On 2

In cattle, RR = red, Rr = roan, and rr = white. What are the predicted color phenotypes for the offspring from crosses between:

timofeeve [1]

Answer:

a) 100% roan

b) 50% red, 50% roan

c) 50% roan, 25% red, 25% white

Explanation:

a) A red bull has the phenotype RR, a white cow has the phenotype rr. Let's do a punnett square. In its gametes, the bull only has R alleles to pass on, and the cow only has r alleles to pass on:

Bull

R R

r Rr Rr

Cow

r Rr Rr

The offspring can only be Rr, so all of them will be roan. (100% roan).

b) A red bull has the phenotype RR, a roan cow has the phenotype Rr. Let's do a punnett square. In its gametes, the bull only has R alleles to pass on, and the cow has an R allele and an r allele to pass on:

Bull

R R

R RR RR

Cow

r Rr Rr

50% of the offspring will be RR, and 50% will be Rr. So half of the offspring will be red, and half will be roan.

c) A roan bull has the phenotype Rr, a roan cow has the phenotype Rr. Let's do a punnett square. In its gametes, the bull has an R allele and an r allele to pass on, and the cow also has an R allele and an r allele to pass on:

Bull

R r

R RR Rr

Cow

r Rr rr

25% of the offspring will be RR, 50% will be Rr, and 25% will be rr. So half of the offspring will be roan, a quarter will be red, (RR), and a quarter will be white (rr)

6 0

3 years ago

Carbon – containing energy sources formed from the decomposing a prehistoric organism or called

Anastaziya [24]

Answer:

Fossil fuels

explanation: Coal is one type of fossil fuel. This is a nonrenewable energy source whose extraction often damages the environment. Fossil fuels are made from decomposing plants and animals. These fuels are found in the Earth's crust and contain carbon and hydrogen, which can be burned for energy

:)

5 0

2 years ago

How many molecules are in 3Hg(SO)4

Vlad1618 [11]

Answer:

1.806 × 10²⁴ molecules

Explanation:

From the chemical compound given in this question, there are 3 moles of Hg(SO)4.

To calculate the number of molecules in 3 moles of Hg(SO)4, we multiply the number of moles by Avagadro number (6.02 × 10²³)

number of molecules = 3 × 6.02 × 10²³

= 18.06 × 10²³

= 1.806 × 10²⁴ molecules

6 0

3 years ago

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

class 6to 8 students come here I will give you notes class is started come plz plz and it is free class no fee need fast com

IceJOKER [234]

Answer:

......

Explanation:

BLA BLA BLA......

8 0

3 years ago