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lara [203]
2 years ago
9

There are two kinds of behavior that all organisms are capable of doing. If Inez blinks her eyes because a gnat flies close to t

hem, that's ________. But if she then swats at the gnat, that's ________. Group of answer choices instrumental; classical involuntary; voluntary voluntary; involuntary operant; instrumental
Biology
1 answer:
Kryger [21]2 years ago
6 0

Answer:

involuntary; voluntary

Explanation:

Involuntary behaviors in organisms are unlearned behaviors exhibited as a reflex response to a stimuli around their environment. They are usually unconscious behaviors. On the other hand, voluntary behaviors are deliberate behaviors exhibited by an organism consciously. They are actually under the control of the organism, and can be learned.

If Inez blinks her eyes because a gnat flies close to them, that's involuntary. But if she then swats at the gnat, that's voluntary.

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4) A homozygous groucho fly ( gro, bristles clumped above the eyes) is crossed with a homozygous rough fly (ro, eye abnormality)
docker41 [41]

Answer and Explanation:

  • A homozygous groucho fly ( gro, bristles clumped above the eyes) is crossed with a homozygous rough fly (ro, eye abnormality).
  • The F1 females are testcrossed, producing these offspring: groucho 518 rough 471 groucho, rough 6 wild-type 5 1000 a) What is the linkage distance between the two genes? B) Plot the genes on a map c) If the genes were unlinked and the F1 females were mated with the F1 males, what would be the offspring in the F2 generation?

1st cross:

Parental) grogro ro+ro+ x  gro+gro+ roro

F1) gro+gro ro+ro

2nd cross:

Parental)  gro+gro ro+ro   x  grogro roro

Gametes) gro+ro+                       gro ro

                gro+ro                         gro ro

                gro ro+                        gro ro

                gro ro                          gro ro

Punnet square)  

                   gro+ro+             gro+ro              gro ro+            gro ro  

gro ro    gro+gro ro+ro   gro+gro roro    grogro ro+ro    grogro roro

gro ro    gro+gro ro+ro   gro+gro roro    grogro ro+ro    grogro roro

gro ro    gro+gro ro+ro   gro+gro roro    grogro ro+ro    grogro roro

gro ro    gro+gro ro+ro   gro+gro roro    grogro ro+ro    grogro roro

F2)

0.518 grogro ro+ro (518 individuals)

0.471 gro+gro roro (471 individuals)

0.006 grogro roro (6 individuals)

0.005 gro+gro ro+ro (5 individuals)

Total number of individuals 1000

<u><em>Note</em></u>: These frequencies were calculated dividing the number of individuals belonging to each genotype by the total number of individuals in the F2.

To know if two genes are linked, we must observe the progeny distribution. <em>If individuals, whos </em><em>genes assort independently,</em><em> are test crossed, they produce a progeny with equal </em><em>phenotypic frequencies 1:1:1:1</em>. <em>If</em> we observe a <em>different distribution</em>, that is that <em>phenotypes appear in different proportions</em>, we can assume that<em> genes are linked in the double heterozygote parent</em>.  

In the exposed example we might verify which are the recombinant gametes produced by the F1 di-hybrid, and we can recognize them by looking at the phenotypes with lower frequencies in the progeny.  

By performing this cross we know that the phenotypes with lower frequencies in the progeny are groucho, rough and wild-type. So the recombinant gametes are <em>gro+ro+</em> and <em>gro ro</em>, while the parental gametes are <em>gro+ro</em> and <em>gro ro+.</em>

So, the genotype, in linked gene format, of the double heterozygote individual in the <u>F1</u> is gro+ro/gro ro+.

To calculate the recombination frequency we will make use of the next formula: P = Recombinant number / Total of individuals. The genetic distance will result from multiplying that frequency by 100 and expressing it in map units (MU). One centiMorgan (cM) equals one map unit (MU).

The map unit is the distance between the pair of genes for which one of every 100 meiotic products results in a recombinant product.

The recombination frequency is:

P = Recombinant number / Total of individuals

P = 6 + 5 / 1000

P = 11 / 1000

P = 0.011

The <u>genetic distance between genes,</u> is 0.011 x 100= 1.1 MU.

<u>Genetic Linkage Map:</u>

Parental Phenotypes)  

-----gro+------ro----              -----gro------ro+----

----- gro ------ro----               ---- gro------ ro ----

Recombinant phenotypes)

-----gro+------ro+----              -----gro------ro----

----- gro ------ ro----                -----gro------ro----

<u>If the genes were unlinked</u> and the F1 females were mated with the F1 males, the offspring in the F2 generation would have been

4/16 = 1/4 gro+gro ro+ro  

4/16 = 1/4 gro+gro roro  

4/16 = 1/4 grogro ro+ro    

4/16 = 1/4 grogro roro

Their phenotypic frequencies would be 1:1:1:1 related.                                                  

7 0
3 years ago
Learning Task 1: True or False. Write TRUE if the statement is correct and FALSE if it is incorrect.
vladimir2022 [97]
Answers
1. True
2. True
3. True
4. True
5. True
4 0
2 years ago
Read 2 more answers
The disease cystic fibrosis (CF) is caused by a mutation to the CFTR gene which affects the respiratory, endocrine, reproductive
Lady bird [3.3K]

Answer: DF508 mutation. A Genetic, Hereditary, Autosomal and Recessive Mutation.

Explanation:

Cystic fibrosis (CF) is a recessive autosomal lethal disease, it is most common on Caucasoid populations. Its diagnosis is suggested by the clinical features of chronic obstructive pulmonary disease, persistent pulmonary colonization (particularly with mucoid Pseudomonas strains), meconium ileus, pancreatic insufficiency with or familiarity history of the disease. The FC gene is large, with about 250 Kb of genomic DNA, 27 exons representing about 5% of genomic DNA; encodes a 6.5 kb transcribed mRNA. This mRNA is transcribed into a protein of 1480 amino acid called CFTR (Regulator Transmembrane Conductance Cystic Fibrosis). When a three-base pair deletion, adenosine-thymine-thymine (ATT) identified in the CFTR gene, exon 10, it results in the loss of a single amino acid phenylalanine at position 508 of the protein. This mutation is called DF508; “D” stands for deletion and “F” for phenylalanine amino acid.

4 0
3 years ago
miRNAs can control gene expression by what action?A. degrading proteins as soon as they are formedB. inhibiting the catalytic ac
levacccp [35]

Answer:

C) binding to mRNAs and degrading them or blocking their translation

Explanation:

<u>miRNAs:</u>

miRNAs is the abbreviation of MicroRNAs. These are the small noncoding RNAs of ∼22 nucleotides which can not code for peptides. miRNAs are responsible for gene expression regulation at the level of post transcription. They can do so by forming complementary base pairing with target mRNA and inhibiting their translation.

They silenced mRNA by the following processes:

(1) Cleavage of the mRNA strand into pieces,

(2) stopping mRNA from translation into proteins by ribosomes.

(3) Shortening of mRNA poly(A) tail and destabilizing it.

5 0
3 years ago
DNA,RNA and protein synthesis
Brut [27]

Answer:

DNA = Genetic material of cell

RNA = Transcribed from DNA

Protein synthesis= mRNA is translated into functional protein to perform different functions

3 0
3 years ago
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