Which of the following are examples of eukaryotic cells? (check all that apply)

two pure breeding lines of pea, one tall and hairy, the other short and smooth, are crossed. the f1 hybrid is tall and hairy. th

wolverine [178]

The genotypes A B_, A bb, aaB_, and aabb are matched to the four phenotypic groups.

<h3>How do phenotypic classifications work?</h3>

The genotypes A B_, A bb, aaB_, and aabb are matched to the four phenotypic groups. If any of the single homozygous recessive genotypes (i.e., A bb or aaB_) exhibits the same phenotype as the double homozygous recessive (aabb), then a 9:3:4 phenotypic ratio will result.

<h3>What does phenotypic classification entail?</h3>

phenotype, all of an organism's observable traits that emerge from the interaction of its genotype (total genetic inheritance) and environment. Observable features include things like behavior, biological characteristics, colour, shape, and size.

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8 0

1 year 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

Why do scientists use the scientific method

Arte-miy333 [17]

The scientific method is a process of steps in order to get a conclusion. First, a scientist will observe something. It could be really anything. Then he/she will ask themselves a question. Like how does it do that? Or why does it do that?

Then they will form a hypothesis. This is where you will ask yourself what will be the outcome of the experiment.

Next you actually perform the experiment. If the experiment comes out as your hypothesis predicted, then you accept your hypothesis.

If not, you reject your hypothesis.

Of course, in science you just don't do one experiment. You do it multiple times and even with different circumstances.

5 0

3 years ago

In cats, black fur color is caused by an X-linked allele; the other allele at this locus causes orange color. The heterozygote i

allochka39001 [22]

Answer:

A) tortoiseshell female; black male

Explanation:

Females have two X chromosomes and males have an X and a Y chromosome.

The possible genotypes and phenotypes are:

XᵒXᵒ: black female
XᵒY: black male
XᴼXᴼ: orange female
XᴼY: orange male
XᵒXᴼ: tortoiseshell female

Cross of a black female and an orange male

The female only produces Xᵒ gametes. The male produces Xᴼ and Y.

The possible offspring therefore is: XᵒXᴼ (tortoiseshell females) and XᵒY (black males). The answer is A.

5 0

3 years ago

_______ is a method of growing or controlling forest growth for the health and value of the environment.

Kisachek [45]

Answer:

I think its reforestation

Explanation:

8 0

3 years ago

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