Answer:
Write the amount similar dominant and different squares as one phenotypic group. Count the amount of similar recessive squares as another group. Write the result as a ratio of the two groups. A count of 3 from one group and 1 from the other would give a ratio of 3:1.
Explanation:
hope this helps and why you said thx
A mutation may result change in the way you look, you could lose sight, could be born with some genetic issues.
The answer is 4 diploid cells.
Meiosis is a cell division which results in the reduction of chromosome number by half - from diploid to haploid - in daughter cells. It consists of meiosis I and meiosis II. Meiosis I produces <em>two haploid cells</em>.<span> Meiosis II is analogous to mitosis, so in total, meiosis results in four haploid cells. This is achieved through suppression of DNA replication between two meiotic divisions.
If there were no </span><span>suppression of DNA replication, then meiosis I would produce <em>two diploid cells</em>, and after meiosis II there will be four diploid cells.</span>
biotic- living things
abiotic- non living things
Answer/ Explanation:
a. The genotype of a homozygous white eyed long winged female would be Vg+Vg+XrXr. We denote the white allele as recessive (r) because the XY male only has one copy and yet has red eyes, so the red eye trait (R) must be dominant. A homozygous red eyed vestigial winged male would have be VgVgXRY. The possible gametes for the female are Vg+Xr only. For the male, the possible gametes are VgXR or VgY
The attached punnett square shows the results of the cross. The females will all be Vg+VgXRXr. The males will all be Vg+VgXRY (must inherit Y from father). That means they will all have normal length wings, the males will have white eyes and the females will have red eyes.
b. The F2 flies arise from intercrossing the F1, so the cross will be Vg+VgXRXr x Vg+VgXRY. The possible gametes for the mother are: Vg+XR, Vg+Xr, VgXR or VgXr. The possible gametes for the father are Vg+Xr
, Vg+Y
, VgXr
, VgY
. The attached punnet square shows this cross. The ratio of the phenotypes will be 6:6:2:2, or 3:3:1:1 (long-winged red eye: long-winged white eye: vestigial wing red eye: vestigial wing white eye), genotypes shown in the attachment.
c. F1 cross back to the mother would be Vg+VgXRY x Vg+Vg+XrXr. The genotypes are shown in the attached punnet square. The offspring will all be long-winged with white eyes. The F1 to the father would be Vg+VgXRXr x VgVgXRY. The ratio would be 3:3:1:1 long-winged red eye: long-winged white eye: vestigial wing red eye: vestigial wing white eye
