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Chapter 4: Modification of Mendelian Ratios Honors Genetics 2012-2013.

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Presentation on theme: "Chapter 4: Modification of Mendelian Ratios Honors Genetics 2012-2013."— Presentation transcript:

1 Chapter 4: Modification of Mendelian Ratios Honors Genetics 2012-2013

2 Chapter Focus While alleles are transmitted from parent to offspring according to Mendelian principles, they sometimes fail to display the clear- cut dominant-recessive relationship observed by Mendel. In many cases, contrast to Mendelian genetics, two or more genes are known to influence the phenotype of a single characteristic. Another exception to Mendelian inheritance is the presence of genes on sex chromosomes, where males only receive a single copy of genes on that chromosome. 1 2 3

3 Chapter Focus Phenotypes are the combined result of both genetics and the environment within which genes are expressed. The result of the various exceptions to Mendelian principles is the occurrence of phenotypes that differ from those resulting from mono-, di-, and tri-hybrid crosses. 5 4 6 Extranuclear inheritance, resulting from the expression of DNA found in mitochondria (and chloroplasts) can modify Mendelian inheritance patterns. These genes are transmitted through the female gamete.

4 MENDEL’S POSTULATES #1: Unit factors come in pairs. #2: Unit factors have either a dominant or recessive form. #3: Unit factors segregate/ separate during gamete formation. #4: Unit factors assort independently from one another. Chapter 3 Lessons #1: Chromosomes come in pairs. #2: GENES have either a dominant or recessive form. #3: Chromosomes segregate/ separate during gamete formation. #4: Chromosomes assort independently from one another.

5 Mendel’s postulates for OTHER INHERITANCE PATTERNS do NOT hold true in all respects These both hold TRUE for other types of inheritance. #3: Unit factors segregate/ separate during gamete formation. #4: Multiple unit factors assort independently from one another. These postulates DO NOT. #1: Unit factors come in pairs. #2: Unit factors have either a dominant or recessive form. Chapter 3 Lessons

6 Alleles are alternative forms of the same gene. Wild-Type Allele Appears most frequently in a population Arbitrary designation of NORMAL Often DOMINANT Used as the standard which all alterations/mutations are compared. Mutant Allele Contains modified genetic information. Specifies an altered gene product. Loss-of-Function Mutation that results in reduced function of a protein Null Allele Mutation that results in COMPLETE loss of function in proteins Gain-of-Function Mutation that results in increased function of a protein 4.1: Alleles Alter Phenotypes in Different Ways

7 4.2: Geneticists Use a Variety of Symbols for Alleles Mendel Abbreviations Dominant allele = capital letter of trait of interest Recessive allele = lowercase letter of trait of interest Work with Drosophila melanogaster (fruit fly) Mutant allele = lowercase letter if recessive; capital letter if dominant. Wild type allele = uses same letter designation with superscript + A slash (/) between the letters designates the location of the allele on homologous chromosomes.

8 Example Recessive body color Ebony = e Normal Wild Type body color Gray = e + Possible genotypes for diploid fly: e + /e + = gray homozygous (wild type) e + /e = gray heterozygous (wild type) e/e = ebony homozygous (mutant)

9 Cross between parents with contrasting traits may produce offspring with intermediate phenotypes. Occurs when the phenotype is controlled by a single gene with two alleles, neither of which is dominant. Because there is no dominant trait, abbreviations can vary: Red = R 1 / White = R 2 R = Red White = W 1 / Red = W 2 W = White Red = C R / White = C W C = Color 4.3: Neither Allele is Dominant in Incomplete (Partial) Dominance

10 Snapdragons: Red + White = Pink Red = C R / White = C W Incomplete/Partial Dominance

11 Humans: Carriers have 50% activity of affected enzyme.

12 QUESTION #1, PAGE 87 IN SHORTHORN CATTLE, COAT COLOR MAY BE RED, WHITE, OR ROAN. ROAN IS AN INTERMEDIATE PHENOTYPE EXPRESSED AS A MIXTURE OF RED AND WHITE HAIRS. THE FOLLOWING DATA ARE OBTAINED FROM VARIOUS CROSSES: RED X RED = ALL RED WHITE X WHITE = ALL WHITE RED X WHITE = ALL ROAN ROAN X ROAN = ¼ RED; ½ ROAN, ¼ WHITE HOW IS COAT COLOR INHERITED? WHAT ARE THE GENOTYPES OF PARENTS AND OFFSPRING FOR EACH CROSS?

13 RED X RED = ALL RED C R /C R X C R /C R = C R /C R WHITE X WHITE = ALL WHITE C W /C W X C W /C W = C W /C W RED X WHITE = ALL ROAN C R /C R X C W /C W = C R /C W ROAN X ROAN = ¼ RED; ½ ROAN, ¼ WHITE C R /C W X C R /C W = C R /C R, C R /C W, C W /C W CRCR CRCR CWCW CWCRCWCR CWCRCWCR CWCW CWCRCWCR CWCRCWCR CRCR CWCW CRCR CRCRCRCR CWCRCWCR CWCW CRCWCRCW CWCWCWCW

14 Incomplete: Phenotype expression different than either parent. Mixture Codominance: Phenotype expression that is equal to BOTH parent’s phenotypes. Incomplete/Partial vs. Codominance

15 4.4: Codominance and MN Blood Groups Joint expression of BOTH alleles. In humans, 2 forms/alleles for the glycoprotein are present on the red blood cell surface, M and N The gene for the glycoprotein is located on chromosome #4. The 2 alleles are designated L M and L N GenotypePhenotype LMLMLMLM M LMLNLMLN MN LNLNLNLN N

16 4.5: ABO Blood Groups Identified by Landsteiner in 1901. A and B antigens are located on chromosome 9.

17 4.5: ABO Blood Groups 3 alleles I = isoagglutinogen; agglutination means to clump. I A = A antigens; B antibodies I B = B antigens; A antibodies I O = NO antigens; A and B antibodies GenotypeAntigenPhenotype IAIAIAIA A A IAIOIAIO A IBIBIBIB B B IBIOIBIO B IAIBIAIB A and BAB IOIOIOIO NO antigensO

18 PAGE 87 QUESTIONS #3 AND #11

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21 A man is suing his wife for divorce on the grounds of infidelity. Their first child and second child, whom they both claim, are blood groups O and AB, respectively. The third child, whom the man disclaims, is blood type B. (a)Can this information be used to support the man's case? MN/ABO Practice

22 (a)Can this information be used to support the man's case? Child #1: I O I O Child #2: I A I B Child #3: I B I B or I B I O MOTHER’S GENOTYPE: I A I O FATHER’S GENOTYPE: I B I O

23 A man is suing his wife for divorce on the grounds of infidelity. Their first child and second child, whom they both claim, are blood groups O and AB, respectively. The third child, whom the man disclaims, is blood type B. (b) Another test was made using the M-N blood group system. The third child was group M, the man was group N. Can this information be used to support the man's case? MN/ABO Practice

24 (b) Another test was made using the M-N blood group system. The third child was group M, the man was group N. Can this information be used to support the man's case? Child 3: L M L M Father: L N L N Impossible for the man to be the child’s father due to the difference in M and N antigen on the red cell surface.

25 4.5: ABO/Bombay Phenotype Bombay phenotype results from the incomplete production of an enzyme needed to express the A or B antigen. Although they may genetically be I A or I B, the antigen is not expressed and the phenotype is type O blood. See pedigree, page 64.


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