Three linked autosomal loci were studied in smurfs. Locus Dominant Allele Recessive Allele
Color Blue Pink
Height Dwarf Tall
Mood Happy Gloomy
Pure breeding pink, tall, happy smurfs were mated to pure breeding blue, dwarf, gloomy smurfs. The F1 were testcrossed and the testcross progeny are given below.
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 1614
Determine the order of the loci and the map distances between adjacent loci. Then label the map below. Round your map distances properly to two decimal digits.
When labeling the loci on the map, the two outside loci must be in alphabetical order. For example, if you have the following loci: Albinism, Bone length, Cranial size, and Albinism is in the center, you would label the map Bone length - Albinism - Cranial size.
AnswerHeightColorMood Answer cM AnswerHeightColorMood Answer cM AnswerHeightColorMood
Calculate the coefficient of coincidence and interference. Place your final answers in the spaces provided. Use three decimal places in your answer.
Final answers: C = Answer; I = 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

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)

Simple recombinant)

Double Recombinant)  

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 Pinkhappy  (Parental) 580 individuals

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

Dwarf--- PinkGloomy (Double Recombinant) 6 individuals

DwarfBlueGloomy (Parental) 601 individuals

Tall Blue Gloomy (Simple Recombinant)  98 individuals

Tall Blue Happy   (Double Recombinant) 8 individuals  

Region II

Tall Pinkhappy (Parental) 580 individuals

TallPink Gloomy (Simple Recombinant) 113 individuals

DwarfPink Gloomy (Double Recombinant) 6 individuals

DwarfBlueGloomy (Parental) 601 individuals

Dwarf BlueHappy (Simple Recombinant) 107 individuals

Tall Blue Happy   (Double Recombinant) 8 individuals

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

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

P1 = 213/1614

P1 = 0.132    

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:

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=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

correct answer is option a.

tropopause, stratopause and mesopause have all in common aspect that is point of maximum temperature in its layer of the atmosphere. all layers of atmosphere observe abrupt change in temperature. tropopause is the boundary between troposphere and stratosphere. stratopause is the boundary between stratosphere and mesosphere. mesopause is the boundary between mesosphere and thermosphere. these all boundaries are ranges between altitude of 0 to 80 km.

answer; the tunica intima also called tunica interna, consists of elastic fibers,connective tissue and smooth muscle cells;