GENETICS PRACTICE Watch videos BOZEMAN- Beginner’s Guide to Punnett Squares BOZEMAN- Beginner’s Guide to Punnett Squares BOZEMAN- Mendelian genetics BOZEMAN-

GENETICS PRACTICE Watch videos BOZEMAN- Beginner’s Guide to Punnett Squares BOZEMAN- Beginner’s Guide to Punnett Squares BOZEMAN- Mendelian genetics BOZEMAN- Advanced genetics

MENDEL’S PEA EXPERIMENTS http://hus.yksd.com/distanceedcourses/YKSDbiology/lessons/FourthQuarter/Chapter11/11-1/images/MendelExperiment.gif P 1 = Parental F 1 = Filial (pffspring) F 2 = Filial (pffspring)

Refresh your “Bio Brain” about: GENETICS VOCAB DOMINANT: RECESSIVE: HOMOZYGOUS (pure-breeding): HETEROZYGOUS (hybrid): GENOTYPE: PHENOTYPE:

Refresh your “Bio Brain” about: GENETICS VOCAB DOMINANT: RECESSIVE: gene that is hidden by another represented by lower case letter gene that hides another represented by capital letter

Refresh your “Bio Brain” about: GENETICS VOCAB HOMOZYGOUS (pure-breeding): HETEROZYGOUS (hybrid): Organism with two identical alleles for a gene TT OR tt Organism with two different alleles for a gene Tt

Refresh your “Bio Brain” about: GENETICS VOCAB GENOTYPE: PHENOTYPE: Genetic make up of an organism “What genes it has” Appearance of an organism “Way it looks”

It can be written as a: Fraction ____ Percent ____ Ratio ____ Probability is the likelihood that an event will occur 1/4 25% 1:3

COIN FLIP There are 2 possible outcomes: HEADS TAILS http://www.arborsci.com/CoolStuff/CoinFlip.jpg The chance the coin will land on either one is: ____ ____ ____ Alleles segregate randomly just like a coin flip... So can use probability to predict outcomes of genetic crosses. 1/2 50% 1:1 NOT 1:2!

IN PEAS: R = round T = tall Y = yellow peas P = purple flowers r = wrinkled t = short y = green peas p = white flowers

rrrr R GENOTYPE PHENOTYPE ____________ ___________ _______ show dominant phenotype _______ show recessive phenotype Make a cross between a HOMOZYGOUS ROUND pea plant X PURE-BREEDING WRINKLED pea plant Show probabilities for: genotypes and phenotypes of possible offspring Rr ROUND 100% 4/4 0% 0/4

tttt T GENOTYPE PHENOTYPE ____________ ___________ _______ show dominant phenotype _______ show recessive phenotype Make a cross between a HOMOZYGOUS TALL pea plant X HOMOZGYOUS SHORT pea plant Show probabilities for: genotypes and phenotypes of possible offspring Tt TALL 100% 4/4 0% 0/4

WHAT’s THE PATTERN? HOMOZYGOUS DOMINANT pea plant X HOMOZGYOUS RECESSIVE pea plant DO ONE IN YOUR HEAD HOMOZYGOUS PURPLE FLOWER X HOMOZGYOUS WHITE FLOWER pea plants ALL Pp 100% PURPLE FLOWERS 0% WHITE FLOWERS

IN PEAS: R = round T = tall Y = yellow peas P = purple flowers r = wrinkled t = short y = green peas p = white flowers

YyYy Y y GENOTYPE PHENOTYPE ____________ ___________ _______________________ _______________________ _______ show dominant phenotype _______ show recessive phenotype _________ phenotypic ratio Make a cross between a HETEROZYGOUS YELLOW pea plant X HYBRID GREEN pea plant Show probabilities for: genotypes and phenotypes of possible offspring YY Yy Yy yy YY YELLOW 75% 3/4 25% 1/4 Yy yyGREEN YELLOW ____________ genotypic ratio 1:2:1 3:1

TtTt T t GENOTYPE PHENOTYPE ____________ ___________ _______________________ _______________________ _______ show dominant phenotype _______ show recessive phenotype _________ phenotypic ratio 3:1 Make a cross between a HETEROZYGOUS TALL pea plant X HETEROZYGOUS SHORT pea plant Show probabilities for: genotypes and phenotypes of possible offspring TT Tt Tt tt TT TALL 75% 3/4 25% 1/4 Tt ttSHORT TALL ____________ genotypic ratio 1:2:1

WHAT’s THE PATTERN? HOMOZYGOUS DOMINANT pea plant X HOMOZGYOUS RECESSIVE pea plant DO ONE IN YOUR HEAD HOMOZYGOUS PURPLE FLOWER X HOMOZGYOUS WHITE FLOWER pea plants ______ PURPLE? ______ WHITE? _____ PP? _____ Pp? _____ pp? _________ genotypic ratio _______ phenotypic ratio 75% 25% 50% 25% 1:2:13:1

POSSIBLE PARENT GAMETES? INDEPENDENT ASSORTMENT __________________________

POSSIBLE PARENT GAMETES? TRTR TRTR TRTR TRTR INDEPENDENT ASSORTMENT TTRRTTRR ________________________

POSSIBLE PARENT GAMETES? trtr trtr trtr trtr INDEPENDENT ASSORTMENT ttrrttrr ________________________

TR tr GENOTYPE _____________ PHENOTYPE ____________ TtRr TtRr TtRr TtRr TtRr TtRr TtRr TtRr TtRr TtRr TtRr TrRr TrRr TrRr TtRr TtRr TrRr TALL & ROUND

POSSIBLE PARENT GAMETES? TRTR trtr TrTr tRtR INDEPENDENT ASSORTMENT TtRrTtRr ________________________

POSSIBLE PARENT GAMETES? RYRY ryry RyRy rYrY INDEPENDENT ASSORTMENT RrYyRrYy ________________________

RYRyrYry RY Ry rY ry ____ Round & Yellow ____ Round & green ____ Wrinkled & yellow ____ wrinkled & green

RYRyrYry RY Ry rY ry ____ Round & Yellow ____ Round & green ____ Wrinkled & yellow ____ wrinkled & green RRYY RRYy RrYYRrYy RRYy RRyy RrYy Rryy RrYY RrYy rrYY rrYy RrYy Rryy rrYy rryy

RYRyrYry RY Ry rY ry Sign of HETEROZYGOUS DIHYBRID cross What’s the pattern? ____ Round & Yellow ____ Round & green ____ Wrinkled & yellow ____ wrinkled & green 9 3 3 1 RRYY RRYy RrYYRrYy RRYy RRyy RrYy Rryy RrYY RrYy rrYY rrYy RrYy Rryy rrYy rryy

__________ratio is a clue that it’s a ____________________________cross ____ ____________ TRAIT 1 ; ____________ TRAIT 2 ____ ____________ TRAIT 1; _____________ TRAIT 2 9 3 3 1 9:3:3:1 HETEROZYGOUS TWO gene dominant dominant recessive recessive dominant recessive

NON-MENDELIAN INHERITANCE INCOMPLETE DOMINANCE http://www.emc.maricopa.edu/faculty/farabee/BIOBK/BioBookTOC.html COMPLETE DOMINANCE F 2 generation- Not a 3:1 ratio Heterozygote= blended intermediate phenotype

HETEROZYGOTE: Both traits are expressed together (NO BLENDING) Roan horse has BOTH red & white hair (NOT A PINK HORSE! CO-DOMINANCE NON-MENDELIAN INHERITANCE An A allele AND a B allele make BOTH A and B GLYCOPROTEINS = AB Blood type

RWRW R W GENOTYPE PHENOTYPE ____________ ___________ _______________________ _______________________ _________ phenotypic ratio 1:2:1 Make a cross between TWO HETEROZYGOUS ROAN HORSES Show probabilities for: genotypes and phenotypes of possible offspring RR RW RW WW RRRED 25% 1/4 50% 1/2 RW WWWHITE ROAN ____________ genotypic ratio 1:2:1 25% 1/4

WWWW R GENOTYPE PHENOTYPE ____________ ___________ _________ ROAN offspring Make a cross between a HOMOZYGOUS RED HORSE and HOMOZYGOUS WHITE HORSE Show probabilities for: genotypes and phenotypes of possible offspring RW 100% RW ROAN

BLOOD TYPE MULTIPLE ALLELE TRAIT & CODOMINANT TRAIT ALLELES: ____ ____ ____ NON-MENDELIAN INHERITANCE _____ allele is dominant to _____ allele. _____ allele is dominant to _____ allele. _____ and _____ are CODOMINANT. BOTH SHOW TOGETHER

BLOOD TYPE MULTIPLE ALLELE TRAIT & CODOMINANT TRAIT ALLELES: ____ ____ ____ NON-MENDELIAN INHERITANCE _____ allele is dominant to _____ allele. _____ allele is dominant to _____ allele. _____ and _____ are CODOMINANT. BOTH SHOW TOGETHER A B O AO B O A B

SEX-LINKED GENES Some genes are carried on SEX chromosomes X-linked recessive disorders: Hemophilia Color blindness Duchenne Muscular Dystrophy XX= ________________ XY = ________________ X-linked traits show up more frequently in males No backup X to cover for the “broken” gene FEMALES MALES

X-LINKED Cross a colorblind male with a normal vision (non-carrier) female X c Y XCXC XCXC X C X c X C Y PHENOTYPE 100% normal vision 50% normal males 50% carrier females

X-LINKED Cross a hemophilia male with a carrier female X h Y XHXh XHXh X H X h X H Y X h X h X h Y PHENOTYPE 25%- hemophilia female 25%- hemophilia male 25%- normal male 25% - normal (carrier) females

TEST CROSSES

TEST CROSSES Dominant looking parent could have these genotypes: ________ OR __________ Can ’ t tell which by looking. Test cross used to determine which it is. ALWAYS TESTCROSS WITH A _________________________________ Offspring provide clue about genotype of unknown parent. HOMOZYGOUS RECESSIVE Tt TT

TEST CROSSES t T T T t t Tt tt Tt All offspring will be TALL 50% will be TALL 50% will be SHORT If any offspring show the recessive trait... unknown parent genotype was ________. If all offspring show the dominant trait... still don’t know. BOTH genotypes could produce offspring that look dominant! Tt

Deafness in dogs is caused by a recessive allele. Deaf dogs have the genotype dd. You have a hearing dog. Do a test cross to determine its genotype. d D D D d d d Dd dd Dd dd Dd Dd An actual test cross results in a litter with: 5 hearing puppies 0 deaf puppies What can you conclude?

Firebreathing (F) in dragons is dominant over NON-firebreathing (f). You have a fire-breathing dragon. What possible alleles could the fire-breathing parent have? _______ OR ________ EXPLAIN how you could use a TEST CROSS to help determine the parental genotype. Show the results of test crossing BOTH OF THE POSSIBLE PARENT GENOTYPES: An actual test cross results in a litter with: 6 firebreathing dragons 1 NON-firebreather EXPLAIN how you could use these results to determine the correct parental genotype.

TEST CROSS Used to determine genotype of unknown DOMINANT LOOKING parent Always cross with HOMOZYGOUS RECESSIVE (EX: tt) Observe offspring- If any offspring show recessive trait… know parent was HETEROZYGOUS If all offspring show DOMINANT trait... Still don’t know genotype. Do another test cross. Both TT and Tt can produce tall offspring with tt cross

DRAGONS An actual test cross results in a litter with: 6 firebreathing dragons 1 NON-firebreather ffff F F F f f f Ff ff Ff Ff What can you conclude? Unknown parent genotype is Ff Only way you can get a NON-FIREBREATHER

Write a NULL hypothesis that describes the mode of inheritance for the trait (purple eyes) I would expect this pattern in the F 1 offspring _______________________________ I would expect this pattern in the F 2 offspring ________________________________ THERE IS NO DIFFERENCE BETWEEN THE OBSERVED DATA AND THE EXPECTED DATA IF PURPLE EYES IS A(N) _________________ ____________ TRAIT

++++ + p p p +p+p F 1 100% wild typeF 2 75% wild type 25% purple eyed +p ++ +p +p pp IF YOU IGNORE SEX: If 1000 flies, expect 750 to be wild type and 250 to be purple eyed If you DON’T IGNORE SEX: 375 WT MALES 125 purple eyed MALES 375 WY FEMALES 125 purple eyed FEMALES Calculate Chi-square DO YOU ACCEPT OR REJECT THE NULL HYPOTHESIS?

There is no difference between the frequencies observed and the frequencies expected if PURPLE eyes is an autosomal recessive trait. H0-H0-

m m + +m IF GENE is AUTOSOMAL and RECESSIVE TO + MALES:FEMALES 1:1 + m + m ++ +m +m mm http://www.exploratorium.edu/exhibits/mutant_flies/curly-wings.gif F 1 All = +m (wildtype) F 2 ¼ = ++ 75% wildtype ½ = +m ¼ = mm 25% mutant 3:1

M M + +M IF GENE is AUTOSOMAL and DOMINANT TO + MALES:FEMALES 1:1 + M + M ++ +M +M MM F 1 All = +M (mutant) F2 ¼ = ++ 25% wildtype ½ = +M ¼ = MM 75% mutant 1:3

X + y X m X + X m X m y IF GENE is X-linked recessive Different pattern if gene is inherited from mom or dad Mutant mom X wild type dad X m y X + X m 50% normal females 50% mutant males X + X m X+y X m X m X m y When using Virtual fly lab Choose ignore sex and see if it changes the ratios 25% normal females 25% mutant females 25% mutant males 25% normal males F1

POSSIBLE PARENT GAMETES? RYRY ryry RyRy rYrY INDEPENDENT ASSORTMENT RrYyRrYy ________________________

RYRyrYry RY Ry rY ry Sign of HETEROZYGOUS DIHYBRID cross What’s the pattern? ____ Round & Yellow ____ Round & green ____ Wrinkled & yellow ____ wrinkled & green 9 3 3 1 RRYY RRYy RrYYRrYy RRYy RRyy RrYy Rryy RrYY RrYy rrYY rrYy RrYy Rryy rrYy rryy

If CLOSE TOGETHER on homologous chromosomes, STAY TOGETHER and end up together 100% of time Act like one gene

CROSSING OVER If far apart on homologous chromosomes, end up together 50% of time http://image.slidesharecdn.com/meiosisnotes-100204185918-phpapp02/95/meiosis-notes-23-728.jpg?cb=1265310007 RECOMBINANTS- Put different maternal/paternal alleles together on different chromosomes

INDEPENDENT ASSORTMENT http://image.slidesharecdn.com/meiosisnotes-100204185918-phpapp02/95/meiosis-notes-23-728.jpg?cb=1265310007 If on different chromosomes, END UP TOGETHER 50% of time.

__________ratio is a clue that it’s a ____________________________cross ____ ____________ TRAIT 1 ; ____________ TRAIT 2 ____ ____________ TRAIT 1; _____________ TRAIT 2 9 3 3 1 9:3:3:1 HETEROZYGOUS TWO gene dominant dominant recessive recessive dominant recessive

A large ear of corn has a total of 433 kernals, including 271 purple & starchy, 73 purple & sweet, 63 yellow & starchy, and 26 yellow & sweet. HYPOTHESIS: This ear of corn was produced by a dihybrid cross (PpSs X PpSs) involving two pairs of heterozygous genes resulting in a theoretical (expected) ratio of 9:3:3:1. Test your hypothesis using Chi-square and probability values. SHOW YOUR WORK!

H 0 - There is no difference between the frequencies observed and the frequencies expected for a HETEROZYGOUS DIHYBRID (9:3:3:1) cross. TOTAL = 433 offspring IF 9:3:3:1 then expect: Purple & Starchy = 433 X 9/16 = 243.56 Purple & Sweet = 433 X 3/16 = 81.19 Yellow & Starchy = 433 X 3/16 = 81.19 Yellow & Sweet = 433 X 1/16 = 27.06 271 73 63 26 243.56 81.19 27.06 27.44 -8.19 -18.19 -1.06 752.95 67.08 330.88 1.12 3.09 0.83 4.08 0.04 8.04 4-1=3 8.04 is larger than 7.82 REJECT THE NULL There is a difference between observed and expected for 9:3:3:1 = NOT A 9:3:3:1 cross

What is the probability? AaBbCcDd parent genome What is the probability of producing a gamete with this gene combination? ABCD _____________________________ aBcD _______________________________

What is the probability? AaBbCcDd parent genome What is the probability of producing a gamete with this gene combination? ABCD _____________________________ aBcD _______________________________ ½ X ½ X ½ X ½ = 1/16 ½ X ½ X ½ X ½ = 1/16

What is the probability? AaBbCcDD parent genome What is the probability of producing a gamete with this gene combination? abcD _____________________________ abcd _______________________________ AbcD ______________________________

What is the probability? AaBbCcDD parent genome What is the probability of producing a gamete with this gene combination? abcD _____________________________ abcd ______________________________ AbcD ______________________________ ½ X ½ X ½ X 1 = 1/8 ½ X ½ X ½ X 0 = 0 ½ X ½ X ½ X 1 = 1/8

What is the probability? AaBbCcDd X AaBbCcDd parents What is the probability of producing a offspring with this gene combination? aabbccDd _____________________________ AaBBccDD______________________________ AaBBCCDd______________________________ _

What is the probability? AaBbCcDd X AaBbCcDd parents What is the probability of producing a offspring with this gene combination? aabbccdd _____________________________ AaBBccDD______________________________ AaBBCCDd______________________________ _ ¼ X ¼ X ¼ X ¼ ½ X ¼ X ¼ X ¼ ½ X ¼ X ¼ X ½ = 1/256 = 1/128 = 1/64

What is the probability? AaBbCcDd X AaBbCcDd parents What is the probability of producing a offspring with this gene combination? AABbCcDd _____________________________ AaBBccdd______________________________ AaBbCcDd_______________________________

What is the probability? AaBbCcDd X AaBbCcDd parents What is the probability of producing a offspring with this gene combination? AABbCcDd _____________________________ AaBBccdd______________________________ AaBbCcDd_______________________________ ¼ X ½ X ½ X ½ ½ X ¼ X ¼ X ¼ ½ X ½ X ½ X ½ = 1/32 = 1/128 =1/16

PEDIGREES = male; doesn’t show trait = female; doesn’t show trait = shows trait = carrier; doesn’t show trait

Circle all males that show the trait BLUE Circle all females that show the trait RED Circle all carrier females GREEN Circle all carrier males PURPLE If this is an autosomal recessive trait, what is the genotype of individual #1? If this is an autosomal recessive trait, what is the genotype of individual # 2? If this is an autosomal recessive trait, what is the genotype of individual #3? 12 34

Circle all males that show the trait BLUE Circle all females that show the trait RED Circle all carrier females GREEN Circle all carrier males PURPLE If this is an autosomal recessive trait, what is the genotype of individual #1? If this is an autosomal recessive trait, what is the genotype of individual # 2? If this is an autosomal recessive trait, what is the genotype of individual #3? 12 3 4

Is it possible that this pedigree is for an autosomal recessive trait? Is it possible that this pedigree is for an X-linked recessive trait? Write the genotype of each individual next to the symbol.

Is it possible that this pedigree is for an autosomal recessive trait? Is it possible that this pedigree is for an X-linked recessive trait? Write the genotype of each individual next to the symbol. Aa aa Aa or AA YES

Is it possible that this pedigree is for an autosomal recessive trait? Is it possible that this pedigree is for an X-linked recessive trait? Write the genotype of each individual next to the symbol. X B Y X B X b X b X B Y NO

The inheritance of the disorder in II-3 from his father rules out what form of inheritance? http://www.mybookezzz.org/ebook.php?u=aHR0cDovL21jYi5iZXJrZWxleS5lZHUvY291cnNlcy9tY2I0MS9wcmFjdGljZV9wcm9ibGVtc19hbnN3LnBkZgpQcmFjdGljZSBwc m9ibGVtcyAod2l0aCBhbnN3ZXJzKSBUaGlzIGlzIHRoZSBkZWdyZWUgb2YgZGlmZmljdWx0eSBvZiAuLi4= Can’t be X-linked recessive Males get their X-linked allele from their mother If dad passed to son it must be AUTOSOMAL X b Y

Is this trait inherited as AUTOSOMAL RECESSIVE? http://www.mansfield.ohio-state.edu/~sabedon/biol1128.htm#A1

Is this trait inherited as AUTOSOMAL RECESSIVE? http://www.mansfield.ohio-state.edu/~sabedon/biol1128.htm#A1 aa a Aa AUTOSOMAL RECESSIVE POSSIBLE

Is this trait inherited as AUTOSOMAL DOMINANT ? http://www.mansfield.ohio-state.edu/~sabedon/biol1128.htm#A1 Aa aa a aa AUTOSOMAL DOMINANT IMPOSSIBLE

Is this trait inherited as X-LINKED RECESSIVE? http://www.mansfield.ohio-state.edu/~sabedon/biol1128.htm#A1 XaXaXaXa X A Y X-LINKED RECESSIVE IMPOSSIBLE X a Y X a XaYXaY

Is this trait inherited as X-LINKED DOMINANT? http://www.mansfield.ohio-state.edu/~sabedon/biol1128.htm#A1 XAXaXAXa X a Y X-LINKED DOMINANT IMPOSSIBLE XAYXAY X a XaYXaY XAYXAY X A X ? X a Y

PATTERNS ARE THE KEY Image modified from: http://www.laskerfoundation.org/rprimers/gnn/timeline/1866.html http://www.accessexcellence.org/AB/GG/mendel.html

MAKING PUNNETT PREDICTIONS http://www.exploratorium.edu/exhibits/mutant_flies/mutant_flies.html

USING CHI-SQUARE TO ANALYZE GENETICS DATA http://www.exploratorium.edu/exhibits/mutant_flies/mutant_flies.html CROSS BETWEEN A RED EYED (WILD TYPE) FEMALE AND PURPLE EYED MALE DATA: F 1 - 570 WT females; 615 WT males F 2 - 460 WT females; 147 purple eyed females; 451 WT males; 138 Purple eyed males IGNORE SEX : 911 WT flies; 285 purple eyed flies pur pur + + p + p + + + p + p p p

USING CHI-SQUARE TO ANALYZE GENETICS DATA http://www.exploratorium.edu/exhibits/mutant_flies/mutant_flies.html CROSS BETWEEN A RED EYED (WILD TYPE) FEMALE AND SEPIA EYED MALE DATA: F 1 - 529 WT females; 497 WT males F 2 - XXXX WT females; XXXX sepia eyed females; XXXXX WT males; XXXXX sepia eyed males se se + + se se + se + se se + se + se se se

There is no difference between the frequencies observed and the frequencies expected if sepia eyes is an autosomal recessive trait. H0-H0-

http://www.exploratorium.edu/exhibits/mutant_flies/mutant_flies.html P1P1 F 1 622 male WT 590 female WT F 2 608 female WT 326 male WT 281 male Yellow body Female WT Male yellow body t + t + X t+ Y t + t + X t+ X t+

y y + +y IF GENE is AUTOSOMAL and RECESSIVE TO + MALES:FEMALES 1:1 + y + y ++ +y +y yy http://www.exploratorium.edu/exhibits/mutant_flies/curly-wings.gif F 1 All = +y (wildtype) F 2 ¼ = ++ 75% wildtype ½ = +y ¼ = yy 25% yellow body 3:1

Y Y + +Y IF GENE is AUTOSOMAL and DOMINANT TO + MALES:FEMALES 1:1 + Y + Y ++ + Y +Y YY F 1 All = +Y (yellow body) F 2 ¼ = ++ 25% wildtype ½ = +Y ¼ = YY 75% mutant 1:3

X + y X m X + X m X m y IF GENE is X-linked recessive Different pattern if gene is inherited from mom or dad Mutant mom X wild type dad X m y X + X m 50% normal females 50% mutant males X + X m X+y X m X m X m y When using Virtual fly lab Choose ignore sex and see if it changes the ratios 25% normal females 25% mutant females 25% mutant males 25% normal males F1

X t y X t+ X t+ X t+ X t+ y X t+ X t X t+ y IF GENE is X-linked recessive Different pattern if gene is inherited from mom or dad Wild type mom X yellow body dad X t+ y X t+ X t 50% normal females 50% normal males X t+ X t+ X t+ y X t+ X t X t y When using Virtual fly lab Choose ignore sex and see if it changes the ratios 50% Wild type FEMALE 25% tan MALES 25% Wild type males F1F1 F2F2

X + y X M X + X M X M y IF GENE is X-linked dominant MUTANT MOM X WILD TYPE DAD X M y X + X M 50% mutant females 50% mutant males X + X M X + y X M X M X M y When using Virtual fly lab Choose ignore sex and see if it changes the ratios F1 F2 50% mutant females 25% wild type males 25% mutant males

X M y X + X + X M X + y IF GENE is X-linked dominant WILD TYPE MOM X MUTANT DAD X + y X + X M 50% mutant females 50% normal males X + X + X + y X + X M X M y When using Virtual fly lab Choose ignore sex and see if it changes the ratios 25% wild type females 25% mutant females 25% wild type males 25% mutant males F1 F2

If CLOSE TOGETHER on homologous chromosomes, STAY TOGETHER and end up together 100% of time Act like one gene

CROSSING OVER If far apart on homologous chromosomes, end up together 50% of time http://image.slidesharecdn.com/meiosisnotes-100204185918-phpapp02/95/meiosis-notes-23-728.jpg?cb=1265310007 RECOMBINANTS- Put different maternal/paternal alleles together on different chromosomes

INDEPENDENT ASSORTMENT http://image.slidesharecdn.com/meiosisnotes-100204185918-phpapp02/95/meiosis-notes-23-728.jpg?cb=1265310007 If on different chromosomes, END UP TOGETHER 50% of time.

A Wild type fruit fly (heterozygous for gray body and normal wings) is mated with a black fly with vestigial wings. OFFSPRING: 778- wild type 785- black-vestigial 158- black- normal wings 162- gray body-vestigial wings Is it 9:3:3:1? (2 genes on 2 different chromosomes) Is it 3 Wild type : 1 black vestigial? (linked on homologous chromosomes)

A Wild type fruit fly (heterozygous for gray body and normal wings) is mated with a black fly with vestigial wings. OFFSPRING: 778- wild type 785- black-vestigial 158- black- normal wings 162- gray body-vestigial wings What is the recombination frequency between these genes?

A Wild type fruit fly (heterozygous for gray body and normal wings) is mated with a black fly with vestigial wings. OFFSPRING: 778- wild type 785- black-vestigial 158- black- normal wings 162- gray body-vestigial wings Recombinants = Total 320 = 17% 1883

A Wild type fruit fly (heterozygous for gray body and normal wings) is mated with a black fly with vestigial wings. OFFSPRING: 778- wild type 785- black-vestigial 158- black- normal wings 162- gray body-vestigial wings What is the recombination frequency between these genes?

A Wild type fruit fly (heterozygous for gray body and normal wings) is mated with a black fly with vestigial wings. OFFSPRING: 778- wild type 785- black-vestigial 158- black- normal wings 162- gray body-vestigial wings What is the recombination frequency between these genes? Recombinants = Total 314 = 16.7% 1877

A Wild type fruit fly (heterozygous for gray body and red eyes) is mated with a black fly with purple eyes. OFFSPRING: 721- gray body/red eyes 751- black body/purple eyes 49- gray body/purple eyes 45- black body/red-eyes What is the recombination frequency between these genes?

A Wild type fruit fly (heterozygous for gray body and red eyes) is mated with a black fly with purple eyes. OFFSPRING: 721- gray body/red eyes 751- black body/purple eyes 49- gray body/purple eyes 45- black body/red-eyes What is the recombination frequency between these genes? Recombinants = Total 94 = 6 % 1566

A Wild type fruit fly (heterozygous for normal bristles and red eyes) is mated with a spineless bristle fly with sepia eyes. OFFSPRING: 648- normal bristles/red eyes 681- spineless bristles/sepia eyes 72- normal bristles/sepia eyes 83- spineless bristles/red-eyes What is the recombination frequency between these genes?

A Wild type fruit fly (heterozygous for normal bristles and red eyes) is mated with a spineless bristle fly with sepia eyes. OFFSPRING: 648- normal bristles/red eyes 681- spineless bristles/sepia eyes 72- normal bristles/sepia eyes 83- spineless bristles/red-eyes What is the recombination frequency between these genes? Recombinants = Total 155 = 10.4% 1484

Determine the sequence of genes along a chromosome based on the following recombination frequencies A-C 20% A-D 10% B-C 15% B-D 5%

Determine the sequence of genes along a chromosome based on the following recombination frequencies A-C 10% A-D 30% B-C 24% B-D 16%

Determine the sequence of genes along a chromosome based on the following recombination frequencies A-C 10% A-D 30% B-C 24% B-D 16% CABD

Determine the sequence of genes along a chromosome based on the following recombination frequencies A-B 8% A-C 28% A-D 25% B-C 20% B-D 33%

GENETICS PRACTICE Watch videos BOZEMAN- Beginner’s Guide to Punnett Squares BOZEMAN- Beginner’s Guide to Punnett Squares BOZEMAN- Mendelian genetics BOZEMAN-

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