At the end of transcription it's either DNA or RNA try googling it
Answer:
Aristotle first wrote about the practice of animal testing in ancient
Greece during the third century BC.
Explanation:
Procreation comes from pairing. In different phrases you have 2 codes for each little thing. you have the code from the father and the code from the mother. evaluate the code for blood. you're attentive to that there are 4 varieties of blood variety O, variety A and variety B and variety AB. If mom has 2 codes for her blood, enable us to declare O and O and father has 2 codes for his blood enable us to declare O and A There are a sequence variety of combinatins the 1st mom O can combine with father O, OO The 2d mom O can combine with father A, OA. the 1st mom O can combine with father A, OA The 2d mom O can combine with father O, OO the baby is going to have the two OO or OA. If the baby has OO the the baby has no selection different than to have blood variety O because of fact the expressed phenotype. If the baby has OA the A is dominate and it takes expression. subsequently the only way a recessive allele could be expressed is that if and on condition that there are 2 recessive allele as interior the case of OO. Now do the different combinations mom OO mom OA mom AA mom OB mom BB mom OO can combine with father OO making OO mom OO can combine with father OA making OA mom OO can combine with father AA making OA mom OO can combine with father OB making OB mom OO can combine with father BB making BB The recessive allele is expressed if and on condition that the two codes are recessive. What if mom is dominant A. mom AA mom OA can a recessive allele be produced particular If and on condition that the recessive O of the mother combines with a recessive O of the father. subsequently a mom dominant A might desire to be coded as OA and a father dominant A might desire to be coded as OA. the baby can get carry of mom O blended with the father O and be OO, the recessive. the baby is recessive even whilst the mothers and dads are dominant because of fact the mothers and dads carried the recessive allele. this might ensue if and on condition that the two mothers and dads carry the recessive allele. pass via the mixture mom OA Father AA you will by no capacity get a recessive phenotype expressed
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
Lysosomes : Small organelles filled with enzymes which are used to break down lipids, proteins, and carbs
