Answer:
b. It controls movements on the right side of the body.
Explanation: is correct
Answer:
There is 1/8 of the parent C-14 left. 1/8 means that 3 half-lives have elapsed. Multiply the half-life of C14 by 3: 5,730 years x 3 = 11,460 years old. The sample is only 11,460 years old, to the claim is not valid.
Answer:
The gene of interest from another source.
Explanation:
A plasmid is a small, circular double-stranded extrachromosomal DNA molecule within a cell particularly a bacteria cell that is physically separated from chromosomal DNA and can also replicate independently.
Expression vector is usually a plasmid (or a virus can also be utilized) constructed to allow for gene expression in cells. The vector is used to introduce a gene of interest into a target cell, and can drive the cell's mechanism for protein synthesis to produce the protein encoded by the gene, thus giving rise to the expression of the gene.
This gene of interest is integrated into the plasmid that is cut by endonucleases and then ligated using DNA ligase. Thus the expression differs from the normal plasmid in that it contains an external gene sequence.
Answer:
4. Infrared Light 5. Ultra-Violet Light 6. ...damage...sunburn.
Explanation:
Looking at a Electromagnetic Spectrum will tell you the the visible and non visible. Hope this helps!
Answer:
0.035
Explanation:
<u>cv+ is the wild-type dominant allele over cv, therefore:</u>
- cv+cv+ and cv+cv cause wild-type phenotype for crossveinless
- cv cv causes the crossveinless phenotype
<u>Sb is a dominant mutant allele over wild-type Sb+, therefore:</u>
- Sb Sb and Sb Sb+ cause Stubble phenotype
- Sb+ Sb+ causes wild type phenotype for Stubble
<h3><u>Test cross</u></h3>
It's the cross between the heterozygous female with a homozygous recessive male. Remember that cv and Sb+ are the recessive alleles.
X 
-The male produces only 1 type of gamete: cv Sb+
-The female produces 4 types of gametes:
- cv Sb+ ] Parental
- cv+ Sb ] Parental
- cv Sb ] Recombinant
- cv+ Sb+ ] Recombinant
The genes are linked and separated by 7 map units. A distance of 7 mu means that 7% of the resulting gametes will be recombinant. Because there are 2 possible recombinant gametes, each of them will appear in 3.5% of the cases.
The genotypes and proportions of the offspring resulting from the test cross can be seen in the Punnett Square. The phenotypically wild-type individuals will have the genotype cv+ Sb+ / cv Sb+ (heterozygous for crossveinless and homozygous recessive for Stubble) and a 0.035 proportion.