______ is exemplified by a new starfish developing out of a single ray, or arm.

The ratio of the offspring of 9:3:3:1 is for the:

Bess [88]

The correct answer is the Phenotype.
The phenotypic ratio of 9:3:3:1 is the ratio associated with a dihybrid cross. It is a classical Mendelian example of independent assortment. It is a result of crossing two organisms that are heterozygous for two traits.

8 0

4 years ago

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In Drosophila melanogaster, vestigial wings (vg) is recessive to normal wings (vg+), black body (b) is recessive to gray body (b

yan [13]

Correct progeny phenotype:

1779 vestigial wings, black body and purple eyes, vg b pr
1665 normal wings, a grey body and red eyes, vg+ b+ pr+
252 normal wings, an black body and purple eyes, vg+ b pr
241 vestigial wings, a gray body and red eyes, vg b+ pr+
131 normal wings, an black body and red eyes, vg +b pr+
118 vestigial wings, a gray body and purple eyes, vg b+ pr
13 vestigial wings, an black body and red eyes, vg b pr+
9 normal wings, a gray body and purple eyes, vg+ b+ pr

Answer:

The order of these genes is vg --- pr --- b
Map distances between the genes vg/pr = 12.2 MU
Map distance between the genes pr/b = 6.4 MU
Map distances between the genes vg/b = 18.6 MU

Explanation:

We know that

•Normal wings expressed by vg+ is dominant over vestigial wings, vg

•Gray body b+ is dominant over black body

•Red eyes, pr+, is dominant over purple ayes, pr

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

•1779 vestigial wings, black body and purple eyes vg b pr

•1665 normal wings, a grey body and red eyes vg+ b+ pr+

•252 normal wings, an black body and purple eyes vg+ b pr

•241 vestigial wings, a gray body and red eyes vg b+ pr+

•131 normal wings, an black body and red eyes vg +b pr+

• 118 vestigial wings, a gray body and purple eyes vg b+ pr

•13 vestigial wings, an black body and red eyes vg b pr+

• 9 normal wings, a gray body and purple eyes vg+ b+ pr

The total number of individuals is 4208.

In a tri-hybrid cross, it can occur that the three genes assort independently or that two of them are linked and the third not, or that the three genes are linked. In this example, in particular, the three genes are linked on the same chromosome.

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 genotypes of the parental gametes with the ones of the double recombinants. We can recognize the parental gametes in the descendants because their phenotypes are the most frequent, while the double recombinants are the less frequent. So:

Parental)

1779 vestigial wings, black body and purple eyes vg b pr

1665 normal wings, a grey body and red eyes vg+ b+ pr+

Double recombinant)

13 vestigial wings, an black body and red eyes vg b pr+
9 normal wings, a gray body and purple eyes vg+ b+ pr

Simple recombinant)

252 normal wings, an black body and purple eyes vg+ b pr
241 vestigial wings, a gray body and red eyes vg b+ pr+
131 normal wings, an black body and red eyes vg +b pr+
118 vestigial wings, a gray body and purple eyes vg b+ pr

Comparing parental with the double recombinants we will realize that between

vg b pr (parental)

vg b pr+ (double recombinant)

and

vg+ b+ pr+ (Parental)

vg+ b+ pr (double recombinant)

They only change in the position of the alleles pr/pr+. This suggests that the position of the gene pr is in the middle of the other two genes, vg and b, because in a double recombinant only the central gene changes position in the chromatid.

So, the order of the genes is:

---- vg ---- pr -----b ----

Now we will call Region I to the area between vg and pr and Region II to the area between pr and b.

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 vg and pr genes, and P2 to the recombination frequency between pr and b.

P1 = (R + DR) / N

P2 = (R + DR)/ N

Where: R is the number of simple recombinants in each region, DR is the number of double recombinants in each region, and N is the total number of individuals.

So:

Parental)

• 1779 vestigial wings, black body and purple eyes vg pr b

• 1665 normal wings, a grey body and red eyes vg+ pr+ b+

Double recombinant)

• 13 vg pr+ b

• 9 vg+ pr b+

Simple recombinant)

• 252 vg+ pr b

• 241 vg pr+ b+

• 131 vg+ pr+ b

• 118 vg pr b+

P1 = (R + DR) / N

P1 = (252+241+13+9)/4208

P1 = 515/4208

P1 = 0.122

P2= (R + DR) / N

P2 = (131+118+13+9)/4208

P2 = 271/4208

P2 = 0.064

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

P1 + P2= Pt

0.122 + 0.064 = Pt

0.186=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. Every 100 meiotic products, one of them 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 = 0.122 x 100 = 12.2 MU

GD2= P2 x 100 = 0.064 x 100 = 6.4 MU

GD3=Pt x 100 = 0.186 x 100 = 18.6 MU

---- vg ---------------------- pr ---------------------b ----

R1 R2

-----vg----12.2MU---------pr—

----pr--------6.4 MU----b—

-----vg ----------------18.6 MU--------------------b----

3 0

3 years ago

12. Se obtienen 500 mL de CO2 en el laboratorio a una temperatura de 18 °C y una presión de 560 mmHg. El volumen que ocupa a con

vfiekz [6]

Answer:

A. 345,6 mL

Explanation:

Tenemos un gas definido en ciertas condiciones (de presión, temperatura y volumen) y queremos saber el volumen que ocupará en condiciones normales de presión y temperatura, es decir, 1 atm y 0ºC.

Para resolver este problema utilizaremos la ecuación de los gases ideales:

pV=nRT

Ya tenemos todos los datos, y sólo nos resta calcular n, el número de moles. R es la constante de los gases, la cual tiene el valor redondeado de 0.082 L.atm/mol.K, por lo cual es importante convertir las unidades para que éstas se cancelen y den como resultado los moles de gas.

Dividimos 560 en 760 para pasar de mmHg a atm, 500 mL se dividen por 1000mL para pasar a litros, y a la temperatura de 18ºC se le suman 273 para pasarla a Kelvin.

Despejamos n de la ecuación:

n=p.V ÷ R.T

n= 0.740 atm × 0.50 L ÷ 0.082 L.atm/mol.K × 291 K

n= 0.0155 mol

Ahora que tenemos el número de moles del gas, podemos usar la ecuación nuevamente para calcular el volumen que tendrá el gas en condiciones normales (1 atm y 0ºC, o 273 K).

Despejamos V de la ecuación:

V= n.R.T ÷ p

V= 0.0155 mol × 0.082 L.atm/mol.K × 273 K ÷ 1 atm

V= 0.346 L ≈ 346 mL

El volumen que ocupará el mismo gas a condiciones normales será 346 mL, por ende la respuesta correcta es la A. Hay una pequeña discrepancia pero ésta es normal, se debe al redondeo del resultado al final (acá redondeo 345.6 a 346).

4 0

3 years ago

FlowersWhat do the pollen grain and the egg becomes during fertilization? They form a sepal. They become honey. They form a new

sdas [7]

Hi hi I am so glad that you’re still there for a couple hours and you have a lot to say hi

6 0

3 years ago

The roots nodules of leguminous plants contain a parasitic bacterium it is true or false

Rasek [7]

Answer:

false

Explanation: