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Nadusha1986 [10]
2 years ago
15

Two frog populatioes of the same species, living in two neighboring lakes sing slightly different courtship songs. Now make a di

fferent Assumption: that courtship song differences have a genetic basis and that fitness differences are linked to the song tynes i.c. cach population has highest fitness in its home pond Now predict what likely happens to the songs of the two frog populations after increased irrigation makes the land between the two lakes wetter, and therefore leads to increased connectedness between the two lakes
a) There is no change in the frog courtship songs of the two lakes
b) Female preferences for males from her own population will decrease
c) The songs will become more similar to each other
d) The songs will become more differeet from each other.
Biology
1 answer:
Likurg_2 [28]2 years ago
6 0

Answer:

The songs will become more similar to each other.

Explanation:

Speciation may be defined as the process of formation of new species by the external or internal force. Two main types of speciation are allopatric speciation and sympatric speciation.

Two frog population belongs to the same species that lives two neighboring lakes. The irrigation makes the pond wetter and connect the two species. The  songs of the two species tends to be more similar in both species s singing is genetic basis and allows the mating between two species.

Thus, the correct answer is option (c).

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Genetic linkage mapping for a large number of families identifies 4% recombination between the genes for Rh blood type and ellip
Cloud [144]

Answer:

0.2404

Explanation:

The genes R/r and E/e are linked and there is 4% recombination between them.

<u>The possible genotypes and phenotypes are:</u>

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  • rr: Rh- blood type
  • EE or Ee: elliptocytosis
  • ee: normal red blood cells

Tom and Terri each have elliptocytosis (they are E_), and each is Rh+ (they are R_).

Tom's mother has elliptocytosis (E_) and is Rh- (rr), so she has the genotype Er/_r. His father is healthy (ee) and has Rh+ (R_), so he has the genotype eR/e_. Tom must have inherited his E allele from his mother and his R allele  from his father, so he has the genotype eR/Er.

Terri's father is Rh+ (R_) and has elliptocytosis (E_), while Terri's mother is Rh- (rr) and is healthy (ee) with the genotype er/er. Terry could only receive the chromosome <em>er </em>from her mother, and because she is heterozygous for both genes the dominant alleles were both received from her father. Terri's genotype is ER/er.

The frequency of recombination is 4%, so 4% of the produced gametes will be recombinant. There are two possible recombinant gametes, so each will appear 2% of the times (a frequency of 0.02).

<u />

<u>Tom will produce the following gametes:</u>

  • eR, parental (0.48)
  • Er, parental (0.48)
  • er, recombinant (0.02)
  • ER (recombinant (0.02)

<u>Terri will produce the following gametes:</u>

  • ER, parental (0.48)
  • er, parental (0.48)
  • Er, recombinant (0.02)
  • eR, recombinant (0.02)

A child Rh- with elliptocytosis has the genotype rrE_. This can happen from the independent combination of the following gametes from Tom and Terri respectively:

  • Er (0.48) × er (0.48) = 0.2304 Er/er
  • Er (0.48) × Er (0.02) = 0.0096 Er/Er
  • er (0.02) × Er (0.02) = 0.0004 er/Er

And the total probability of having a rrE_ child will be 0.2304 + 0.0096 + 0.0004 = 0.2404

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Answer:

Explanation:

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5'-AGC GGG AUG AGC GCA UGU GGC GCA UAA CUG-3'
    SER  GLY  MET SER ALA  CYS GLY ALA  STOP LEU 

Note that the protein would stop being made at the stop codon and the LEU wouldn't matter at the end...

Now, I will remove one bp...(I bolded it up top). Rewrite the mRNA and find the corresponding AA...

NEW
5'-AGC GGG AUG GCG CAU GTG GCG CAU AAC UG-3'
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