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19
EXTRANUCLEAR
INHERITANCE
CHAPTER SUMMARY QUESTIONS
2. The Mitochondrial Eve Hypothesis proposes that all human mitochondrial genomes
evolved from a “single” original genome approximately 200,000 years ago. The
human mitochondrial genome is maternally inherited. Therefore, the original genome
must have been present in the first Homo sapiens female—hence, the name “Eve.”
4. The mitochondrion is the site of electron transport, a process that produces high
levels of ATP via oxidative phosphorylation. A mitochondrion possesses two main
features that make it ideal for ATP production: (1) It contains membrane-enriched
cristae which provide a large surface area for ATP production. (2) Its inner
membrane is impermeable to ions and small molecules.
The process of energy production begins when sugars are broken down in the
cytoplasm to produce pyruvate and a small amount of ATP. The pyruvate enters the
mitochondria, where it is degraded to CO2 via the Krebs cycle. The Krebs cycle also
produces reduced NADH and FADH2 molecules which are the source of the
electrons that pass through the respiratory electron transport chain, with oxygen as
the terminal electron transport. This electron transfer creates a proton gradient across
the inner mitochondrial membrane that drives ATP biosynthesis.
6. Both mitochondria and chloroplasts contain DNA and ribosomes with prokaryotic
affinities. The two organelles are not generated de novo when a eukaryotic cell
divides. Rather, they arise from pre-existing chloroplasts and mitochondria via
binary fission. Defects in either organelle can produce similar phenotypic effects of
reduced growth.
8. The endosymbiotic theory proposes an explanation for the origin of mitochondria
and chloroplasts. The theory holds that ancestral eukaryotic cells were archaeal cells
that grew anaerobically and lacked mitochondria and chloroplasts. These early cells
engulfed by endocytosis an aerobic -proteobacterium, which evolved into a
mitochondrion. Chloroplasts were formed when this cell (or its descendants)
subsequently endocytosed a photosynthetic cyanobacterium.
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151
Data that supports this theory include: (1) Each organelle is surrounded by two
membranes—the inner membrane is likely derived from the proteobacterium or
cyanobacterium, and the outer membrane from the endocytosing host cell. (2) Like
bacteria, both organelles divide by binary fission. (3) Many mitochondrial genomes
and all chloroplast genomes are circular, just like the vast majority of bacterial
genomes. (4) Mitochondrial gene sequences are similar to related genes from proteobacteria. (5) Many -proteobacteria form symbiotic relationships with
eukaryotic cells. (6) Mitochondrial ribosomes are similar in structure to bacterial
ribosomes (even though the mitochondrial ribosome is constructed of imported
cellular proteins, it is sensitive to antibiotics that inhibit translation from bacterial
ribosomes but not from eukaryotic ribosomes). (7) Like bacterial cells, mitochondria
use tRNAfMet for translation initiation (it is charged with N-formyl methionine
instead of regular methionine).
10. The two terms refer to the presence of multiple mitochondria in a single cell and more
than one genome per organelle. Homoplasmy is the presence of a single common
genotype in the organellar genome, while heteroplasmy is the presence of a mixture
of organellar genomes in a cell.
12. The shrinking of mitochondrial genomes has occurred through two basic mechanisms:
(1) transfer of genes to the nuclear genome, and (2) loss of genes that were no longer
required (degenerative evolution).
EXERCISES AND PROBLEMS
14. By looking at different species, it is clear that few genes for oxidative
phosphorylation are found in all mitochondrial genomes.
16. A mating between an affected male and normal female would produce normal
children. The children inherit the mitochondria of their normal mother and not those
of their affected father. In contrast, the children of a normal father and affected
mother could develop blindness. If the mother is homoplasmic, all children will be
blind. If she is heteroplasmic, the children may or may not develop blindness. Their
exact phenotype would depend on how many defective mitochondria they inherit
from their mother.
18. When a trait is inherited maternally, genetic transmission is only through the
maternal cytoplasm. If we crossed sensitive females with resistant males, we would
expect all of the offspring to be sensitive. In contrast, a cross of resistant females
with sensitive males would produce offspring that are resistant.
20. In this cross, we see that the F1 and F2 phenotypes are the same as the parental
female, which suggests maternal inheritance. If we had looked at the F1 phenotypes
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only, we probably would have concluded that the trait was likely inherited as
autosomal with calm dominant to nervous. This problem illustrates that it is
important to examine multiple generations resulting from a cross before drawing
conclusions.
22. In X-linked inheritance both parents can transmit the trait to their children, while in
cytoplasmic inheritance only one does.
24. Nuclear genes would show a Mendelian segregation pattern, whereas chloroplast
genes would be inherited strictly from the female.
a. CT NTNt—normal leaves.
b. Ct NTNt—normal leaves.
c. The cross is: female CT NTNt male Ct NTNt. The offspring would be
1/4 CT NTNT, 2/4 CT NTNt, 1/4 CT NtNt. All would have normal leaves.
d. The cross is: female Ct NTNt male CT NTNt. The offspring would be
1/4 Ct NTNT (normal leaves), 2/4 Ct NTNt (normal leaves), 1/4 Ct NtNt (twisted
leaves).
26. Y-linked inheritance shows a strict male-to-male transmission, while paternal
inheritance occurs from male to all offspring (male and female).
28. Segregation can occur in one of two ways: (1) wild-type (A) mitochondria to the
same pole and mutant (a) mitochondria to the opposite pole, or (2) a wild-type
(A) and a mutant (a) mitochondrion migrate to each of the two poles. Therefore, the
probabilities are 1/4 AA:2/4 Aa:1/4 aa.
30. In strain 1, both crosses produced offspring that have the same prescottine phenotype
as the mt+ parent. Therefore, the gene for prescottine resistance is located in the
chloroplast genome. On the other hand, the reciprocal crosses involving strain 2
yielded a 1:1 phenotypic ratio in the offspring. Because this is a Mendelian
segregation pattern, eduardomycin resistance must be encoded by a nuclear gene.
Both strain 3 crosses produced offspring with the same brownicillin phenotype as the
mt– parent. Therefore, the gene for brownicillin resistance is found in the
mitochondrial genome.
32. a.
The labeled rbcL probe hybridized to chloroplast but not nuclear DNA. This
indicates that the large subunit of RuBisCO is encoded in the chloroplast genome. The
rbcS probe on the other hand hybridized to nuclear but not chloroplast DNA. Indeed,
two bands appeared in the Southern blot, suggesting the presence of two rbcS genes.
Therefore, the small subunit of RuBisCO is encoded by two nuclear genes.
b. It could be that the large and small subunit polypeptides have to undergo
posttranslational modification before they can form the RuBisCO holoenzyme. E. coli
cells do not have the proper proteins for this modification.
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CHAPTER INTEGRATION PROBLEM
a.
Autosomal dominance is possible, but not an attractive explanation. Let A = disease allele
and a = normal allele. All unaffected individuals are homozygous recessive (aa), while all
affected individuals, with the possible exception of I-1, are heterozygous (Aa). The mating
between individuals II-1 and II-2 is Aa aa and is expected to produce offspring in a
1:1 phenotypic ratio. However, the mating actually produced children in an 8 affected:1
unaffected ratio. Therefore, individual II-1 would have to pass the A allele to eight of her
nine children. Girls III-2, III-7, and III-13 would then go on to transmit the A allele to all of
their children, while boys III-4 and III-10 would transmit the normal a allele to their
children. Therefore, although possible, autosomal dominant inheritance is highly unlikely.
The same goes for autosomal recessive inheritance. Let A = normal allele and a =
disease allele. All affected individuals are homozygous recessive (aa). Individuals I-2, II-2,
III-1, III-6, and III-12 would all have to be carriers. This is not likely since the disease is
rare. Moreover, II-2 (genotype Aa) would have to transmit the a allele to eight of his nine
children. The probability of this happening is (1/2)8 = 1/256. Therefore, this disease is
unlikely to be inherited in an autosomal recessive manner.
Y-linkage is impossible because the disease is found in females.
X-linkage is also impossible. The disease cannot be inherited in a dominant fashion
because affected males III-4 and III-10 should have transmitted the disease to all of their
daughters. Furthermore, the disease cannot be inherited recessively because males I-2 and
III-1, who are normal, have daughters with the disease.
Sex-limited inheritance is impossible because both males and females are affected.
Finally, sex-influenced inheritance can also be excluded. Let A = disease allele
and a = normal allele. If the trait was dominant in males, female I-1 and male I-2
would have to be homozygous dominant (AA) and homozygous recessive (aa),
respectively. Their daughter (II-1) would have to be Aa, which means she should not
have the disease. Therefore, the disease cannot be sex-influenced dominant in males.
On the other hand, if the trait was dominant in females, males III-4 and III-10 would
have to be homozygous dominant (AA), because males would not express the trait in
the heterozygous state. These two fathers would transmit the A allele to all of their
daughters, who will be affected with the disease. The presence of five normal
daughters (IV-5 and IV-8,9,10,11) rules out this mode of inheritance.
b. The findings eliminate the possibility of autosomal recessive inheritance because two
unaffected individuals cannot have affected offspring (in the absence of mutations).
Autosomal dominant inheritance is still possible. The disease could be incompletely
penetrant, which would explain why individual III-14 was unaffected, and it could show
variable expressivity, which would explain the varying levels of disease expression in
different individuals.
c.
Affected female II-1 produced only affected offspring (with the exception of daughter
III-14). In addition, she had three daughters (III-2, III-7, and III-13) who got married and all
of them had affected children. However, her two married sons (III-4 and III-10) had only
normal children—who inherited the normal alleles of their mothers III-3 and III-9,
respectively. So it is clear that the gene responsible for this neurodegenerative disease is
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transmitted only by female parents. Because human mitochondrial DNA is transmitted by a
mother to all her children, mitochondrial inheritance is very likely for this disease.
The variable expressivity of the disease can be explained by the fact that affected
individuals are heteroplasmic, with both mutant and wild-type mitochondria. These
organelles are randomly distributed during cell division, resulting in cells with varying
proportions of mutant and wild-type mitochondria. The severity of the symptoms is
dependent on the proportion of mutant mtDNAs in the zygote and cells derived from it—the
higher the proportion, the more severe the symptoms.
d. The mutant phenotype is expressed only when the proportion of mutant
mitochondrial DNA reaches a threshold level. In individual III-14, this threshold was
clearly not reached. However, because she has two affected daughters, individual
III-14 must have inherited mutant mitochondria from her mother. Therefore, just by
chance female II-1 must have transmitted predominantly normal mitochondria to her
daughter III-14. On the other hand, if random mitochondrial segregation in III-14
was heavily slanted in the “opposite” direction of that of her mother, III-14 would
transmit predominantly mutant mitochondria to both of her children.
e.
Let’s first write the complementary strand to produce the double-stranded DNA region:
AATGATCTGC TGCAGTGCTC TGAGCCCTAG GATTCATCTT TCTTTTCACC GTAGGTGGCC
TTACTAGACG ACGTCACGAG ACTCGGGATC CTAAGTAGAA AGAAAAGTGG CATCCACCGG
The polarity is 5' 3' for the top strand and 3' 5' for the bottom strand.
Type II restriction endonucleases recognize and cut at palindromic sequences.
Palindromes in DNA are sequences that are identical when read in the 5' 3' direction on
each strand. Careful scanning of this double-stranded DNA reveals the following four
restriction sites:
AATGATCTGC TGCAGTGCTC TGAGCCCTAG GATTCATCTT TCTTTTCACC GTAGGTGGCC
TTACTAGACG ACGTCACGAG ACTCGGGATC CTAAGTAGAA AGAAAAGTGG CATCCACCGG
f.
The restriction digest does validate the conclusions of parts (c) and (d). First the restriction
patterns clearly rule out nuclear inheritance because there is no evidence of equal
segregation.
All normal individuals have two fragments of size 14.5 and 2.1 kb, while all affected
individuals possess a 16.6-kb DNA molecule. Individuals IV-1 and IV-2 exhibit the
restriction pattern of their affected mother (and not that of their normal father), while the
restriction pattern of individuals IV-3, IV-4, and IV-5 is identical to that of their normal
mother (and not that of their affected father). Furthermore, individual III-14 appears to have
very little mutant mitochondrial DNA (based on the relative intensity of the bands), which
would explain why she was not affected with the disease. However, her children have a
large proportion of mutant mitochondrial DNA, and hence are affected.
g.
Wild-type human mitochondrial DNA is a double-stranded circle of 16,569 bp. When the
DNA of normal individuals was subjected to the 6-bp cutter used in this experiment, two
fragments of sizes 14.5 and 2.1 kb were produced. This indicates that there are two
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recognition sites for this restriction enzyme. (Remember that a circular molecule that
contains n sites for restriction enzymes will produce n fragments when digested.) The
location of one of these two restriction sites can be obtained from part (e). The 6-bp
palindromic sequence is situated 10 nucleotides from the beginning of the given DNA
region. Therefore, one of the two restriction sites for the 6-bp cutter is at nucleotide 6910 of
the wild-type mitochondrial DNA sequence. Because one of the two digestion fragments is
2.1 kb in size, the second restriction site must be about 2.1 kb away from the first. This can
be visualized by the following map:
First site
2.1 kb
Second site
14.5 kb
The disease appears to correlate with a mutation that eliminated one of the two restriction
sites. The remaining wild-type restriction site was digested by the 6-bp cutter leading to a
single molecule of 16.6 kb. It is not clear from this experiment which of the two sites was
mutated.
Furthermore, we cannot deduce the exact nature of this mutation. The gel is not
sensitive enough to detect the actual size of this mutant 16.6-kb fragment. The intact wildtype mitochondrial DNA is 16,569 bp in length. A base-pair substitution could have
eliminated one of the restriction sites and still produced no change in the overall length of
the mitochondrial genome. Similarly, a small (one or a few base pairs) deletion or even
insertion could eliminate the restriction site without a detectable change in the length of the
mitochondrial genome. However, we can deduce that the region encompassing the mutated
restriction site must be crucial to mitochondrial function.
h. The restriction pattern of individual III-8 shows only the 16.6-kb fragment,
indicating that she is homoplasmic for mutant mitochondria. Therefore, all of her
children would be expected to be affected by the disease.