Heredity
Heredity, Meiosis & Independent Assortment
Classwork
1. Describe what is meant by the terms: dominant and recessive traits.
2. In genetics we often use the terms homozygous and heterozygous. Describe the
meaning of the terms and provide an example of each.
3. Describe the term alleles in terms of the sequence of DNA.
4. Geneticists often use the terms genotype and phenotype in the description of an
organism’s characteristics. Describe the meaning of each and provide an example.
5. Mendel used a testcross to determine the genotype of an unknown plant. Describe the
process of performing a testcross. Describe the possible results.
6. In pea plants the allele T produces tall plants and allele t produces dwarf. A tall plant is
crossed with a dwarf plant, producing offspring with equal numbers of both tall and dwarf
plants. What are the genotypes of the parents?
7. Brown eyes are dominant to blue eyes. A brown eyed couple, both of whom had one
blue eyed parent and one brown eyed parent are expecting their first child. What is the
probability that the child will have blue eyes?
8. Describe Mendel’s law of independent assortment.
9. One way of predicting the outcome of a cross is to use a Punnett square. Describe what
is meant by a monohybrid cross and a dihybrid cross. Show examples of each.
10. People who are able to taste PTC paper have a dominant allele. A homozygous
recessive person cannot taste the PTC paper. Brown eyes are dominant to green eyes.
Neither trait is sex-linked. A brown-eyed male taster marries a female green-eyed nontaster. Describe the kinds of children and proportions they could expect if one of the
father’s parents was a green eyed non-taster.
11. Describe the disease known as Huntington disease and explain how it is inherited.
Homework
12. Brown eyes are dominant to blue eyes. Two brown-eyed parents have four blue-eyed
children. What is the chance that their next child will have brown eyes? What is the
chance it would have blue eyes?
13. In jack rabbits the allele for smooth fur (S) is dominant to the allele for rough fur (s). A
smooth fur parent mates with a rough fur parent and produce offspring with the
proportion ½ smooth and ½ rough. If two of the smooth offspring are mated what would
be the ratio of their offsprings’ phenotypes?
14. A novice gardener decides to cross lima bean plants that are true breeding for green
pods with another lima bean plant that is true breeding for yellow pods. After collecting
the seeds and planting, she notices all the F1 have yellow seed pods. Which plant is the
dominant phenotype? Show a Punnett square to prove you answer.
15. The ability to roll your tongue in a complete circle is controlled by a dominant gene. Two
parents that can roll their tongue have a child that cannot roll her tongue. Explain why
she cannot. List the genotypes and phenotypes for the parents and the child.
16. Suppose the homozygous recessive genotype is lethal for a particular trait. What ratios
would you expect to be living in the following crosses?
Dd x DD and Dd x Dd
17. A researcher observes that when two blue-flowered plants are mated together they
always produce blue flowers. When two pink-flowered plants are mated together 75 %
are pink and 25% are blue. Which flower color is dominant? What is the genotype of
the pink flowered plants that were mated together?
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18. You have two rose plants the each produce different color petals, one red and one
yellow. You want to cross them for next year’s flower show. How could you determine
which plant has the dominant flower color?
19. A grey rabbit is crossed with a Chinchilla rabbit and produces 3 grey rabbits and 1 white
rabbit. Determine the genotypes of the parents.
20. In chickens, pea combs are produced by a dominant allele, P. The recessive allele, p,
produces a single comb. Feathered legs are produced by a dominant allele, F.
Assuming no gene linkage, a farmer performs the following crosses with four chickens:
#1, 2, 3, and 4. Using the results below, determine the genotypes of chickens #1, 2, 3,
and 4.
Rooster
#1
#1
#1
Hen
#2
#3
#4
Results
All feathered and all pea comb
75% feathered, all pea comb
56.25% feathered, pea comb; 18.75% featherless pea comb;
18.75 feathered, single-comb; 6.25% featherless, single comb
21. In humans, the gene for brown hair, B, is dominant to the gene for blond hair, b. A
brown-haired man, whose mother had blond hair, and father had brown hair marries a
woman having blond hair, both of whose parent were brown-haired. Determine the
genotype of the man. Determine the possible genotypes for the children produced from
this marriage.
22. Sweet corn comes in two varieties, white and yellow kernels. Yellow kernels, Y is
dominant to white kernels, y. What kinds of gametes world be produced by the parent
pants in the following crosses?
A. YY x Yy B. Yy x Yy
C. Yy x yy
What Mendel Didn’t Know
Classwork
23. Describe incomplete dominance and provide an example.
24. Four o’clock flowers show incomplete dominance. Red flowers are homozygous for the
dominant allele R, white flowers are homozygous for the recessive allele, r.
Heterozygotes have pink flowers. Determine the expected offspring from the following
crosses: RR x Rr, RR x rr, and Rr x Rr.
25. Describe the genetic terminology of pleiotropy and polygenic inheritance.
26. Cats produce many different coat colors. Tabby fur color can be produced by AA or Aa,
black is aa. Another gene pair is epistatic to the gene for fur color. When present in its
dominant form (WW or Ww), this gene blocks the formation of fur color resulting in
offspring with white fur (ww offspring have normal fur). Determine the proportions of
offspring from the following cross: AaWw x AaWw. (Assume no gene linkage).
27. Explain why people with type O blood are known as universal donors. Likewise, explain
why people with type AB blood are known was universal recipients.
28. Red-green color blindness is a sex-linked recessive trait. A man with normal color vision
and a color-blind woman have a son. Determine the probability that their son will be
color-blind. If they have a daughter, what is the probability that she will be color-blind?
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29. Use the pedigree below to determine the probability that a son of the 3rd generation
female will be color-blind if she marries a normal male. What if she marries a color-blind
male? Explain your answer.
30. Describe what is meant by linkage, relating to genes.
31. Discuss the relationship between frequency of recombination between genes and their
positions on the chromosomes.
Homework
32. Describe polygenic inheritance and explain how it differs from the inheritance of traits by
a single gene.
33. Describe epistasis and provide an example.
34. A woman with normal color vision whose father was color-blind marries a man with
normal color vision whose father was also color-blind. Determine the proportion of
probable offspring.
35. Suppose you are working on a genetics lab in your college biology class. You cross the
assigned parental generation of Drosophila flies. You notice that the F1 generation has
twice as many females as males. Explain what might be the cause of the unusual ratio
of sexes. Show a genetic cross that supports your explanation.
36. In Drosophila, normal wings are dominant to vestigial wings. Wing version is not sexlinked. However, the gene for eye color is sex linked. Wild type flies have red eyes,
which is dominant to black eyes. A red eyed male with normal wings is mated with a
black-eyed female with vestigial wings. Determine the expected phenotypic ratio in the
F1 generation.
37. Explain what is meant by recombination in terms of chromosomes and chromatids.
38. Place the following genes that are located on the same chromosome in the correct
order. There are three genes: X, Y, and Z. In a genetic recombination experiment, 12%
recombination is found between genes X and Y. Eight percent recombination is found
be X and Z, and 3% recombination between Y and Z.
39. Determine the sequence map of the following data showing crossing-over frequency.
Genes
B and A
C and D
C and A
B and D
C and B
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Frequency
5%
8%
9%
4%
4%
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Heredity
Probability, Statistics & Genetics
Classwork
40. When performing a chi-squared analysis a null hypothesis is necessary. Explain what a
null hypothesis is and why it is necessary.
41. A group of biology students wanted to see if there was a color preference towards the
blue trays in class. They set up 4 separate piles of trays consisting of blue, green,
orange and yellow. Using chi-squared analysis, can they reject the null hypothesis that
these trays are selected at random?
Tray Color
Blue
Green
Orange
Yellow
Number of Trays Chosen
22
17
54
27
42. The laws of probability can simplify many genetic problems. A 6-sided die is tossed 5
times. Determine the probability of rolling a 4, five times in a row.
43. A woman has the following combination of alleles for the genes H, I, J, K, L, M and N:
HhIIJjkkLlMmNn. How many different kinds of gametes can be produced?
Homework
44. A biology student is trying to determine if his genetic corn was derived from a true
dihybrid cross. He obtained the following data regarding two traits. Use the chi-square
test to determine if this is a true dihybrid cross.
Purple and
smooth
Purple and
wrinkled
Yellow and
smooth
Yellow and
wrinkled
Observed
283
Expected
X2
81
79
24
467 total
45. A researcher believes she has invented a new drug to lessen tooth decay. Two groups
totaling 1000 participants were asked to rinse with an unknown solution after brushing
twice a day for 1 year. One group was given a placebo (group A, 500 members) and the
other the drug (group B, 500 members). Both groups were assessed for cavities. Use
the data to complete the chi-squared analysis to determine if there is a significant
difference between the placebo and the new drug.
Group A
Group B
Observed cavities
227
186
X2
46. Suppose we have 4 gene pairs, each of which affect a different characteristic. Tt, Gg,
Hh, and Mm are the alleles for these genes. Determine the probability of obtaining the
genetic combination of TtGGHhmm from the parental cross of TfGgHhMm x
TTGgHhMm
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47. Using the same genes as the previous question, determine the probability of obtaining
the genetic combination of TtGGHhmm from the parental cross of ttGGhhMm x
TTGGHhmm
48. Describe the advantage to an organism reproducing through sexual reproduction.
Free Response
Calculated distances and percent crossovers.
Shaw, K. & Miko, I. (2008) Chromosome theory and the Castle and Morgan debate. Nature Education 1(1):142
1. If traits occur together 50% of the time, then they also occur apart 50% of the time. If traits
occur together more often than they do separately, then they show crossing-over events less
than 50% of the time, and the traits are predicted to be more closely associated on the
chromosome. This crossing-over percentage is therefore a measure of their degree of linkage
on the same chromosome. Specifically, for genes on the same chromosome, the following
statements hold true:
Crossing-over rate < 50% = Linkage (for genes and traits on same chromosome)
Crossing-over rate = 50% = No linkage for genes and traits on separate chromosomes)
A. H. Sturtevant, an undergraduate student working with Morgan, crossed Drosophila and
looked at the linkage of 5 traits (P,R,M,C,O) around a single locus or trait B. The percent
crossovers and calculated distance from B are shown in the above table.
a. Draw a plausible genetic map using the crossover percent data in above table.
b. Describe why these traits would not follow Mendelian phenotypic ratios between
homozygous Parental genotypes or F1 generation crosses.
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Tall
Short
Expected
300
100
Observed
305
95
Lobo, I. (2008) Genetics and statistical analysis. Nature Education 1(1):109
2. When Mendel crossed a pure breed Tall with a pure breed Short pea plant, the progeny was
always Tall, but when he crossed the Tall progeny- he got a 3 Tall:1 Short ratio. For four
hundred plants, the expected and observed numbers for a heterozygous cross is shown in the
above table.
a. Perform a Chi-square test for the above data. The null hypothesis for this test will be that
observed data will differ from the expected 3:1 Mendel ratios by chance at a significance
level of 95%. Does your test confirms or rejects the null hypothesis? Show your
calculations.
b. If you did a test cross with the parent generation and F1 generation, how would
probabilities of the offspring differ?
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http://en.wikipedia.org/wiki/Mitochondrial_DNA
Miko, I. (2008) Non-nuclear genes and their
inheritance. Nature Education 1(1):135
3. Right: In humans, mitochondrial
DNA is the smallest chromosome,
coding for 37 genes and containing
approximately 16,600 base pairs.
Most of the genes are involve in
respiration or protein synthesis. In
most species, including humans,
mtDNA (mitochondrial DNA) is
inherited solely from the mother.
Left: Mitochondria that have wildtype mtDNA are shown in red
(lighter shade); those having
mutant mtDNA are shown in blue
(darker shade).
a. In a heterozygous
population of cells or tissue, how
would the distribution of mutant
alleles present in the mitochondrial
DNA differ from the distribution of mutant alleles present in chromosomal DNA?
b. Would how the distribution of mutations in mtDNA be affected in mitosis versus meiosis?
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4. In 1909, not long after Mendel's
principles of inheritance became well
accepted, Carl Correns noticed some
strange patterns of inheritance in fouro'clock plants, Mirabilis jalapa. He
crossed female plants (seed) of three
different phenotypes with male plants
(pollen) having the same three
phenotypes: white, green, and
variegated stem and leave colors.
Correns’s results are shown in the
table on the left.
Miko, I. (2008) Non-nuclear genes and their
inheritance. Nature Education 1(1):135
a. Identify the dominant
phenotype, and describe whether or
not the sex of the plant matters.
b. How do Correns’s results differ
from the expected Mendelian patterns
of inheritance?
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5. Microsatellites are repeated sequences that usually consist of two, three or
four nucleotides (di-, tri-, and tetranucleotide repeats respectively), and can be repeated 3 to
100 times, with the longer loci generally having more alleles due to the greater potential for
slippage, a type of replication error. Microsatellites, such as CA nucleotide repeats, are very
frequent in human and other genomes and can be present every few thousand base pairs. If
human DNA is cut with one or more enzymes, the fragments will be of varying lengths for
individuals; therefore, theses difference to can be used to identify a specific person with near
perfect accuracy. The diagram above shows an electronic readout for the electrophoresis of
DNA fragments from a potential father, mother, and child. The numbered peaks represent
specific DNA fragments that are specific to the respective individual.
Adams, J. (2008) Paternity testing: blood types and DNA. Nature Education 1(1):146
a. How do you know the child is the son of this father and mother?
b. Using the following chromosome representations or models, show how meiosis would
produce the gametes that through fertilization generated the genetic combination seen in
the DNA electrophoregram.
9.3
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6
7
9
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Heredity-Answer Key
1. Dominant traits are alleles that show up more often compared to the recessive trait. The
dominant allele trait can show up in the homozygous dominant and heterozygous. The
recessive trait will only show up in the homozygous recessive.
2. Heterozygous means two different alleles, such as Rr. Homozygous means two of the
same alleles such as RR or rr.
3. An allele is alternate for of a gene. Alleles can be either dominant or recessive and are
represented as a capital letter or lowercase letter respectively.
4. Genotype is the genetic makeup of an organism. Phenotype is the outward appearance
of the organism.
5. A test cross is performed by mating a plant that is homozygous recessive with an
unknown individual. If the organism is homozygous recessive all the offspring will look
like the known recessive. If the results are half like one parent and half like the other
parent, than the unknown is heterozygous. If the traits exhibited are like the unknown
individual than it is homozygous dominant.
6. Tt x tt
7. Both parents would be heterozygous, therefore the chance would be ¼.
8. The law of independent assortment is when different genes that control different traits
separate independently in the formation of gametes.
9. Monohybrid cross involves genetic crosses that involve a singe trait from both parents.
A dihybrid cross involves two different traits.
10. The children would be expected to be ¼ brown eyed tasters, ¼ brown eyed non-tasters,
¼ green eyed tasters and ¼ green-eyed nontasters.
11. Huntington’s Disease is a degenerative brain disorder, that eventually takes away a
persons ability to walk, think, talk and reason. Huntington’s disease is a result of a
dominant allele.
12. Chance of brown eyes is ¾, blue eyes is ¼.
13. ¼ SS, ½ Ss, ¼ ss therefore the phenotypes would be ¾ smooth and ¼ rough
14. Yellow would be dominant.
15. Both parents are heterozygous and the child received the recessive allele from each
parent. Parent’s genotype Tt, phenotype tongue roller: child genotype tt and phenotype
non-tongue roller.
16. ½ DD and ½ Dd for the first cross. ¼ DD and ½ Dd for the second cross. ¼ will die.
17. Pink is dominant. Parents are both heterozygous for pink.
18. Complete a test cross. Cross two red and two yellow to determine if both are
homozygous. Cross the red with the yellow to create the heterozygous. Cross the one
whose color does not show up with the heterozygous, if the result is half red and half
yellow you know the yellow is recessive and the red is dominant.
19. The grey rabbit is Ggw and the chinchilla is Ggw
20. Chicken #1 – FfPp, Chicken #2, FFPP, Chicken # 3 FfPP, Chicken #4 – FfPp
21. The man would be Bb. Therefore, the possible offspring would be ½ Bb and ½ bb.
22. A. ½ YY, ½ Yy
B. ¼ YY, ½ Yy, ¼ yy
C. ½ Yy, ½ yy
23. The effects of both alleles can be detected in a diploid heterozygous with two different
alleles of a gene. The heterozygous gives a different phenotype than homozygous
dominant or homozygous recessive. One example is Red four o’clock flower being
mated with a white four o’clock gives a pink four o’clock.
24. ½ will have red flowers; RR, ½ will have pink flowers; Rr. Next cross: all will have pink
flowers Rr. Last cross ¼ will have red RR, ½ will have pink Rr, and ¼ will have white rr.
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25. Pleiotropy involves a situation where a single gene influences more than one phenotypic
characteristic. Polygenic inheritance involves a pattern of inheritance in which the
interaction of two or more similar genes determines the phenotype.
26. White, tabby and black in 12:3:1 ratio
27. Type O blood type lack the A and B antigens they can be given to any one. Type AB
blood does no have the neither anti-A nor anti-B antibodies, therefore they can receive
any blood because they don’t have the antibodies to reject them.
28. For the son, 100%. For the daughter none will be colorblind.
29. The chance is ½ for both.
30. Genes that are located on the same chromosome and are near one another tend to be
inherited together.
31. Genes that are closer together on the same chromosome have a lesser chance of
recombination.
32. In polygenic inheritance the same phenotype is controlled by two or more different genes
with various alleles. Inheritance by a single gene is controlled by one set of alleles.
33. Epistasis is where the alleles of one gene overrides the expression of one or more
different genes. In guinea pigs hair color produces black or brown. The expression of
the B – black gene is controlled by a different gene C for color. For any color, black or
brown to be expressed they must have a CC or Cc, otherwise the guinea pig will be
albino.
34. All females will have normal vision, ½ males will be color blind, ½ normal color vision.
35. A recessive allele carried on the X chromosome is lethal in males. XLXl x XLY = ¼ XLX ,
¼ X LXl, XLY, ¼ XlY (lethal)
36. 3/8 red eyes/ normal, 3/8 black eye/ normal, 1/8 red /vestigial, 1/8 black /vestigial
37. Recombination occurs when homologues on each chromatid cross over, break of and
exchange places.
38. X – 8% -- Z -3% -Y
----------12%------39. ADBC
40. A null hypothesis is no difference between what is being tested and what is accepted
and any difference is due to chance.
41.
Observed
22
17
54
27
Total = 120
Blue
Green
Orange
Yellow
Expected
30
30
30
30
Chi = 27.2
Df = 4-1 = 3. At 0.05 we see a p value of 7.82 our value of 27.2 is much greater so we reject
our null hypothesis meaning there is a color preference.
42. 1/6 x 1/6 x 1/6 x 1/6 x 1/6 = 1/7776
43. 32 different combinations.
44.
Purple and smooth
Purple and
wrinkled
Yellow and smooth
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Observed #
Expected #
283
81
262
88
(observedexpected)2/expected
1.68
0.56
79
88
0.92
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Yellow and
wrinkled
24
29
0.86
467 total
Total =4.03
Df = n – 1 or 4 – 1 = 3
Chi-square = 7.82 our value of 4.03 is less indicating that there is a good fit between our
data and the expected values.
45.
Observed with
cavities
227
186
Group A
Group B
Expected with
cavities
250
250
Sum of (observedexpected)2/expected
2.12
16.38
18.50
Df = n-1 probability at 0.05 would be 3.84; since our value is greater we reject the null
hypothesis.
46. ½ x 1/4 x ½ x ¼ = 1/64
47. 1x1x1/2x1/2 = ¼
48. Sexual Reproduction provides more opportunity for genetic variability in the offspring.
There is also a greater chance of producing more fit offspring do to sexual cycles limiting
when a species can produce offspring.
1. Recombination Frequency
a.
b. Since these genes and the traits they contribute to are situated on the same
chromosomes, the traits would not be able to independently assort and
segregate during meiosis; therefore, the number of possible alleic gametes and
combinations between these genes would be diminished and the phenotypic and
genotypic ratios would differ from the ratios expected from Mendelian
inheritance.
Learning Objectives:
LO 3.12 The student is able to construct a representation that connects the process of meiosis
to the passage of traits from parent to offspring. [See SP 1.1, 7.2]
2. Chi Squared
N=1
(305-300)2/300
.083
a.
Since χ2=,333
and is well below the allowed
value for χ2 of 3.841 at a
significance level of 95%, we
Degrees of
Total (χ2)
.333
must confirm the null
freedom=
hypothesis that the
N-1= (2-1)= 1
differences between the
expected and observed results are due to fluctuations in the data that can be
accounted for by chance.
N=2
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(95-100)2/100
.25
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b. A test cross involves the testing of a putative homozygous or heterozygous dominant
individual with a known homozygous recessive. The progeny of a homozygous
recessive individual with a homozygous dominant genotype would give heterozygous
dominant phenotype and genotypes. . The progeny of a homozygous recessive
individual with a heterozygous dominant genotype would give half heterozygous
dominant phenotype and genotypes and half homozygous recessive. The test cross
allows you to discriminate between homozygous and heterozygous genotypes.
Learning Objectives:
LO 3.12 The student is able to construct a representation that connects the process of meiosis
to the passage of traits from parent to offspring. [See SP 1.1, 7.2]
LO 3.14 The student is able to apply mathematical routines to determine Mendelian patterns of
inheritance provided by data sets. [See SP 2.2]
3. mtDNA
a. Since mitochondria do not independently assort and segregate, like
chromosomes, their distribution and thus their allelic distribution will be more
random than alleles in chromosomes. Due to random segregation of
mitochondria during cell division, some tissues may become enriched in cells
with a large number of mutant, possibly defective, alleles that lead to organ
failure or disorders.
b. The distribution of mitochondria would be random in either case and not be
affected by the type of cell division. But in the gametogenesis of some
organisms, like humans, all four cells from meiosis do not mature into oocytes or
egg cells; therefore, if the cell with more mutant mitochondria is selected then the
consequences of those mutations would be more evident in the offspring that is
derived from the oocyte.
Learning Objectives:
LO 3.15 The student is able to explain deviations from Mendel’s model of the inheritance of
traits. [See SP 6.5]
LO 3.16 The student is able to explain how the inheritance patterns of many traits cannot be
accounted for by Mendelian genetics. [See SP 6.3]
LO 3.17 The student is able to describe representations of an appropriate example of
inheritance patterns that cannot be explained by Mendel’s model of the inheritance of traits.
[See SP 1.2]
4. Non Mendelian Inheritance
a. The dominant phenotype depends on the plant that provides the ovule that
becomes the seed after fertilization with pollen. Whatever phenotype the seed
plant is, the progeny show that phenotype, or in the case of variegated- all three
phenotypes.
b. Mendelian inheritance involves the contribution of two alleles to one trait, with
one allele being dominant over another allele. The flower and stem color in
Mirabilis jalapa only matters the contribution of the seed plant without regard to
pollen plant phenotype. At first, the inheritance looks Mendelian until the sex of
the plants is reverse and the dominant allele become recessive and vice versa.
Learning Objectives:
LO 3.15 The student is able to explain deviations from Mendel’s model of the inheritance of
traits. [See SP 6.5]
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LO 3.16 The student is able to explain how the inheritance patterns of many traits cannot be
accounted for by Mendelian genetics. [See SP 6.3]
LO 3.17 The student is able to describe representations of an appropriate example of
inheritance patterns that cannot be explained by Mendel’s model of the inheritance of traits.
[See SP 1.2]
5. Paternity
a. The child has DNA fragments with microsatellites that are derived from both the
mother and the father (fragments 6 from the mother and fragment 7 from the
father). Sexually reproducing organisms contribute equally to the genetic makeup
of the offspring.
b.
6 6
9.3 9.3
Meiosis I
6
6
Meiosis I
9.3
6
9 9
7 7
9.3
Meiosis II
6
9 9
7 7
Meiosis II
9.3
9.3
7
7
9
9
Allelic combination that yielded the observed results
Learning Objectives:
LO 3.12 The student is able to construct a representation that connects the process of meiosis
to the passage of traits from parent to offspring. [See SP 1.1, 7.2]
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