Small mineral deposits or cells found in a urine sample are called ____

For an offspring to ___________ a recessive trait, both parents must have at least one

Crazy boy [7]

For an offspring to dominate a recessive trait, both parents must have at least one dominant allele in their genotype.

8 0

3 years ago

Select the correct answer from the drop down menu. Enzymes break down large molecules. In an experiment, a scientist adds lipase

Natalka [10]

Answer:

There are no options to this question, however, it can be answered. The answers to the blank spaces are:

1. Fatty Acids

2. A decrease in the rate of lipid breakdown

Explanation:

Lipids are large biomolecules that are formed from monomeric units called FATTY ACIDS. Digestive enzymes such as lipase as described in this question breaks down lipids into its monomer called FATTY ACIDS.

However, enzymes are proteinous molecules, meaning they are subject to denaturation when exposed to adverse conditions such as heat. According to this question, the scientist heats the enzymes and finds that it can't bind onto the lipids anymore because it has been DENATURED. This situation will result in the DECREASE IN THE RATE OF LIPID BREAKDOWN because the enzyme in charge is no longer functional.

4 0

3 years ago

Read 2 more answers

In a cell undergoing meiosis, during which stage do the sister chromatids separate from each other?

klemol [59]

Answer:

B.) Anaphase II

Explanation:

In metaphase II, the chromosomes line up individually along the metaphase plate. In anaphase II, the sister chromatids separate and are pulled towards opposite poles of the cell. In telophase II, nuclear membranes form around each set of chromosomes, and the chromosomes decondense.

4 0

3 years ago

Please help hurry!

Svet_ta [14]

Answer:

I think the answer is D: Infrographic that shows carbon moving through ecosystems

6 0

2 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

Small mineral deposits or cells found in a urine sample are called _____?