All categories

3 years ago

Explain how Punnett squares are set up

1 answer:

6 0

1. determine the genotypes of the parent organisms
2. write down your "cross" (mating)
3. draw a p-square
4. "split" the letters of the genotype for each parent & put them "outside" the p-square
5. determine the possible genotypes of the offspring by filling in the p-square
6. summarize results (genotypes & phenotypes of offspring)

You might be interested in

Kepler's first law demonstrates that each planet has an elliptical orbit of unique size and shape with the Sun at one focus.

kramer

Answer:

true, Kepler's first law demonstrates that each planet has an elliptical orbit of unique size and shape with the Sun at one focus.

7 0

3 years ago

Why is the sickle cell allele still common in tropical regions?

slavikrds [6]

Sickle-cell disease is common in tropical and sub-tropical regions because being a carrier, carrying a single sickle-cell mutation (sickle cell trait), affords some resistance to malaria.

3 0

3 years ago

Endosymbiotic theory is supported by similarities between chloroplasts and * Cyanobacteria Viruses Yeasts None of the above

stiv31 [10]

Answer:

Endosymbiotic theory is supported by similarities between chloroplasts and * Cyanobacteria.

Explanation:

The endosymbiotic theory proposes that mitochondria and chloroplasts were once free-living bacteria that were phagocytized by another cell but not digested. These bacteria got to adapt to their host, and both cells became interdependent.

Both organelles have many similarities with other free-living bacteria. Chloroplasts probably derivate from cyanobacteria because both cells absorb sunlight, produce ATP, and organic molecules. And mitochondria derivate from rickettsias because they produce ATP in the same way by using the Krebs Cycle and Oxidative Phosphorylation.

From the phagocytosis moment, these two cells became so dependant on each other, they could not survive without the other one.

Chloroplasts and mitochondria share some traits with free-living bacteria, that support the theory.

• Both organelles present their genetic material. This DNI is independent of the cells´ DNA, is bi-catenary and circular, identical to the bacterial DNA, and very different from the one of the eukaryotic cells.

• Both organelles divide by binary fission, not by mitosis, and can synthesize their ribosomes and organelles.

• Both organelles present a double membrane, a characteristic that reinforces the idea of being phagocyted. The internal membrane looks identical to the bacterial membrane, while the external membrane looks like the eukaryotic one.

In fact, in this internal membrane are placed the energy centers, just as it occurs in bacterias membrane.

• Finally, the sizes of the organelles are similar to the size of some procaryotes

6 0

2 years ago

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 <u>color</u><u>.</u> 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

Answer the question please

MrRissso [65]

Answer: the answer is D

Explanation:

8 0

3 years ago

Read 2 more answers