Which statements apply to the orbits of dwarf planets? Check all that apply.

Three linked autosomal loci were studied in smurfs.

cupoosta [38]

Answer:

height -------- color --------- mood

(13.2cM) (14.5cM)

C=0.421

I = 0.579

Explanation:

We have the number of descendants of each phenotype product of the tri-hybrid cross.

Phenotype Number

pink, tall, happy 580

blue, dwarf, gloomy 601

pink, tall, gloomy 113

blue, dwarf, happy 107

blue, tall, happy 8

pink, dwarf, gloomy 6

blue, tall, gloomy 98

pink, dwarf, happy 101

Total number of individuals = 1614 = N

Knowing that the genes are linked, we can calculate genetic distances between them. First, we need to know their order in the chromosome, and to do so, we need to compare the phenotypes of the parental with the ones of the double recombinants. We can recognize the parental in the descendants because their phenotypes are the most frequent, while the double recombinants are the less frequent. So:

Parental)

Pink, tall, happy 580 individuals

Blue, dwarf, gloomy 601 individuals

Simple recombinant)

Pink Tall Gloomy 113 individuals

Blue, Dwarf, Happy 107 individuals

Blue Tall Gloomy 98 individuals

Pink Dwarf Happy 101 individuals

Double Recombinant)

Blue Tall Happy 8 individuals

Pink Dwarf Gloomy 6 individuals

Comparing them we realize that parental and double recombinant individuals differ in the position of the gene codifying for color. They only change in the position of Blue and Pink. This suggests that the position of the color gene is in the middle of the other two genes, height and mood, because in a double recombinant only the central gene changes position in the chromatid.

So, the alphabetic order of the genes is:

---- height ---- color ----- mood ----

Now we will call Region I to the area between Height and Color, and Region II to the area between Color and Mood.

Once established the order of the genes we can calculate distances between them, and we will do it from the central gene to the genes on each side. First We will calculate the recombination frequencies, and we will do it by region. We will call P1 to the recombination frequency between Height and color genes, and P2 to the recombination frequency between color and mood.

P1 = (R + DR) / N

P2 = (R + DR)/ N

Where: R is the number of recombinants in each region (the ones that have an intermediate phenotypic frequency), DR is the number of double recombinants in each region, and N is the total number of individuals. So:

Region I

Tall------ Pink--------happy (Parental) 580 individuals

Dwarf ---Pink------- Happy (Simple Recombinant) 101 individuals

Dwarf--- Pink-------Gloomy (Double Recombinant) 6 individuals

Dwarf----Blue-------Gloomy (Parental) 601 individuals

Tall ------Blue------- Gloomy (Simple Recombinant) 98 individuals

Tall ----- Blue------- Happy (Double Recombinant) 8 individuals

Region II

Tall------ Pink--------happy (Parental) 580 individuals

Tall-------Pink------- Gloomy (Simple Recombinant) 113 individuals

Dwarf----Pink------- Gloomy (Double Recombinant) 6 individuals

Dwarf----Blue-------Gloomy (Parental) 601 individuals

Dwarf ----Blue-------Happy (Simple Recombinant) 107 individuals

Tall ----- Blue------- Happy (Double Recombinant) 8 individuals

In each region, the highlighted traits are the ones that suffered recombination.

P1 = (R + DR) / N

P1 = (101+6+98+8)/1614

P1 = 213/1614

P1 = 0.132

P2= = (R + DR) / N

P2 = (113+6+107+8)/1614

P1 = 234/1614

P1 = 0.145

Now, to calculate the recombination frequency between the two extreme genes, height and mood, we can just perform addition or a sum:

P1 + P2= Pt

0.132 + 0.145 = Pt

0.277=Pt

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 every 100 meiotic products, one results in a recombinant product.

Now we must multiply each recombination frequency by 100 to get the genetic distance in map units:

GD1= P1 x 100 = 0.132 x 100 = 13.2 MU = 13.2 cM

GD2= P2 x 100 = 0.145 x 100 = 14.5 MU = 14.5 cM

GD3=Pt x 100 = 0.277 x 100 = 27.7 MU = 27.7 cM

To calculate the coefficient of coincidence, CC, we must use the next formula:

CC= observed double recombinant frequency/expected double recombinant frequency

Note:

-observed double recombinant frequency=total number of observed double recombinant individuals/total number of individuals

-expected double recombinant frequency: recombination frequency in region I x recombination frequency in region II.

CC= ((6 + 8)/1614)/0.132x0.145

CC=0.008/0.019

CC=0.421

The coefficient of interference, I, is complementary with CC.

I = 1 - CC

I = 1 - 0.421

I = 0.579

8 0

3 years ago

Which of the following terms is used to describe how part or parts of an organism work?

il63 [147K]

Answer:

B. Physiology

Explanation:

The way in which a living organism or bodily part functions.

The branch of biology that deals with the normal functions of living organisms and their parts.

8 0

3 years ago

15

s344n2d4d5 [400]

OPTIONS:

a. When there are just two other people at the party.

b. When everyone is wearing a nametag.

c. When there are more than 50 people attending the same party.

d. When the need for help from the person who is having a heart attack is very clear

Answer:

c. When there are more than 50 people attending the same party.

Explanation:

Bystander effect is a term used in social psychology to describe the tendency of an individual to intervene in the event of an emergency to offer help to the person needing it, when others are present at the scene of the emergency.

In the case of an emergency just like the scenario stated in the question, where someone develops a heart attack at a dance party, if the party has more people, the slimmer the chance of the person getting help from any of the 50 people at the party, as the presence of others would tend to discourage the any individual from attempting to help the victim having the heart attack.

The situation that would more likely show the bystander intervention effect is "c. When there are more than 50 people attending the same party."

5 0

3 years ago

How can carbohydrates, lipids, and proteins be detected in foods?

tino4ka555 [31]

The Benedict’s Test is the test which can detect carbohydrates and glucose in foods. During this process, the sample or the food is placed into a test tube and then some drops of Benedict's solution is added to it,and is then put into a beaker of boiling water. When the mixture of food and the Benedict substance turns reddish orange instead of staying blue, this means that it is positive. The mixture has simple sugar or has carbohydrates or glucose. On the other hand, lipids are tested using the brown paper test. When the paper turns transparent when a drop of food is put into it, it means that is is positive as well. When you want to test if the food has proteins, you have to test it with Biurets. Observe if the color change when you put some Biurets solution to the food sample. It is positive if the food turns light purple. Test if the food is positive with starches through the use of iodine. Food having iodine drops should turn purple instead of staying a yellow color.

6 0

2 years ago

How many bacteria can you have in ten hours if you started with one bacteria?

pav-90 [236]

Bacteria can divide every minute, so given 600 minutes in ten hours it would be an extremely large number

8 0

3 years ago