A hypothesis that does not generate a testable prediction is not useful

____________ is translated by the ribosomes and contains the code that specifies the sequence of amino acids in a polypeptide ch

HACTEHA [7]

mRNA (Messenger RNA) is translated by ribosomes and contains the code that specifies the sequence of amino acids in polypeptide chain.

A single-stranded ribonucleic acid molecule is known as messenger RNA(mRNA) plays a major role in translation.

Translation is the method by which an mRNA codes for a certain protein. mRNA provides the template for protein synthesis.

The ribosome translates the mRNA that is produced from the DNA into a chain of certain amino acids and protein synthesis is facilitated by this amino acid sequence.

The genetic code, which connects the DNA sequence to the amino acid sequence of proteins, is used to "read" the mRNA. Each group of three nucleotides in mRNA forms a codon, and each codon corresponds to a particular amino acid (triplet code).

Thus mRNA contains the code that specifies the sequence of amino acids in a polypeptide chain.

Learn more about different type of RNA here brainly.com/question/21177344

#SPJ4

4 0

2 years ago

Bruce has a genetic disorder. bruce and his wife, kim, have three children, one of which has the genetic disorder. how is this d

laiz [17]

This would be a recessive disease.

Father is dd
Mother is Dd (has to be because one child has the disease)

Parents cross is dd x Dd which gives rise to Dd (0.5) and dd (0.5). Each time they have a child they have a 50% chance of the child having the disease. In their case, only one of their 3 children is dd. The others though are carriers!

5 0

3 years ago

What is an example of a stimulus and a response that humans exhibit

PtichkaEL [24]

A mosquito bite, human response to smack their arm as a way of trying to defend

4 0

3 years ago

4) A homozygous groucho fly ( gro, bristles clumped above the eyes) is crossed with a homozygous rough fly (ro, eye abnormality)

docker41 [41]

Answer and Explanation:

A homozygous groucho fly ( gro, bristles clumped above the eyes) is crossed with a homozygous rough fly (ro, eye abnormality).

The F1 females are testcrossed, producing these offspring: groucho 518 rough 471 groucho, rough 6 wild-type 5 1000 a) What is the linkage distance between the two genes? B) Plot the genes on a map c) If the genes were unlinked and the F1 females were mated with the F1 males, what would be the offspring in the F2 generation?

1st cross:

Parental) grogro ro+ro+ x gro+gro+ roro

F1) gro+gro ro+ro

2nd cross:

Parental) gro+gro ro+ro x grogro roro

Gametes) gro+ro+ gro ro

gro+ro gro ro

gro ro+ gro ro

gro ro gro ro

Punnet square)

gro+ro+ gro+ro gro ro+ gro ro

gro ro gro+gro ro+ro gro+gro roro grogro ro+ro grogro roro

F2)

0.518 grogro ro+ro (518 individuals)

0.471 gro+gro roro (471 individuals)

0.006 grogro roro (6 individuals)

0.005 gro+gro ro+ro (5 individuals)

Total number of individuals 1000

Note: These frequencies were calculated dividing the number of individuals belonging to each genotype by the total number of individuals in the F2.

To know if two genes are linked, we must observe the progeny distribution. If individuals, whos genes assort independently, are test crossed, they produce a progeny with equal phenotypic frequencies 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 double heterozygote parent.

In the exposed example we might verify which are the recombinant gametes produced by the F1 di-hybrid, and we can recognize them by looking at the phenotypes with lower frequencies in the progeny.

By performing this cross we know that the phenotypes with lower frequencies in the progeny are groucho, rough and wild-type. So the recombinant gametes are gro+ro+ and gro ro, while the parental gametes are gro+ro and gro ro+.

So, the genotype, in linked gene format, of the double heterozygote individual in the F1 is gro+ro/gro ro+.

To calculate the recombination frequency we will make use of the next formula: P = Recombinant number / Total of individuals. 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 one of every 100 meiotic products results in a recombinant product.

The recombination frequency is:

P = Recombinant number / Total of individuals

P = 6 + 5 / 1000

P = 11 / 1000

P = 0.011

The genetic distance between genes, is 0.011 x 100= 1.1 MU.

Genetic Linkage Map:

Parental Phenotypes)

-----gro+------ro---- -----gro------ro+----

----- gro ------ro---- ---- gro------ ro ----

Recombinant phenotypes)

-----gro+------ro+---- -----gro------ro----

----- gro ------ ro---- -----gro------ro----

If the genes were unlinked and the F1 females were mated with the F1 males, the offspring in the F2 generation would have been

4/16 = 1/4 gro+gro ro+ro

4/16 = 1/4 gro+gro roro

4/16 = 1/4 grogro ro+ro

4/16 = 1/4 grogro roro

Their phenotypic frequencies would be 1:1:1:1 related.

7 0

3 years ago

Is a desk being pushed by a student

Vsevolod [243]

Answer: a) applied force

Explanation: the student is applying force onto the desk when they push it.

5 0

3 years ago