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Introduction to Genetic Analysis (12th Edition, 2021) – Solutions Manual – by Griffiths

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INSTANT PDF DOWNLOAD – Complete Solutions Manual for Introduction to Genetic Analysis (12th Edition, 2021) by Anthony J.F. Griffiths, John Doebley, Catherine Peichel, and David Wassarman. Covers all 20 chapters with clear, step-by-step solutions to problems on Mendelian inheritance, gene mapping, molecular genetics, genomics, and population genetics. Ideal for biology and genetics students seeking verified answers, detailed explanations, and problem-solving practice for exams and coursework. genetic analysis solutions, griffiths solutions manual, genetics textbook pdf, mendelian inheritance problems, gene mapping exercises, molecular genetics examples, population genetics workbook, genomics practice solutions, dna and rna problems, biology genetics manual, university genetics course, griffiths 12th edition pdf, molecular biology textbook, inheritance and variation, genetic analysis pdf, solved genetics questions, advanced biology study, genetics problem solving, molecular genetics answers, student solutions pdf, biology workbook download, genetics chapter solutions

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Introduction To Genetic Analysis, Griffiths, 12e
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Introduction to Genetic Analysis, Griffiths, 12e
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Introduction to Genetic Analysis, Griffiths, 12e

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October 14, 2025
Number of pages
445
Written in
2025/2026
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ALL 20 CHAPTERS COVERED




SOLUTIONS MANUAL

,1
The GeneTics RevoluTion
PROBLEMS

In each chapter, a set of problems tests the reader’s comprehension of the concepts in the chapte
their relation to concepts in previous chapters. Each problem set begins with some problems base
the figures in the chapter, which embody important concepts. These are followed by problems of a
general nature.


WORKING WITH THE FIGURES

1. If the white-flowered parental variety in Figure 1-3 were crossed to the first-generation hybrid
in that figure, what types of progeny would you expect to see and in what proportions?

Answer: You would get a 1:1 ratio of purple to white. This is because the first-generation h
plant has one copy of the purple allele and one copy of the white allele, and as a result, 50 perce
the gametes would carry the purple allele and 50 percent of the gametes would carry the white a
The white-flowered parental variety has two copies of the white allele, and all the gametes prod
from the white plant will carry the white allele. Hence, a cross between the two would produce
ratio of purple to white.

Hybrid plant P/p ¥ white plant p/p

Gametes 50% P 50% p ¥ 100% p

50% P/p : 50% p/p
Purple : white

2. In Mendel’s 1866 publication as shown in Figure 1-4, he reports 705 purple (violet) flow
offspring and 224 white-flowered offspring. The ratio he obtained is 3.15:1 for purple:white.
do you think he explained the fact that the ratio is not exactly 3:1?

Answer: This depends on the sample size. When the sample size was large, the proportions
close to 3:1 (e.g., for round and wrinkled seeds the ratio was 2.95:1 and the total population

,2 CHAPTER 1 The Genetics Revolution


was 7324), whereas for a small sample size such as the purple and white petal flowered plants (929
plants), the ratio was not as close to 3:1.


3. In Figure 1-6, the students have 1 of 15 different heights plus there are two height classes (4 ft 11 in
and 5 ft 0 in) for which there are no observed students. That is a total of 17 height classes. If a single
Mendelian gene can only account for two classes of a trait (such as purple or white flowers), how
many Mendelian genes would be minimally required to explain the observation of 17 height classes?

Answer: If a single gene can only account for two classes of a trait, minimum of 9 genes are required
to explain the 17 height classes.


4. Figure 1-7 shows a simplified pathway for arginine synthesis in Neurospora. Suppose you have a
special strain of Neurospora that makes citrulline but not arginine. Which gene(s) are likely mutant or
missing in your special strain? You have a second strain of Neurospora that makes neither citrulline
nor arginine but does make ornithine. Which gene(s) are mutant or missing in this strain?

Answer: If the mutant strain makes citrulline, that means genes A and B must be functional. Therefore,
the only gene that is missing or mutant in the first Neurospora strain must be gene C.

In the second strain, gene A must be functional since it is able to make ornithine. Gene B must be
missing or mutant since it is unable to make citrulline. However, gene C may or may not be missing/
mutant. Enzyme C converts citrulline into arginine (they are in the same sequential pathway), and
enzyme C is dependent on the availability of citrulline for its function.


5. Consider Figure 1-8a.

a. What do the small blue spheres represent?
b. What do the brown slabs represent?
c. Do you agree with the analogy that DNA is structured like a ladder?

Answer:
a. The blue ribbon represents sugar phosphate backbone (deoxyribose and a phosphate group), while
the blue spheres signify atoms.
b. Brown slabs show complementary bases (A, T, G, and C).
c. Yes, it is a helical structure.


6. In Figure 1-8b, can you tell if the number of hydrogen bonds between adenine and thymine is the
same as that between cytosine and guanine? Do you think that a DNA molecule with a high content
of A + T would be more stable than one with a high content of G + C?

Answer: There are two hydrogen bonds between adenine and thymine; three between guanine and
cytosine. No, the molecule with a high content of G-C would be more stable.

, CHAPTER 1 The Genetics Revolution


7. Which of three major groups (domains) of life in Figure 1-11 is not represented by a model organ

Answer: Archaea


8. Figure 1-13b shows the human chromosomes in a single cell. The green dots show the location
gene called BAPX1. Is the cell in this figure a sex cell (gamete)? Explain your answer.

Answer: It is not a sex cell. Cloned BAPXI gene has hybridized to two chromosomes in the
indicating there are two copies of the BAPXI gene. If it were a gamete, it would have only one
of each chromosome and of the BAPXI gene.


9. Figure 1-15 shows the family tree or pedigree for Louise Benge (Individual VI-1) who suffers
the disease ACDC because she has two mutant copies of the CD73 gene. She has four siblings
2, VI-3, VI-4, and VI-5) who have this disease for the same reason. Do all the 10 children of L
and her siblings have the same number of mutant copies of the CD73 gene, or might this numb
different for some of the 10 children?

Answer: All 10 children have one mutant copy of the CD73 gene. Children get one CD73 copy
their mom and one from their dad. Since Louise and her four siblings each carry two defective g
all their children will get one mutant CD73 copy.



BASIC QUESTIONS

10. Below is the sequence of a single strand of a short DNA molecule. On a piece of paper, rewrite
sequence and then write the sequence of the complementary strand below it.

GTTCGCGGCCGCGAAC

Comparing the top and bottom strands, what do you notice about the relationship between them

Answer:
GTTCGCGGCCGCGAAC
CAAGCGCCGGCGCTTG

They are complementary to each other and run in the opposite direction (antiparallel). The seque
are also palindromic; they read the same in either direction.


11. Mendel studied a tall variety of pea plants with stems that are 20 cm long and a dwarf variety
stems that are only 12 cm long.

a. Under blending theory, how long would you expect the stems of first and second hybrids to b
b. Under Mendelian rules, what would you expect to observe in the second-generation hybrids
the first-generation hybrids were tall?

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