Which process makes it possible for all the major organs of the body to be formed by week 10 of human

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

Where does the esophagus lead to in a frog?

irina1246 [14]

The tube that connects the mouth to the stomach.

5 0

3 years ago

Question 3 of 10

Ad libitum [116K]

I think the right answer is b

3 0

2 years ago

What is the probability that the first offspring from the cross listed below will be heterozygous for all loci?

Tanya [424]

Answer and Explanation:

Cross: aa Bb dd Ee x AA bb Dd Ee

We can calculate the probability of getting heterozygous individuals in the progeny by using the product rule. Assuming that these four genes assort independently (events that occur independently from each other), we can infer that the F1 will have the next genotypic proportions for each gene:

1) aa x AA

F1) 4/4=1 Aa

2) Bb x bb

F1) ½ bb

½ Bb

3) dd x Dd

F1) ½ dd

½ Dd

4) Ee x Ee

F1) ¼ EE

2/4 = ½ Ee

¼ ee

So, to know what the probability is that the F1 of being heterozygous for all loci, we must multiply the respective individual probabilities of getting a heterozygous genotype, like this:

1 Aa x ½ Bb x ½ Dd x ½ Ee = 1/8 AaBbDdEe

6 0

3 years ago

Check your understanding. Which items did you

s2008m [1.1K]

RNA nucleotides form complimentary base pairs with DNA bases. Hydrogen and sugar-phosphate bonds form and the mRNA is synthesized.

Explanation:

The process of mRNA formed by DNA is called transcription. The genes coding for the protein is present on DNA as a nitrogenous base or nucleotides.

Transcription takes place inside the nucleus with the assistance of major enzyme RNA polymerase. In transcription 5'to3' single strand of RNA is synthesized which is complementary to the 3'to5' strand of DNA. The sequence of DNA has promoter sites which will be recognized and DNA will unzip to form transcription bubble. Several transcription factors and enzyme will be required to synthesize mRNA.

This mRNA after some modification would become primary transcript and pass on to the cytoplasm where protein synthesis will take place.

8 0

3 years ago