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An additional gene, gene W, was also examined. A test cross was made between true-breeding EEWW flies and EEWW flies. The result

Debora [2.8K]

This question is incorrect but here is the correct question below;

An additional gene,gene W was also examined. a test cross was made between true breeding EEWW flies and eeww flies. The resulting F₁ generation was then crossed with eeww flies. 100 offspring in the F₂ generation were examined and it was discovered that the E and W genes were not linked.

Which is the correct genotype of the F₂ offspring if the genes were linked and if the genes were not linked?

a) Linked: 50% EeWw and 50% eeww; not linked: 25% EeWw, 25% Eeww, 25% eeWw and 25% eeww.

b) Linked: 25% Eeww, 50% eeWw; not linked:parental genotypes EeWw and eeww.

c) Linked genotypes (EeWw and eeww) and recombinant genotype ( Eeww & eeWw) in the F₂ generation are nearly the same irrespective of their linkage.

d) Linked: mostly with parental genotypes, Eeww and eeWw; unlinked: 25% EeWw and eeww with 75% Eeww and eeWw.

Answer:

a) Linked: 50% EeWw and 50% eeww; not linked: 25% EeWw, 25% Eeww, 25% eeWw and 25% eeww.

Explanation:

a test cross was made between true breeding EEWW flies and eeww flies

If EEWW self crossed, we have the following ( EW, EW, EW, EW)

Also, for eeww, we have ( ew, ew, ew, ew)

EW EW EW EW

ew EWew EWew EWew EWew

All offspring are (EWew)

The question goes further by saying "The resulting F₁ generation was then crossed with eeww flies".

And we are asked to find the correct genotype of the F₂ offspring if the genes were linked and if the genes were not linked

∴

To determine the offsprings of the linked genes we need to go by the definition and understand what linked genes are: Linked genes are genes that are physically close together on the same chromosomes. Effect of recombinantion on linked genes, results in gene swaps which occur in chromosomes that are homologous.

Having said that; If EWew × eeww

we have; EW & ew × ew & ew

EW ew

ew EeWw eeww

offspring that

are linked in ⇒ EeWw EeWw & eeww eeww

F₂ will be

$\frac{1}{2}$ = 50% of EeWw of the total 100 offspring in the F₂ cross

$\frac{1}{2}$ = 50% of eeww of the total 100 offspring in the F₂ cross

∴ Linked genes = 50% EeWw and 50% eeww.

For unlinked genes; If EWew × eeww

if rearrangement occurs in EWew and EWew self crossed, we have ( EW,Ew,eW,ew) as the traits needed for the unlinked gene F₂ crossing.

Also ewew will be (ew, ew, ew, ew).

EW Ew eW ew

ew EeWw Eeww eeWw eeww

We have the following results for the unlinked genes

$\frac{1}{4}$ = EeWw 25% of the total 100 offspring in the F₂ cross

$\frac{1}{4}$ = Eeww 25% of the total 100 offspring in the F₂ cross

$\frac{1}{4}$ = eeWw 25% of the total 100 offspring in the F₂ cross

$\frac{1}{4}$ = eeww 25% of the total 100 offspring in the F₂ cross

∴ not linked: 25% EeWw, 25% Eeww, 25% eeWw and 25% eeww.

3 0

3 years ago

Match each description to the correct step. Metaphase 11 Nuclear membranes form around each set of chromosomes. Prophase 11 Chro

zvonat [6]

Answer:

Metaphase ⇒ Chromosomes line up in the middle of the cell

Prophase ⇒ DNA condenses to form chromosomes

Anaphase ⇒ Each chromosome separates and moves to opposite ends of the cell

Telophase ⇒ Nuclear membranes form around each set of chromosomes

3 0

3 years ago

A scientist conducts an experiment to test the effects of light on earthworm activity. Which is a variable in this experiment? l

777dan777 [17]

<span>I believe it is light, but this is kind of a confusing question. Technically there is a independent and dependent variable in the question. The variable being manipulated is light, and the earthwork activity depends on the light. </span>

7 0

3 years ago

Read 2 more answers

What is the explanation of why avery, macleod, and mccarty's in vitro transformation experiment showed that dna, but not rna or

astraxan [27]

The Griffith's experiment, the Avery-MacLeod-McCarty experiment, and the Hershey–Chase experiments were the set of experiments that established DNA as the key hereditary molecule. The Avery-MacLeod-McCarty experiment was an extension to the Griffith's experiment. The heat killed virulent S strain cells of the Griffith's experiment were lysed to form a supernatant containing a mix of RNA, DNA, proteins and lipids from the cell. The supernatent was equally divided into 3 parts after the removal of the lipids. The 3 parts were respectively treated with an RNAase to degrade the RNA, DNAase to degrade the DNA and proteinase to degrade the proteins. The treated supernatant was then added into the culture containing the non-virulent R cells. In case of the supernatant treated with the DNAse, no transformation of R cells into S cells occurred. The transformation of R cells to S cells occurred in the proteinase and the RNAse cases. This indicated that DNA was the hereditary molecule and not protein or RNA.

5 0

4 years ago

How would an x-ray technician position the camera to produce a frontal view of a patient's abdominal cavity?

leva [86]

Answer: Option A. "In front of the patient, pointing at his navel"

Explanation:

The X-ray of abdominal cavity is known as Abdominal x-ray. It has same principle as X-ray and uses ionizing radiation to take pictures of inner parts of abdominal cavity such as intestine, stomach, liver and spleen. it is used to diagnose patients with vomiting, unexplained pain, and nausea.

While diagnosing patient's abdominal cavity, x-ray technician position the camera in front of the patient, pointing at his navel in order to produce a frontal view of abdominal cavity.

Talking about a x-ray machine, it is a compact apparatus which can be taken to the patient in a hospital bed or the emergency room. The x-ray tube connects flexible arm which further extends towards the patient and an x-ray film holder (image recording) plate is placed under the patient.

The technologist will ask the patient to take position on the x-ray table or bed and will position x-ray machine over the abdominal area of the patient, in order to take the frontal view of abdominal cavity. The patient will be asked to hold their breathe to get a clear picture. The technician will turn on the x-ray machine that allows x-ray machine to produce a small burst of radiation which passes through the abdominal area and records the frontal view of patient's abdominal cavity. As a result, soft tissue shows up in shades of gray and air appears black while bones appear white on the x-ray, through which patient can be diagnosed.

Hence, the correct option is A "In front of the patient, pointing at his navel".

8 0

4 years ago