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3 years ago

Polysaccharides are made of?

Biology

2 answers:

stellarik [79]3 years ago

7 0

Answer:

b. Many glucose molecules

Explanation:

poly means many

Tatiana [17]3 years ago

3 0

D definetly
Hopefully this helps you
May God bless you

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When crossing over occurs between two linked genes in an F1 dihybrid testcross, the frequency of the parental phenotypes will al

NeX [460]

Answer:

Greater.

Explanation:

Linked genes may be defined as the genes that are present on the same chromosome or may close enough that can pass in the next generation together.

The results of the crossing over in linked genes result in the formation of recombinant progeny. The frequency of the recombinant progeny will always be lesser than the parental progeny and cannot be more than 50%. If it is more than 50% these genes are considered as linked genes.

Thus, the answer is greater.

4 0

3 years ago

Q: Those are called homolgous structures because the skeletons are very similar and imply a common ancestor!!!! HOW DO I ANSWER

katrin2010 [14]

The skeletons may be similar but it usually focuses on one part. Meaning that the homologous structure would be a monkey paw, or the flipper of a whale.

They would be in the same classification. Such as humans are called homo sapien sapien. The family would be homo if I'm correct, meaning that if scientists are able to use the DNA of that structure to relate it to a common ancestor.

It would be true. Or if it's a multiple choice question, we need to know the choices.

6 0

4 years ago

I need help with this

antoniya [11.8K]

Answer:

don't know what do

Explanation:

3 0

3 years ago

Which of the following best explains what would happen if there were no decomposers in an ecosystem?

azamat

Answer: if there were no decomposers in an ecosystem then, wastes and the remains of dead organisms would pile up and the nutrients within the waste and dead organisms would not be released back into the ecosystem.

Explanation:

5 0

3 years ago

22. A plant geneticist is investigating the inheritance of genes for bitter taste (Su) and explosive rind (e) in watermelon (Cit

mylen [45]

Answer:

See the answer below

Explanation:

Using the formula for calculating Chi square ( $X^2$ ):

$X^2$ = $\frac{(O - E)^2}{E}$ where O = observed frequency and E = expected frequency.

The observed frequencies for the four phenotypes are 88, 62, 62, and 81 respectively.

For the expected frequency, the phenotypes are supposed to assort in 9:3:3:1 according to Mendel ratio.

Hence, expected frequencies are calculated as:

phenotype (1) = 9/16 x 293 = 164.81

phenotype (2) = 3/16 x 293 = 54.94

phenotype (3) = 3/16 x 293 = 54.94

phenotype (4) = 1/16 x 293 = 18.31

The $X^2$ is calculated thus:

Phenotype O E $X^2$

1 88 164.81 $\frac{(88 - 164.81)^2}{164.81}$ = 35.80

2 62 54.94 $\frac{(62 - 54.94)^2}{54.94}$ = 0.91

3 62 54.94 $\frac{(62 - 54.94)^2}{54.94}$ = 0.91

4 81 18.31 $\frac{(81 - 18.31)^2}{18.31}$ = 214.64

Total $X^2$ = 35.80 + 0.91 + 0.91 + 214.64 = 252.26

To the nearest tenth = 252

Degree of freedom = n - 1 = 4- 1 = 3

$X^2$ tabulated = 7.815

<em>The calculated </em> $X^2$ <em> exceeds the critical value, hence, the null hypothesis is rejected. The two genes did not assort independently. </em>

4 0

3 years ago