What is the process shown above?

The _________ is the largest, best-known savanna.

nika2105 [10]

Hey there :)

Question- The _________ is the largest, best-known savanna.

☆ Serengeti plain

☆ savanna

☆ short grass prairie

☆ tallgrass prairie

Answer- Serengeti plain

In-Line answer- The Serengeti Plains is the largest, best-known savanna.

A little about it-

The Serengeti plains is best known for the huge herds of plain's animals, and it is the only place in Africa where vast land-animal migrations still take place. The park is an international tourist attraction, was added to the UNESCO World Heritage List. It spans about 12,000 square miles (30,000 square kilometers).

~ Benjemin360

5 0

3 years ago

Consider a hypothetical scenario in which the genes that determine body color, eye color, and wing type in fruit flies occur on

saveliy_v [14]

Answer:

The most probable location on the chromosome of each gene is:

wing shape gene ------- eye color gene -------- body color gene

Explanation:

To know if two genes are linked, we must observe the progeny distribution. In a tri-hybrid cross, If individuals, whos genes assort independently, are test crossed, they produce a progeny with equal phenotypic frequencies 1:1:1:1:1:1:1:1. If we observe a different distribution, that is that phenotypes appear in different proportions, we can assume that genes are linked in the heterozygote parent.

We can recognize the parental gametes in the descendants because their phenotypes are the most frequent, while the double recombinants are the less frequent. And simple recombinant gametes produced by the cross, which frequencies are intermediate.

By comparing the parental and the double recombinant we will realize which is the allele that changes in position. The position of this gene is in the middle of the other two genes because in a double recombinant only the central gene changes position in the chromatid.

So, according to this knowledge and the available information in the problem we can estimate the position of the genes in the chromosome.

Phenotypic proportions:

90% of flies with red eyes have curly wings (Parental)

75% of flies with red eyes have black bodies (Simple recombinant)

25% of flies with red eyes have brown bodies (Simple recombinant)

1% of flies have curly wings, white eyes, and black bodies (Double recombinant)

90% of individuals are red-eyed and curly-winged. This is a high frequency, which means that this is the parental phenotype. The other parental phenotype must be white eyes and straight wings.

Flies with curly wings, white eyes, and black bodies represent the lower proportion (1%) so this phenotype must be the double recombinant.

Flies with red eyes and black bodies, and flyes with red eyes and brown bodies are intermediate in proportion, so they must be simple recombinant phenotypes.

Parental)

Red, Curly

White, Straight

Double recombinant)

White, Curly, Black

Red, Straight, Brown

Simple recombinant)

Red, Black

Red, Brown

In the chromatid, there are two regions, Region 1 (from the first gene to the gene in the middle), and region 2 (from the gene in the middle to the gene of the other extreme).

The fact that both simple recombinants are red, tells us that the eye color gene and the body color gene are together, and the recombination occurred either in region 1 or 2.

We know that the double recombinant is White, Curly, Black. We also know that the eye color gene and the body color gene are together, so one of these two must be in the middle.

Parentals are red and curly, which means that the eye color gene and wing shape gene are also together. This leads us to assume that the eye color gene is in the middle.

Finally, the fact that the double recombinant is White, Curly, Black, and knowing that the eye color is in the middle and that both the extreme alleles change position, we can assume that the parental Red, Curly is also black, while the other parental is white, straight and brown. So:

Parental)

Curly ------ Red ------Black (90%)

Straight ----- White ----- Brown

Simple recombinant)

Curly -----White-----Brown

Straight-----Red -----Black (75%)

Curly----Red-----Brown (25%)

Straight ----White-----Black

Double recombinant)

Curly-----White------Black (1%)

Straight----Red------Brown

7 0

3 years ago

What is the sample variance of the following data, which describe the height growth of 10 foxtal pine trees grown in a common ga

maksim [4K]

Answer:

Although the data in the question is not clearly stated, the correct date is stated in the image attached to this answer and it is also stated below

What is the sample variance of the following data, which describe the height growth of 10 foxtail pine trees grown in a common garden ( give your answer to one decimal place)?

Tree Height(cm)

1 25.6987

2 19.6587

3 11.4780

4 9.11130

5 16.8747

6 14.7771

7 33.0001

8 24.0000

9 19.4441

10 14.1010

answer 51.9

Explanation:

what is a sample variance?

A sample variance (denoted as $S^{2}$ ) is used to calculate how far a set of numbers are from the mean value of the set. Mathematically, it is defined as the average of the squared deviations from its mean value.

Let the individual value of the trees be represented as x₁ to x₁₀

Let the total number of Foxtail pine trees be N

To calculate the standard deviation, we will first calculate the mean or average of the height of the trees

1. Mean (μ) = $\frac{x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}+x_{7}+x_{8}+x_{9}+x_{10} }{N}$

= $\frac{25.6987+19.6587+11.4780+9.11130+16.8747+14.7771+33.0001+24.0000+19.4441+14.1010}{10}$ = $\frac{188.1437}{10} = 18.8144$

2. we need to calculate the sum of the squares of the individual deviations.

∑[x-μ]²= $(25.6987-18.8144)^{2}+(19.6587-18.8144)^{2}+(11.4780-18.8144)^{2}+(9.11130-18.8144)^{2}+(16.8747-18.8144)^{2}+(14.7771-18.8144)^{2}+(33.0001-18.8144)^{2}+(24.0000-18.8144)^{2}+(19.4441-18.8144)^{2}+(14.1010-18.8144)^{2}=466.8786$

3. Now the variance is ; (∑[x-μ]²) ÷ (N-1) and since N=10, then N-1 = 10-1 = 9

= $\frac{466.8786}{9} = 51.8754.$

we were asked the give our answer to one decimal place

∴ Variance( $S^{2}$ )= 51.9

3 0

3 years ago

All the offspring of a cross between a tall pea plant and a dwarf pea plant are tall. This means that the allele for tall plants

Olenka [21]

Answer:

All the offspring of a cross between a tall pea plant and a dwarf pea plant are tall. This means that the allele for tall plants is DOMINANT OVER the allele for dwarf plants.

Explanation:

Mendel's law of dominance state that the heterozygous genotype expresses only one allele of a gene out of the two alleles present together. The allele that is expressed in a heterozygous genotype is said to be the dominant one as it does not allow the expression of another allele of the gene.

The allele of a gene that is not expressed in the presence of its dominant allele is said to be the recessive one. Under such conditions, a heterozygous genotype expresses the dominant phenotype. When a tall and dwarf pea plants are crossed, all the F1 progeny is tall. This means that the allele for the tallness is dominant over the allele for the dwarfism.

7 0

3 years ago

Direction: List down nine measuring tools on you answer sheet.

GarryVolchara [31]

Answer:

Calipers.

Micrometer.

Laser Measure.

Ruler.

Compass.

Square.

Measuring Tape.

Tailor's Measuring Tape.

4 0

3 years ago