17

Your reasoning is sound and correct. The answer key is wrong. An unclotted blood sample needs something to prevent clotting. Extracellular calcium is required for both the coagulation cascade and platelet activation. It even has its own name in this context, Factor IV. This why EDTA, a calcium chelator, is used in some blood collection tubes to delay ...


10

There is no answer to your question The doubling time of a bacterium is dependent on the conditions: primarily on the temperature and the availability of nutrients, but other factors can apply. Because of this, the doubling time of any particular bacterial strain cannot be considered in the abstract, thus to give a single number we can consider the optimum ...


7

Short answer Sedges have edges, and they're in different families. See Minnesota Wildflowers for a great summary with images. Long answer Both are in the order Poales, but they are in different families: Grasses = Poaceae (of the graminid clade) Sedges = Cyperaceae (of the [non-monophyletic]1 cypirid lineage) Some anatomical differences: ...


6

The simple answer Under the assumption that each mRNA molecule is translated at least once, by necessity translation will happen more often than transcription. This is because the only way to get a protein is to translate an mRNA. In other words, as long as there are more protein molecules (translation products) than mRNA molecules in a cell, then the ...


5

I think the question's phrasing is arguably a little wobbly, typically as to the nature of what someone would mean by the term "reptile". But I think you understood what your teacher was expecting. I agree with you, answer is B. Reptilia is a clade (clade = monophyletic group) that encompasses all species that we traditionally classify as reptiles + all ...


3

Yes, Agrobacterium is indeed a very widely used vector in plants. So it wouldn't be wrong to consider bacteria as vectors. Just to add, it's worth noting that in recent times, bacterial vector options have been explored in the case of humans also, especially in the case of gene therapy for cancer treatment, though its success hasn't been demonstrated yet. ...


3

TL;DR No, it won't die. Lizards/snakes are usually immune to their own venom. Since I couldn't find a good answer to this recurring question on this site, I will try to summarize it here. I found more research on snakes than on lizards, but for now we will just assume that similar mechanisms can be found in lizards. First, the question is how the venom ...


2

Hormones have antigens. Usually they are antigens our immune systems recognize as self. Pretty much any big biological molecule will have antigens. FSH is no exception. The binding of FSH to anti-FSH antibodies (in a laboratory system) is one way to measure FSH levels in a human. Definition and Measurement of Follicle Stimulating Hormone B. High-...


2

Welcome to Biology.SE! I think you've come to a reasonable conclusion, but not for exactly the right reasons. The genomes of several strains of Escherichia coli have been sequenced (e.g. https://www.ncbi.nlm.nih.gov/nuccore/U00096) so you could predict the sizes of the fragments you would expect from your digest. The genome size of E. coli is about 4.6 ...


2

The individual values in the OP's onion label are not out of line with the U.S. Department of Agriculture's reference data. There are two of the latter, one for "Onions, raw" and one for "Onions, sweet, raw". Their figures would lead to expecting carbohydrates of 13.8g or 11.2g, respectively, for a 148g sample. The negative difference can be explained by ...


2

The values on the nutrition label from the question (11 g total carbs, 9 g sugars, 3 g fiber) could be actually explained by rounding: Total carbohydrates: 11.45 (rounded to 11) Sugars: 8.55 g (rounded to 9 g) Fiber: 2.55 g (rounded to 3 g) ...which would mean there are additional 0.35 grams of non-sugar, non-fiber carbohydrates... ...BUT... ...using ...


2

To answer this question, first, we must define the terms. Major primary plant organs can be characterized as either roots, stems, or leaves. A stem is a structure that contains nodes and internodes (as a previous answer explains), and axillary buds. These buds mean that stems can give rise to more stems, roots, or leaves. Branches can be defined as stems ...


1

Agrobacterium tumefaciens and Agrobacterium rhizogenes are soil-based plant pathogenic bacterial strains containing plasmid. This plasmid is known as Ti plasmid and is responsible for inducing tumor. Part of this plasmid called T-DNA can be integrated into the host chromosomes. So, this bacterium plasmid act as vector, but not the whole bacterial cell. (...


1

I will try to help. But, in the future, please post a picture of the confusing diagram to help us help you. edit: now that I see your diagram, I'm not sure if this answer is helpful. Will re-consider and possibly edit this response. Thanks for including it! Grana lamellae refer to the lamellae (membranes) in direct contact with photosynthetic stacks (...


1

I said that there isn't enough information to tell since the parents genotypes are unknown. It seems like you are misreading the diagrams since you say the parent's genotypes are unknown. All 8 individuals in the two pedigrees have the same information given for each, appearing in the eight columns directly beneath the eight individual's symbols.


1

Based on the information given in the pedigree, you can actually be certain that #2's mate has a genotype of CdCa because of the following: Two of 5's siblings are albinos. Since albino is a recessive trait (lowest in hierarchy of dominance), both parents (4 and 4's mate) must have the Ca allele. Let's call 4's mate "8". Since 8 has the Ca allele, at least ...


1

As alluded to in the comments above, remember that you are amplifying double-stranded DNA, so two new reverse-complementary strands should be generated from each cycle (so that they can hybridize with eachother). As written in your answer, both of your primers would hybridize with your given strand, and the amplicons would not overlap (one binds at each end, ...


1

If that is the case, why would C4 plants make PEPcarboxylase the primary acceptor instead of RuBisCO? Since it is not so, C4 plants were forced to take this alternative and make RuBisCO confined only to the bundle sheath cells. The problem in C3 was photorespiration, since oxygen used to competitively inhibit the uptake of carbon dioxide by RuBisCO. Hence it ...


1

I am going to post an answer that is maybe not helpful for the exact goals, but which will maybe help restate the problem to be easier. As commenters noted it is super big and complex, many whole careers have been devoted to solving very small parts of this problem. Possible problems with premise I think that this question starts from this statement as a ...


Only top voted, non community-wiki answers of a minimum length are eligible