Activity: Alien DNA

Scenario: Alien DNA

NASA’s ‘Exobiology Branch’ (http://exobiology.nasa.gov/) supports research to increase knowledge on the origin, evolution, and distribution of life in the universe. An area of current research seeks to understand what life might look like beyond Earth.  It is possible that the biochemistry and biological processes of extraterrestrial life are quite different than those on Earth.  One of the areas of interest to researchers in this field is the genetic information system that extraterrestrial life may have evolved.  Imagine that you are a member of a research team focusing on genetic information systems in samples collected from Mars by a probe that has just returned to Earth.

Initial analysis reveals the presence of eukaryotic, unicellular life in a sample.  Additional analysis indicates the presence of double-stranded nucleic acids similar, although not identical, to DNA found in eukaryotic organisms on Earth.  The alien nucleic acids consist of two strands of nucleotides bonded together, with the sugar-phosphates on the outside of the structure.  Six nitrogen bases located on the interior of the double-stranded nucleic acid are identified as:  isoAdenine (iA), Uracil (U), isoCytosine (iC), isoThymine (iT), Adenine (A) and betaGuanine (bG). Data also suggests a complementary base-pairing between nitrogen bases of the two strands with the following base paring rules:

• iA complementary with U
• iT complementary with A
• bG complementary with iC

For items 1-3 utilize the following information:

A double-stranded piece of alien nucleic acid has been isolated.  Each individual strand of the nucleic acid is 160 nucleotides long; for a total of 320 nucleotides in the isolated piece of nucleic acid.  An initial analysis shows that of the 320 nucleotides, 95 contain the base uracil.

1. How many nucleotides containing the base isoAdenine (iA) would be found in the isolated piece of double-stranded nucleic acid?

1. 0
2. 95
3. 190
4. 225
5. cannot be determined

1. How many nucleotides in the piece of alien DNA would contain the base adenine?

1. 40
2. 70
3. 95
4. 140
5. cannot be determined

1. If it were determined that the sugar-phosphates making the nucleotides in alien nucleic acid were similar in nature to those found on Earth, how many phosphate groups would be found in the piece of DNA?

1. 40
2. 80
3. 120
4. 160
5. 320

With great anticipation, your team conducts tests on the alien nucleic acids.  Media outlets from around the world are interested in reporting the results of your team’s research. The most fundamental questions to be answered include: Is the structure of the material similar to genetic systems of Earth?  What is the mechanism of replication?

Initial analysis finds that the alien genetic material is a double-helix composed of nucleotides, with sugar-phosphates on the outside of the molecule and nitrogenous bases on the interior.  The bases in the nucleotides of the alien genetic material are similar, but not identical, to those found in genetic systems on Earth.  The alien nitrogenous bases are: isoThymine (iT), betaGuanine (bG), Adenine (A), isoAdenine (iA), uracil (U) and isoCytosine (iC).  Studies reveal that the process of replication results in exact copies of the original nucleic acid.  Below are results of tests to elucidate the nature of the alien replication process.

Below is a section of original alien genetic material showing the sequence of nucleotides in its two strands.

5’                                   5’

iT         –           A

U         –          iA

A        –          iT

iA        –           U

iC        –          bG

U        –          iA

bG       –           iC

iT         –           A

iC         –          bG

3’                                  3’

Below are the two molecules that are produced by replication of the alien genetic material above (*Note: ‘Cursive Font’ represents new nucleotides, ‘Print Font’ represents original/parental nucleotides).

5’                                   5’                               5’                                   5’

iT         –           A                                              iT         –           A

U         –         iA                                             U          –          iA

A          –          iT                                              A         –          iT

iA        –          U                                              iA         –           U

iC         –          bG                                            iC        –          bG

U          –          iA                                             U         –          iA

bG       –          iC                                             bG         –          iC

iT         –          A                                             iT         –           A

iC        –          bG                                             iC         –           bG

3’                                   3’                               3’                                   3’

* Note: A / A   = Adenine

U / U   = Uracil

bG / bG  = betaGuanine

iC / iC             = isoCytosine

iT / iT  = isoThymine

iA / iA = isoAdenine

Analyze the alien genetic material and answer the questions regarding its structure and method of replication.

1. Examine the structure of the genetic material of the alien sample. Explain the similarities in the structures of the alien genetic information system with those found in eukaryotic life on Earth.

2. Examine the structure of the genetic material of the alien sample. Explain the differences in the structures of the alien genetic information system with those found in eukaryotic life on Earth.

3. Compare and contrast the process of replication between the alien sample and eukaryotic life found on Earth.

For items 1-4, utilize the following information:

A piece of human double-stranded DNA is isolated.  The piece of isolated DNA contains 410 nucleotides and 88 of those nucleotides contain the base adenine.

1. How many nucleotides in the piece of DNA would contain the base guanine?
2. 88
3. 117
4. 205
5. 410
6. cannot be determined

2. How many phosphate groups would be found in the piece of DNA?
3. 0
4. 88
5. 205
6. 410
7. cannot be determined

3. How many molecules of uracil would be found in the piece of DNA?
4. 0
5. 88
6. 205
7. 410
8. cannot be determined

4. In the DNA molecule, which molecules are not bonded together?
5. adenine and thymine
6. deoxyribose and phosphate groups
7. cytosine and guanine
8. deoxyribose and adenine
9. phosphate groups and cytosine

5. Given the original piece of DNA show to the right, which of the sets of DNA below reflects the process of replication as it occurs in eukaryotes on Earth?

Note: ATGC = parental DNA

ATGC = new DNA

5′          3′

T – A

G – C

C – G

A – T

G – C

A – T

3′            5′

1. 5′ 3′                    5′          3′

T – A                           T –  A

G – C                          G – C

C – G                          C – G

A – T                           A – T

G – C                          G – C

A – T                           A – T

3′           5′                     3′         5′

1. 5′           3′                     5′          3′

T – A                           T –  A

G – C                           G –  C

C – G                           C – G

A – T                           A – T

G – C                           G – C

A – T                           A – T

3′          5′                     3′         5′

1. 5′           5′                     3′          3′

T – A                           T  –  A

G – C                           G  –  C

C – G                           C  –  G

A – T                           A  –  T

G – C                           G  –  C

A – T                           A –  T

3′                      3′                    5′          5′

1. 5′           3′                     5′          3′

T – A                           T  –  A

G – C                           G  –  C

C – G                           C  –  G

A – T                           A  –  T

G – C                           G  –  C

A – T                           A –  T

3′                      5′                    3′          5′

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