1 molecule of glucose and 2 ATPs
Answer:
The correct option is B) The homologous structures are evidence of descent with modification from a common ancestor.
Explanation:
Homologous structures can be defined as similarities between organs or skeletal of organisms which suggest that they might have a common ancestor in the past. Homologous structures will have alike structures although they may be present in animals of different texa.
For example, The front legs of a frog, which is an amphibian is similar to alligator which is a reptile. This common structure show that they have homologous structures and hence might have had a common ancestor in the past.
Answer:Framed By Your Own Cells: How DNA Evidence Imprisons The Innocent ... But due to the touch-transfer properties of DNA, determining how those cells ... If we were to swab the man's hand for DNA, we might find the man's DNA, his ... Touch-transfer DNA "could falsely link someone to a crime" and forensic ...
Explanation:
Following are the steps to be taken while handling chemicals with precautions:
1. Application of droppers to transfer small amounts of chemicals into another glassware, as it will inhibit the unwanted overflow of the chemical.
2. Check the test tubes for chips before using them, otherwise, it may leak the chemicals within them.
3. Application of tongs to move the hot components as they may even result in extreme burns.
4. Washing the hands after concluding the lab experiment is necessary, as it will make sure that the food does not get intoxicated with the chemicals present in the hands.
Answer:
1. heterozygous yellow and star
2. 37
3. 1/8
4. 168
5. 1/4
Explanation:
Given ,
In f1 generation a cross is made between a true breeding black star bellied sneetch mated with a true breeding yellow starless sneetch
yySS x YYss
It is taken as - Y (yellow) is dominant over y (black)
and S (star) is dominant over s (starless)
1. F1 Generation
Genotype of parents yySS X YYss
gametes - yS, yS, Ys, Ys
All 16 offspring will have genotype YySs
phenotype would be heterozygous yellow and star
2. F2 generation cross
YySs X YySs
YS Ys yS ys
YS YYSS YYSs YySS YySy
Ys YYSs YYss YySs Yyss
yS YySS YySs yySS yySs
ys YySs Yyss yySs yyss
Genotype of offspring are –
YYSS – 1
YYSs – 2
YySS – 2
YySs – 4
YYss- 1
Yyss- 2
yySS – 1
yySs- 1
yyss- 1
2. Out of 16, 2 are black star bellied sneetches . Which means only 1/8 are black star bellied sneetches
So out of 300, 37 are black star bellied sneetches
3. Only 2 out of 16 are true breeding. i.e 1/8
4. 9 out of 16 are yellow star bellied sneetches, so out of 300, 168 are yellow star bellied sneetches
5. 4 out of 16 are true breeding yellow. Thus, ¼ are true breeding
