1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Semmy [17]
3 years ago
5

What is the answer with explaining

Biology
1 answer:
ch4aika [34]3 years ago
3 0

Answer:

EeWw

Explanation:

This question involves two genes; one coding for earlobes and the other for hairline. According to the question, one parent is homzygous recessive for earlobe but homzygous dominant for hairline i.e. (eeWW) while the other parent is homzygous dominant for unattached earlobes but homzygous recessive for hairline i.e. (EEww).

The cross between the two parents is as follows: eeWW × EEww. The following gamete combination will be produced by each parent:

eeWW - eW, eW, eW, eW

EEww - Ew, Ew, Ew, Ew

Using these gametes in a punnet square, the following proportions of offsprings will likely be produced:

All EeWw offsprings

You might be interested in
A diffuse, nontoxic goiter is usually due to a lack of _____ in the diet.
prohojiy [21]

Answer;

Iodine

A diffuse, nontoxic goiter is usually due to a lack of Iodine in the diet.

Explanation;

-A diffuse non-toxic goiter is an enlargement of the thyroid gland without nodularity. It occurs in an endemic and sporadic distribution. It does not result from an inflammatory or neoplastic process and is not associated with abnormal thyroid function.

-Simple nontoxic goiter, which may be diffuse or nodular, is noncancerous hypertrophy of the thyroid without hyperthyroidism, hypothyroidism, or inflammation. Except in severe iodine deficiency, thyroid function is normal and patients are asymptomatic except for an obviously enlarged, non tender thyroid.

5 0
3 years ago
Which of the following are similarities between animal cells and plant cells?
gavmur [86]
Nucleus, Mitochondria, vacuole
8 0
3 years ago
Matching questions match each description with the appropriate phase of mitosis:
strojnjashka [21]
Hello!!
A: Anaphase — 1 chromatids move towards opposite poles. I always remember that “Ana” moves to different places on the sides of town. This is where the chromatids begin to move.
B: Telophase — 4 Cytokinesis may occur. Cytokinesis is the last and final step. The sister chromatids finish moving towards the poles and then cytokinesis occurs.
C: Metaphase — 3 Chromatids line up in the middle of the cell. I always remember since they line up in the middle, they “met” there.
D: Prophase — 2 and 5 Disintegration of the nuclear membrane and the spindle forms. Both of these have to happen first in order for the rest of the processes to occur.
**The order of mitosis goes prophase, prometaphase, Metaphase, Anaphase, Telophase, and Cytokinesis.**
For the bottom:
A: Algae 6 and 10. Both diatoms and kelps (plant related) are a part of the Algae general type.
B: Fungi 7 and 9. Deuteromycetes and Ascomycetes.
C: Protozoa 8. It is ciliates because they are a major group of Protozoa from cilia.
I hope I helped!! Have a great day!! :)
3 0
3 years ago
Read 2 more answers
Why do hurricanes form in tropical latitudes
Bess [88]
In order for a hurricane to form, a temperature of 80 degrees Fahrenheit is needed 
areas close  the equator provide this ideal temperature for the formation of hurricanes
7 0
3 years ago
7. Pyrethrums are chemicals used to kill insects. In bed bugs, the mutant-type allele, r, confers resistance, but only in homozy
natita [175]

Answer:

Allele frequency

Normal allele  = 0.9

Mutant r allele = 0.1

Genotype frequency

Homozygous normal bugs = 0.81

Homozygous mutant bug = 0.01

Heterozygous normal bug with one mutant r allele and one normal allele = 0.18

Explanation:

It is given that 99% of the bugs were killed after the spray of  pyrethrum. This suggests that 1% of the bugs that were not killed must be homozygous for the mutant type allele "r"

Thus, the frequency of homozygous "rr" species i.e q^2 = 0.01

From this we can evaluate the frequency of mutant "r" allele.

Thus, q = \sqrt{0.01} \\

q = 0.1

As per Hardy-Weinberg first equilibrium equation, p + q = 1

Substituting the value of q in above equation, we get

p = 1 - q\\p = 1 - 0.1\\p = 0.9

Thus, the frequency of homozygous normal bug is equal to

p^2 = 0.9^2 = 0.81

As per Hardy-Weinberg second equilibrium equation-

p^2 + q^2 + 2pq = 1\\

Substituting all the available values we get -

0.81 + 0.01+ 2pq = 1\\2pq = 0.18

Allele frequency

Normal allele  = 0.9

Mutant r allele = 0.1

Genotype frequency

Homozygous normal bugs = 0.81

Homozygous mutant bug = 0.01

Heterozygous normal bug with one mutant r allele and one normal allele = 0.18

3 0
3 years ago
Other questions:
  • The answer to the questions
    12·1 answer
  • What is the definition of oceanography?
    10·1 answer
  • Sophia is observing slides under a microscope while adjusting its mirror so that a circle of light can be seen.which type of mic
    11·2 answers
  • ____ 15. which excretory organ eliminates water and some chemical wastes in perspiration?
    8·1 answer
  • What conclusions can be drawn about the existence of carbon-12, carbon-13, and carbon-14? A) These are forms of carbon and they
    5·2 answers
  • Trace evidence is a type of circumstantial evidence, examples of which include:
    14·1 answer
  • Which statement about the pancreas is not true?
    14·1 answer
  • 1.An attraction between molecules can be called a _______.
    14·1 answer
  • The anticodon sequence is located on
    12·1 answer
  • A black hat is made of matter that absorbs light energy and changes it into heat energy. * 5 points True False
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!