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4 years ago

SOMEONE HELP ME PLEASEEEE!!!!!!!

Biology

2 answers:

Aleksandr [31]4 years ago

7 0

Chromosphere C. Layer of the sun that is known for its red color and can only be seen during eclipses

Convective zone B. Layer of the sun where energy is transferred by convection

Core D. The central and hottest section of the sun

Radiative zon A. Layer of the sun where energy is transferred by thermal conduction

givi [52]4 years ago

3 0

Answer:

with what?

Explanation:

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A cat is discovered with a very short tail. What would prove that this change was caused by a genetic mutation and not because t

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A cat is discovered with a very short tail. Checking offspring for short tails will <span>prove that this change was caused by a genetic mutation and not because the cat lost most of its tail at some point in its life.</span>

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A codon is a set of three nucleotides that correspond to a specific amino acid. The table below shows various DNA codons and the

Finger [1]

Answer:

1a. The template DNA strand, from which the mRNA is synthesized, is 5’ CAAACTACCCTGGGTTGCCAT 3’

(RNA synthesis proceeds in a 5’ à 3’ direction, so the template strand and the mRNA will be complementary to each other)

b. The coding DNA strand, which is complementary to the template strand, is 5’ ATGGCAACCCAGGGTAGTTTG 3’

c. The sequence of the mRNA is 5’ AUGGCAACCCAGGGUAGUUUG 3’

(the sequence of the mRNA is complementary to the template strand and identical to the coding strand with U substituted for T)

d. The third codon is 5’ ACC 3’. Therefore, the corresponding anti-codon is 5’ GGU 3’

2. Below is a table for the genetic code:

TTT Phe (F)

TTC "

TTA Leu (L)

TTG "

TCT Ser (S)

TCC "

TCA "

TCG "

TAT Tyr (Y)

TAC "

TAA Stop

TAG Stop

TGT Cys (C)

TGC "

TGA Stop

TGG Trp (W)

CTT Leu (L)

CTC "

CTA "

CTG "

CCT Pro (P)

CCC "

CCA "

CCG "

CAT His (H)

CAC "

CAA Gln (Q)

CAG "

CGT Arg (R)

CGC "

CGA "

CGG "

ATT Ile (I)

ATC "

ATA "

ATG Met (M)

ACT Thr (T)

ACC "

ACA "

ACG "

AAT Asn (N)

AAC "

AAA Lys (K)

AAG "

AGT Ser (S)

AGC "

AGA Arg (R)

AGG "

GTT Val (V)

GTC "

GTA "

GTG "

GCT Ala (A)

GCC "

GCA "

GCG "

GAT Asp (D)

GAC "

GAA Glu (E)

GAG "

GGT Gly (G)

GGC "

GGA "

GGG "

a. The following codons can be mutated by one base to produce an amber codon:

CAG Gln

AAG Lys

GAG Glu

TCG Ser

TTG Leu

TGG Trp

TAA Stop

TAT Tyr

TAC Tyr

b. From part a, CAG (Gln) and TGG (Trp) can become amber stop codons through EMS.

c. From part b, both of the resulting amber codons could be suppressed by amber nonsense suppressors generated by EMS.

3a. The codon is the three nucleotide sequence in the mRNA that indicates which amino acid should be incorporated in the growing polypeptide chain. The anticodon is the complementary three nucleotide sequence in the appropriate tRNA.

b. Template strand is the DNA strand off which the mRNA is synthesized. The coding, or non-template, strand is the DNA strand complementary to the template strand; it has the same sequence (except for T for U substitutions) as the mRNA.

c. The Pribnow box is a sequence of six nucleotides (TATAAT) positioned at -10 that signals where transcription initiation should begin in prokaryotic DNA. The Shine-Delgarno sequence is a short, purine-rich region in the mRNA that is complementary to the rRNA within the 16S ribosomal subunit. The sequence signals which AUG acts as the translation start in mRNA.

4a. False, a wobble allows the anticodon in the tRNA to hybridize with different codons in mRNA.

b. False, a frameshift mutation affects all the subsequent amino acids.

c. False, only one codon (AUG) encodes for the start of protein synthesis; three codons signal the end of protein synthesis.

d. False, the wobble is first base (5’ to 3’) in the anticodon.

e. True, RNA can be used as a template for DNA synthesis in a process known as reverse transcription.

f. True. For example, a single base substitution causing CAT to change to AAT would signal a termination.

g. False, the Wobble Hypothesis explains how alternate base pairing can occur with the first nucleotide (going from 5' to 3') in the anticodon.

5a. Digestion of RNA with alkali will cleave the strand after each 3’ phosphate. Therefore, the products remaining will consist of pppNp, Np, and N-OH

b. If RNA was synthesized in the 3’ to 5’ direction (i.e. by adding ribonucleotides to the 5’ end), then the pppNp and Np fragments should be labeled with tritium.

c. If RNA was synthesized in the 5’ to 3’ direction (i.e. by adding ribonucleotides to the 3’ end), th

Explanation:

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You are studying a population of ground squirrels. You trap 100 individuals and find 9 that have a light colored coat, and the r

Genrish500 [490]

The question is incomplete. Here is the complete question.

You are studying a population of ground squirrels. You trap 100 individuals and find 9 that have a light colored coat and the rest have a dark coat. You know that light coat color is recessive in ground squirrels. What is the allele frequency for the light coat allele and the dark coat allele? How many of the dark coated individuals do you expect to be heterozygotes at the coat color gene? Assume the population is at H-W equilibrium.

Answer: Allele frequency for light coat: q = 0.3;

Allele frequency for dark coat: p = 0.7;

There will be 42 heterozygote;

Explanation: A population in H-W equilibrium is a population which doesn't change from one generation to the next if there are no disturbing factors, as migration or mutation.

It follows the equation: $p^{2}+2pq+q^{2}=1$ .

p represents the frequency of the dominant allele;

q represents the frequency of the recessive allele;

2pq represents heterozygote individuals;

The sum of frequencies of both alleles must be 1, i.e.: $p+q=1$

For the population of squirrels:

Frequency of recessive allele

$q^{2} = \frac{9}{100}$

$q = \sqrt{\frac{9}{100}}$

q = 0.3

Frequency of dominant allele

$p+q=1$

$p=1-q$

$p=1-0.3$

p = 0.7

Frequency of heterozygote

2pq = 2*0.7*0.3 = 0.42

In a population of 100 individuals: 0.42*100 = 42

The allele frequency for the light coat is 0.3; for the dark coat is 0.7.

It is expected to have 42 heterozygote individuals in this population.

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