Pls answer all the questions

If chromosomes fail to separate during cell division ________________ can occur. This can result in cells with the incorrect num

storchak [24]

Answer:

1. NONDISJUNCTION

2.Aneuploidy

3.Trisomy

4.Monosomy.

5.3.

6.Trisomy 21.

Explanation:

8 0

3 years ago

4) A homozygous groucho fly ( gro, bristles clumped above the eyes) is crossed with a homozygous rough fly (ro, eye abnormality)

docker41 [41]

Answer and Explanation:

A homozygous groucho fly ( gro, bristles clumped above the eyes) is crossed with a homozygous rough fly (ro, eye abnormality).

The F1 females are testcrossed, producing these offspring: groucho 518 rough 471 groucho, rough 6 wild-type 5 1000 a) What is the linkage distance between the two genes? B) Plot the genes on a map c) If the genes were unlinked and the F1 females were mated with the F1 males, what would be the offspring in the F2 generation?

1st cross:

Parental) grogro ro+ro+ x gro+gro+ roro

F1) gro+gro ro+ro

2nd cross:

Parental) gro+gro ro+ro x grogro roro

Gametes) gro+ro+ gro ro

gro+ro gro ro

gro ro+ gro ro

gro ro gro ro

Punnet square)

gro+ro+ gro+ro gro ro+ gro ro

gro ro gro+gro ro+ro gro+gro roro grogro ro+ro grogro roro

F2)

0.518 grogro ro+ro (518 individuals)

0.471 gro+gro roro (471 individuals)

0.006 grogro roro (6 individuals)

0.005 gro+gro ro+ro (5 individuals)

Total number of individuals 1000

Note: These frequencies were calculated dividing the number of individuals belonging to each genotype by the total number of individuals in the F2.

To know if two genes are linked, we must observe the progeny distribution. If individuals, whos genes assort independently, are test crossed, they produce a progeny with equal phenotypic frequencies 1:1:1:1. If we observe a different distribution, that is that phenotypes appear in different proportions, we can assume that genes are linked in the double heterozygote parent.

In the exposed example we might verify which are the recombinant gametes produced by the F1 di-hybrid, and we can recognize them by looking at the phenotypes with lower frequencies in the progeny.

By performing this cross we know that the phenotypes with lower frequencies in the progeny are groucho, rough and wild-type. So the recombinant gametes are gro+ro+ and gro ro, while the parental gametes are gro+ro and gro ro+.

So, the genotype, in linked gene format, of the double heterozygote individual in the F1 is gro+ro/gro ro+.

To calculate the recombination frequency we will make use of the next formula: P = Recombinant number / Total of individuals. The genetic distance will result from multiplying that frequency by 100 and expressing it in map units (MU). One centiMorgan (cM) equals one map unit (MU).

The map unit is the distance between the pair of genes for which one of every 100 meiotic products results in a recombinant product.

The recombination frequency is:

P = Recombinant number / Total of individuals

P = 6 + 5 / 1000

P = 11 / 1000

P = 0.011

The genetic distance between genes, is 0.011 x 100= 1.1 MU.

Genetic Linkage Map:

Parental Phenotypes)

-----gro+------ro---- -----gro------ro+----

----- gro ------ro---- ---- gro------ ro ----

Recombinant phenotypes)

-----gro+------ro+---- -----gro------ro----

----- gro ------ ro---- -----gro------ro----

If the genes were unlinked and the F1 females were mated with the F1 males, the offspring in the F2 generation would have been

4/16 = 1/4 gro+gro ro+ro

4/16 = 1/4 gro+gro roro

4/16 = 1/4 grogro ro+ro

4/16 = 1/4 grogro roro

Their phenotypic frequencies would be 1:1:1:1 related.

7 0

3 years ago

Make a list of questions you would ask as you observe the cells

Anestetic [448]

Answer:

1. What genes control the growth of cell growth?

2. What is the purpose of this regulation?

3. What happened when the cell growth is not regulated?

Explanation:

5 0

3 years ago

A marine biologist is studying dolphin calls and notices that sometimes the calls sound louder based on the background noise in

julsineya [31]

Answer:

The correct answer is option 1. "Amplitude".

Explanation:

Waves of sound have different properties, among them amplitude is the one that determines the loudness or volume of the sound. The amplitude is the height of the wave, measured in meters, or in other word the maximum displacement of the sound's particle from its position of equilibrium. If a dolphin calling sounds louder because of the background noise, the waves of the sound that the dolphin produces would have a higher amplitude.

5 0

3 years ago

Read 2 more answers

You are required to make 10mL of a 0.5M solution of HCl. Your stock solution of HCl is 90M. Using distilled water (dH2O) as a so

sleet_krkn [62]

Answer:

0.0556 mL of 90 M HCl would be taken and 9.944 mL distilled water added to make 0.5 M, 10 mL HCl solution.

Explanation:

From the dilution equation:

Initial molarity (m1) x initial volume (v1) = final molarity (m2) x final volume (v2).

In this case, m1 = 90 M, v1 = ?, m2 = 0.5 M, and v2 = 10 mL

v1 = m2 x v2/ m1 = 0.5 x 0.01/90 = 0.0000556 L

v1 = 0.0556 mL

This means that the initial volume of 90 M HCl to be taken is 0.0556 mL. Since the final volume to be prepared is 10 mL,

10 - 0.0556 = 9.944 mL

Hence, 0.0556 mL of 90 M HCl would be taken and 9.944 mL distilled water added to make 0.5 M, 10 mL HCl solution.

8 0

3 years ago

Pls answer all the questions​

Pls answer all the questions