Ask question

Ask question

All categories

3 years ago

9

The embryo inside a chicken egg is developing a beak and wings seven days after fertilization. Which statement best describes wh

at is taking place inside the egg?

1 answer:

Eduardwww [97]3 years ago

6 0

Are there any choices to pick from?

You might be interested in

What is the function of the lymphatic system?

mart [117]

The primary function of the lymphatic system is to transport lymph, a fluid containing infection-fighting white blood cells, throughout the body.

3 0

4 years ago

Which is an SI derived unit measurement?<br><br>0.56 kg<br>01.10 m<br>2730K<br>02.5 N

balandron [24]

Answer:

02.5 N

Explanation:

02.5 N is a derived unit of measurement.

A derived unit is one that is obtained by combining two or more fundamental units.

Force = mass x acceleration

The unit of mass = kg

acceleration = m/s²

Combining the two; kg x m/s² = kgms²

1N = 1 kgms²

kg, m, k and N are fundamental particles

5 0

3 years ago

What can many other forces do that friction forces cannot do?

Nitella [24]

B is the answer my dude

7 0

3 years ago

In humans, albinism is caused by loss-of-function mutations in genes involved in the synthesis of melanin, the dark pigment in s

Naya [18.7K]

If the albino individuals are present at a frequency of about 1 in 10,000 (or 0.0001); then;

aa = 0.0001

a = 0.01

Using Hardy-Weinberg principle of p + q = 1

p + 0.01 = 1

p = 1 – 0.01 = 0.99

Using Hardy-Weinberg equation; p2 + 2pq + q2 = 1

Carriers of albino allele are the heterozygous population (2pq);

2*.099*0.01 = 0.0198

<span>The frequency is 0.0198 </span>

6 0

3 years ago

The R and S loci are 35 m.u. apart. If a plant of genotype is selfed, what progeny phenotypes will be seen and in what proportio

Veronika [31]

Complete question:

The R and S loci are 35 m.u. apart. If a plant of genotype RS/rs is selfed, what progeny phenotypes will be seen and in what proportions

Answer:

R-/S- = 0.6052
R-/ss = 0.1443
rr/S- = 0.1443
rr/ss = 0.1056

Explanation:

<u>Available data:</u>

R-S are 35 mu apart
Cross: RS/rs x RS/rs

First, we need to recognize the parental gametes and the recombinant ones.

Cross:

Parentals) RS/rs x RS/rs

Gametes) RS Parental ⇒ Equal to the parental genotype

rs Parental ⇒ Equal to the parental genotype

Rs Recombinant ⇒ Product of recombination

rS Recombinant ⇒ Product of recombination

We know that genes are 35 MU apart from each other. The map unit is the distance between the pair of genes for which every 100 meiotic products, one of them results in a recombinant one.

To calculate the recombination frequency, we have to know that 1% of recombinations = 1 map unit = 1cm. And that the maximum recombination frequency is always 50%.

According to this information, if 1 MU = 1% recombination frequency, then

35MU --------- 35% recombination frequency = 0.35

Now, let us calculate the frequency of each gamete type.

35 map units = 35 % of recombination in total

= % Rs + % rS

= 17.5% Rs + 17.5% rS

Now, if the recombination frequency equals 35% then the parental frequency is 100% - 35% = 65%

65 % parental frequency = % of RS + % rs

= 32.5% RS + 32.5% rs

The frequency of each gamete is

RS ⇒ Parental ⇒ 32.5% (65% / 2) = 0.325
rs ⇒ Parental ⇒ 32.5% (65% / 2) = 0.325
Rs ⇒ Recombinant ⇒ 17.5% (35% / 2) = 0.175
rS ⇒ Recombinant ⇒ 17.5% (35% / 2) = 0.175

F1 Genotypes and Proportions

RS/RS = 0.325 x 0.325 = 0.1056
RS/Rs = 2x 0.325 x 0.175 = 0.1137
RS/rS = 2x 0.325 x 0.175 =0.1137
rs/rs = 0.325 x 0.325 = 0.10562
rs/RS = 2x 0.325 x 0.325 = 0.2112
rs/Rs = 2x 0.325 x 0.175 = 0.1137
rs/rS = 2x 0.325 x 0.175 = 0.1137
Rs/rS = 2x 0.175 x 0.175 = 0.061
Rs/Rs = 0.175 x 0.175 = 0.0306
rS/rS = 0.175 x 0.175 = 0.0306

F1 Phenotypes and Proportions

R-/S- = 0.1056 + 0.1137 + 0.1137 + 0.2112 + 0.061 = 0.6052
R-/ss = 0.1137 + 0.0306 = 0.1443
rr/S- = 0.1137 + 0.0306 = 0.1443
rr/ss = 0.1056

4 0

3 years ago