food or habitat space and they're competing against each other for the food or habitat.

There is cooperation between two organisms if they are helping each other with resources or alarming each other if there is danger.

Predation between two organisms happen when one organism hunts the other organism for food.

6 0

3 years ago

1. Han Solo and Leia fall madly in love and have one child (Ben). What is the chance that they will have a child who is strong w

Andreas93 [3]

1.
Han Solo genotype: ff
Offspring genotypes: Ff, ff
There is a 50% chance the child will have the force.

2.
Let f be webbed feet and F be non-webbed feet
Parental genotypes: ff ; FF
Possibility of webbed feet in offspring: 0%

3.
Yes, it is possible if both parents are heterozygous. 25% of offspring will be grey.

4.
Pheonotypes: Hairy and non-hairy fur
genotypes: HH, Hh, hh

5.
Parental genotypes: RR and RL
Offspring phenotypes: Brown hair and Brown and Black hair
Offspring genotypes: RR, RL

6.
Genotypes: BW
Phenotypes: Grey

7.
Black

8.
None of his children will be bald.

7 0

3 years ago

(b) Identify the dependent variable in the experiments. Identify a control group missing from the second experiment. Justify the

hammer [34]

Hello. You have not submitted the experiments to which this question refers, which makes it impossible for it to be answered. However, I will try to help you in the best possible way.

A dependent variable is one that is influenced by an element, in order to present a result, that is, it is a variable that does not establish itself, being dependent on another variable to establish itself. An example of a dependent variable can be seen in an experiment that seeks to determine the differences in growth rates of bean plants that were fertilized with different organic fertilizers. As you can see, this is an experiment where the different organic fertilizers influence differences in the growth rate of bean plants. Thus, the growth of these plants depends on the fertilizer, this growth being the variable dependent on the experiment.

A control group, in turn, is an element of the experiment that does not receive the element that is being tested in the same experiment. The control group is important, because it allows the researcher to see the results of a system where the tested element does not exist, thus being able to determine the real differences that the tested element is capable of causing. Using the same example presented in the paragraph above, we can see that the different organic fertilizers are the elements tested in the experiment. These fertilizers will be applied to the plants, however it is important that at least one of these plants does not receive any fertilizer, so that the difference between a system with fertilizer and a system without fertilizer can be evaluated. This plant that received no fertilizer is the control group.

4 0

2 years ago