What is an adaption?

a black rooster Mated with a white hen. based on genotypes, the chance that the offspring will be checkered is 100 percent. what

gtnhenbr [62]

Answer:

The genotypes of the rooster and the chicken are homozygous and that of their offspring is heterozygous.

This case is called codominance, where the offspring receives an allele from each parent, from the rooster and the hen, because there is codominance, so neither allele is recessive and the phenotype of both alleles is expressed so the phenotype of the offspring is checkered.

We can say then that the chicken and the rooster have equal strength between their alleles. in the cases of codominance the laws of mendel do not apply.

5 0

3 years ago

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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

2 years ago

When most liquids are frozen, the molecules get closer together. In water, however, hydrogen bonds cause individual water molecu

Andre45 [30]

When water freezes(ice) it expands slightly
think ice cube tray, and why the amount of ice you get is more than the amount of water you put in

5 0

2 years ago

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What does a centromere hold together?

Rudiy27

Answer:

Structure in a chromosome that holds together the two chromatids

Explanation:

8 0

2 years ago

Some species form social groups only with members of the same gender. For example, elephants form groups with only females and t

Ostrovityanka [42]

Answer:

When the females tend to form a solitary group, the adult males have an advantage of mating with females in different social groups thus enhancing on the productivity of the species.

Explanation:

When both the sexes of the species tend to make a group, the males and the females of the group have to mate with the members of that particular group and hence the males have restricted mating options which reduce the size of the herd. When the females have a separate social group, the males can mate with the members of different groups and raise the herd size in a short period of time. This helps in the increase in the population size and also helps in avoiding inbreeding depression which happens among small grouped animals.

5 0

3 years ago

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