Exam 1 Study Guide

What is the diagnostic phenotypic ratios?

I. Chapter 2 — Single-Gene Inheritance

A. diagnostic phenotypic ratios

1. monohybrid cross:

a) 3:1 with 2 heterozygotes (i.e. Aa x Aa)

b) 1:1 with test cross with a heterozygote (i.e. Aa x aa)

2. dihybrid cross:

a) 9:3:3:1 with 2 heterozygotes (i.e. AaBb x AaBb)

b) 1:1:1:1 with test cross with a heterozygote (i.e. AaBb x aabb)

B. diagnostic genotypic ratios

1. monohybrid cross:

a) 1:2:1 with 2 heterozygotes (i.e. Aa x Aa)

b) 1:1 with test cross with a heterozygote (i.e. Aa x aa)

c) 1:2:1:2:4:2:1:2:1 with 2 heterozygotes (i.e. AaBb x AaBb)

d) 1:1:1:1 with test cross with a heterozygotes (i.e. AaBb x aabb)

One of the different forms of a gene that can exist on at a single locus, is what?

C. pedigree analysis

1. autosomal recessive (albinism, PKU)

a) inherited as a recessive allele

b) must be 2 recessive alleles (i.e. aa)

c) patterns:

(1) appears in progeny of unaffected parents

(2) progeny includes both males and females

(3) horizontal with several generations of unaffected and then several siblings in  one generation affected

What is the eukaryotic chromosome mapping?

2. autosomal dominant (dwarfism, Huntington’s disease) Don't forget about the age old question of Language is received or perceived also through what?

a) can be homozygous dominant or heterozygous (i.e. AA or Aa)

b) patterns:

(1) appears in every generation (every affected has 1 affected parent)

(2) both males and females are affected

(3) 2 affected parents can produce unaffected child

3. X-linked recessive (color blindness)

a) must have the X carry recessive trait for males, and two recessive alleles on X  for females (i.e. XaXa or XaY)

b) patterns: Don't forget about the age old question of What is macroscopic anatomy?

(1) skips generation

(2) more males carry than females with all daughters of carrier mothers being  normal, but 1/2 will be carriers and 1/2 of sons will be affected If you want to learn more check out Why is it important to use theory in the practice of health education/promotion?

(3) none of offspring of affected male show phenotype, but daughters are  carriers (heterozygotes), affected males do not transmit to sons If you want to learn more check out What is incrementalism theory?

4. X-linked dominant (excess body hair)

a) not common If you want to learn more check out What are the steps in the perceptual process?
If you want to learn more check out What do you call the assumption that others are paying attention?

b) patterns:

(1) affected males pass condition to all daughters, but none to their sons (2) affected heterozygous females married to unaffected males pass to 1/2 sons  and daughters

(3) does not skip generations and present in males and females

(4) can be either XAXa or XAXA or XAY

5. Y-linked inheritance (hairy ear rims)

a) only males inherit

b) SRY gene connected to maleness

1

Exam 1 Study Guide

D. meiosis stages and chromosomes numbers

1. number of DNA molecules present during:

a) prophase I = 92 chromatids (46 chromosomes with 2 chromatids)

b) metaphase I = 92 chromatids (46 chromosomes with 2 chromatids)

c) anaphase I = 92 chromatids (46 chromosomes with 2 chromatids)

d) telophase I = 46 chromatids (23 chromosomes with 2 chromatids)

e) prophase II = 46 chromatids (23 chromosomes with 2 chromatids)

f) metaphase II = 46 chromatids (23 chromosomes with 2 chromatids)

g) anaphase II = 46 chromatids with 46 chromosomes

h) telophase II = 23 chromosomes in each almost split cell

E. test cross — genetic cross between a homozygous recessive individual and a  

corresponding suspected heterozygote to determine genotype of unknown and potential  heterozygote

F. genetic terms 

1. gene — fundamental physical and functional unit of heredity, which carries  

information from one generation to the next; segment of DNA composed of a  

transcribed region and a regulatory sequence the makes transcription possible

2. allele — one of the different forms of a gene that can exist on at a single locus

3. dominance — gene that is easily shown and expressed

4. recessiveness — gene that is not easily shown and expressed

5. homozygous — individual with 2 of the same alleles

6. heterozygous — individuals with 2 different alleles

7. chromatin — material in which chromosomes or organisms are composed of

8. mendelian first law — law of equal segregation; two members of a gene pair  

segregate from each other in meiosis; each gamete has an equal probability of  

obtaining either member of the gene pair

9. meiosis — results in 4 daughter cells each with half the number of chromosomes of  the parent cell

G. cytoplasmic segregation and maternal inheritance

1. inheritance occurs outside of the nucleus in organelles such as mitochondria an  

chloroplasts

2. usually independent of mendelian patterns shown by nuclear genes

2

Exam 1 Study Guide

II. Chapter 3 — Independent Assortment

A. dihybrid crosses and independent assortment

1. independent assortment shows that one allele does not determine the other and are  expressed independent of one another

2. example of dihybrid cross:

B. chi-square test (calculate chi-square value, p-value, drawing conclusions) 1. how to calculate chi-square value:

sum of

X2 =(observed — expected)2 

Σ expected degrees of freedom = # of phenotypic classes — 1

2. p-value — used to determine certainty, probability that a given result from a chi squared test is due to chance

3. when finding expected, you need to take each phenotypic class and multiply by the  fraction that would result in the cross (i.e. if the ratio is 9:3:3:1, you need to multiply  the total by 9/16, 3/16, and 1/16 respectively)

4. if the p-value is greater than the value found through the chi-square test at 0.5 with  the correct degree of freedom, then you can accept the hypothesis

5. if the p-value is less than the value found, you reject the hypothesis

C. genetic terms 

1. mendelian second law — law of independent assortment; unlinked or distantly  linked segregating gene pairs assort independently at meiosis

III. Chapter 4 — Eukaryotic Chromosome Mapping

A. two-point and three-point test crosses and their mapping distance

1. two-point — used to determine the recombinant frequency between 2 linked genes 2. three-point — used to determine the recombinant frequency between 3 linked  genes; heterozygote and recessive tester

3. unlinked genes may be on difference chromosomes, or so far apart on the same  chromosome that they are often separated by recombination

4. longer regions have more crossovers, thus higher recombinant frequencies 5. how to find mapping distance:

a) 1 map unit = recombinant frequency of 1%

single crossover + single crossover + (double crossover + double crossover

map unit = x 100total observed

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Exam 1 Study Guide

B. parental vs. recombinant genotypes

1. parental — most common

2. recombinant — most rare (double crossover)

3. single crossovers are the genotypes that are not either parental or recombinant C. linkage analysis

1. linked genes are symbolized by: / (i.e. AB/ab)

2. genes on different chromosomes are symbolized by: ; (i.e. A/a; B/b)

3. test for linkage is a dihybrid cross (AaBb x aabb), if no linkage (independent) ratio  would be 1:1:1:1

D. genetic terms 

1. chiasmata — at meiosis when duplicated homologous chromosomes pair with each  other, these cross-shaped structures form between two non sister chromatids 2. linkage — tendency of DNA sequences that are close together on a chromosome to  be inherited together during meiosis

3. centimorgan — map unit, unit for measuring genetic linkage, distance between  chromosome positions

4. SNPs — single nucleotide polymorphisms; nucleotide-pair difference at a given  location in the genomes of two or more naturally occurring individuals

5. molecular markers — DNA sequence variant that can be used to map an  interesting phenotype to a specific region of DNA (examples: SNP, RFLPs, SSLPs,  minisatellites, microsatellites, VNTRs)

example of how to calculate recombinant frequency:

AaBb = 442 = pr+ / pr x vg+ / vg aabb = 458 = pr / pr x vg / vg Aabb = 46 = pr+ / pr x vg / vg aaBb = 54 = pr / pr x vg+ / vg

**Aabb and aaBb are the recombinants

so,

46 + 54

442 + 458 + 46 + 54

100

= x 100 = 10% 1000

example of how to find the map units between genes:

recombinants (double crossover) single crossover

parental

5 b+wx+cn+ 6 bwxcn

69 b+wxcn 67 bwx+cn+ 48 b+wxcn+ 44 bwx+cn 382 b+wx+cn 379 bwxcn+

so with the total being 1000,

wx/cn =

69 + 67 + 5 + 6 1000

69 + 67 + 48 + 44

x 100 = 14.7

b cn wx10.3 14.7

b/cn = b/wx =

1000x 100 = 22.8

48 + 44 + 5 + 6x 100 = 10.3 1000

4

Exam 1 Study Guide

I. Chapter 2 — Single-Gene Inheritance

A. diagnostic phenotypic ratios

1. monohybrid cross:

a) 3:1 with 2 heterozygotes (i.e. Aa x Aa)

b) 1:1 with test cross with a heterozygote (i.e. Aa x aa)

2. dihybrid cross:

a) 9:3:3:1 with 2 heterozygotes (i.e. AaBb x AaBb)

b) 1:1:1:1 with test cross with a heterozygote (i.e. AaBb x aabb)

B. diagnostic genotypic ratios

1. monohybrid cross:

a) 1:2:1 with 2 heterozygotes (i.e. Aa x Aa)

b) 1:1 with test cross with a heterozygote (i.e. Aa x aa)

c) 1:2:1:2:4:2:1:2:1 with 2 heterozygotes (i.e. AaBb x AaBb)

d) 1:1:1:1 with test cross with a heterozygotes (i.e. AaBb x aabb)

C. pedigree analysis

1. autosomal recessive (albinism, PKU)

a) inherited as a recessive allele

b) must be 2 recessive alleles (i.e. aa)

c) patterns:

(1) appears in progeny of unaffected parents

(2) progeny includes both males and females

(3) horizontal with several generations of unaffected and then several siblings in  one generation affected

2. autosomal dominant (dwarfism, Huntington’s disease)

a) can be homozygous dominant or heterozygous (i.e. AA or Aa)

b) patterns:

(1) appears in every generation (every affected has 1 affected parent)

(2) both males and females are affected

(3) 2 affected parents can produce unaffected child

3. X-linked recessive (color blindness)

a) must have the X carry recessive trait for males, and two recessive alleles on X  for females (i.e. XaXa or XaY)

b) patterns:

(1) skips generation

(2) more males carry than females with all daughters of carrier mothers being  normal, but 1/2 will be carriers and 1/2 of sons will be affected

(3) none of offspring of affected male show phenotype, but daughters are  carriers (heterozygotes), affected males do not transmit to sons

4. X-linked dominant (excess body hair)

a) not common

b) patterns:

(1) affected males pass condition to all daughters, but none to their sons (2) affected heterozygous females married to unaffected males pass to 1/2 sons  and daughters

(3) does not skip generations and present in males and females

(4) can be either XAXa or XAXA or XAY

5. Y-linked inheritance (hairy ear rims)

a) only males inherit

b) SRY gene connected to maleness

1

Exam 1 Study Guide

D. meiosis stages and chromosomes numbers

1. number of DNA molecules present during:

a) prophase I = 92 chromatids (46 chromosomes with 2 chromatids)

b) metaphase I = 92 chromatids (46 chromosomes with 2 chromatids)

c) anaphase I = 92 chromatids (46 chromosomes with 2 chromatids)

d) telophase I = 46 chromatids (23 chromosomes with 2 chromatids)

e) prophase II = 46 chromatids (23 chromosomes with 2 chromatids)

f) metaphase II = 46 chromatids (23 chromosomes with 2 chromatids)

g) anaphase II = 46 chromatids with 46 chromosomes

h) telophase II = 23 chromosomes in each almost split cell

E. test cross — genetic cross between a homozygous recessive individual and a  

corresponding suspected heterozygote to determine genotype of unknown and potential  heterozygote

F. genetic terms 

1. gene — fundamental physical and functional unit of heredity, which carries  

information from one generation to the next; segment of DNA composed of a  

transcribed region and a regulatory sequence the makes transcription possible

2. allele — one of the different forms of a gene that can exist on at a single locus

3. dominance — gene that is easily shown and expressed

4. recessiveness — gene that is not easily shown and expressed

5. homozygous — individual with 2 of the same alleles

6. heterozygous — individuals with 2 different alleles

7. chromatin — material in which chromosomes or organisms are composed of

8. mendelian first law — law of equal segregation; two members of a gene pair  

segregate from each other in meiosis; each gamete has an equal probability of  

obtaining either member of the gene pair

9. meiosis — results in 4 daughter cells each with half the number of chromosomes of  the parent cell

G. cytoplasmic segregation and maternal inheritance

1. inheritance occurs outside of the nucleus in organelles such as mitochondria an  

chloroplasts

2. usually independent of mendelian patterns shown by nuclear genes

2

Exam 1 Study Guide

II. Chapter 3 — Independent Assortment

A. dihybrid crosses and independent assortment

1. independent assortment shows that one allele does not determine the other and are  expressed independent of one another

2. example of dihybrid cross:

B. chi-square test (calculate chi-square value, p-value, drawing conclusions) 1. how to calculate chi-square value:

sum of

X2 =(observed — expected)2 

Σ expected degrees of freedom = # of phenotypic classes — 1

2. p-value — used to determine certainty, probability that a given result from a chi squared test is due to chance

3. when finding expected, you need to take each phenotypic class and multiply by the  fraction that would result in the cross (i.e. if the ratio is 9:3:3:1, you need to multiply  the total by 9/16, 3/16, and 1/16 respectively)

4. if the p-value is greater than the value found through the chi-square test at 0.5 with  the correct degree of freedom, then you can accept the hypothesis

5. if the p-value is less than the value found, you reject the hypothesis

C. genetic terms 

1. mendelian second law — law of independent assortment; unlinked or distantly  linked segregating gene pairs assort independently at meiosis

III. Chapter 4 — Eukaryotic Chromosome Mapping

A. two-point and three-point test crosses and their mapping distance

1. two-point — used to determine the recombinant frequency between 2 linked genes 2. three-point — used to determine the recombinant frequency between 3 linked  genes; heterozygote and recessive tester

3. unlinked genes may be on difference chromosomes, or so far apart on the same  chromosome that they are often separated by recombination

4. longer regions have more crossovers, thus higher recombinant frequencies 5. how to find mapping distance:

a) 1 map unit = recombinant frequency of 1%

single crossover + single crossover + (double crossover + double crossover

map unit = x 100total observed

3

Exam 1 Study Guide

B. parental vs. recombinant genotypes

1. parental — most common

2. recombinant — most rare (double crossover)

3. single crossovers are the genotypes that are not either parental or recombinant C. linkage analysis

1. linked genes are symbolized by: / (i.e. AB/ab)

2. genes on different chromosomes are symbolized by: ; (i.e. A/a; B/b)

3. test for linkage is a dihybrid cross (AaBb x aabb), if no linkage (independent) ratio  would be 1:1:1:1

D. genetic terms 

1. chiasmata — at meiosis when duplicated homologous chromosomes pair with each  other, these cross-shaped structures form between two non sister chromatids 2. linkage — tendency of DNA sequences that are close together on a chromosome to  be inherited together during meiosis

3. centimorgan — map unit, unit for measuring genetic linkage, distance between  chromosome positions

4. SNPs — single nucleotide polymorphisms; nucleotide-pair difference at a given  location in the genomes of two or more naturally occurring individuals

5. molecular markers — DNA sequence variant that can be used to map an  interesting phenotype to a specific region of DNA (examples: SNP, RFLPs, SSLPs,  minisatellites, microsatellites, VNTRs)

example of how to calculate recombinant frequency:

AaBb = 442 = pr+ / pr x vg+ / vg aabb = 458 = pr / pr x vg / vg Aabb = 46 = pr+ / pr x vg / vg aaBb = 54 = pr / pr x vg+ / vg

**Aabb and aaBb are the recombinants

so,

46 + 54

442 + 458 + 46 + 54

100

= x 100 = 10% 1000

example of how to find the map units between genes:

recombinants (double crossover) single crossover

parental

5 b+wx+cn+ 6 bwxcn

69 b+wxcn 67 bwx+cn+ 48 b+wxcn+ 44 bwx+cn 382 b+wx+cn 379 bwxcn+

so with the total being 1000,

wx/cn =

69 + 67 + 5 + 6 1000

69 + 67 + 48 + 44

x 100 = 14.7

b cn wx10.3 14.7

b/cn = b/wx =

1000x 100 = 22.8

48 + 44 + 5 + 6x 100 = 10.3 1000

4