Genetics Review Powerpoint

Genetics Review Powerpoint

MENDEL’S PEA EXPERIMENTS
P1 = Parental F1 = Filial (pffspring) F2 = 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 hides another represented by capital letter gene that is hidden by another represented by lower case 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”

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

COIN FLIP 1/2 50% 1:1 There are 2 possible outcomes: HEADS TAILS
COIN FLIP There are 2 possible outcomes: HEADS TAILS 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

R R r Rr Rr Rr ROUND GENOTYPE 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 R R GENOTYPE PHENOTYPE ____________ ___________ _______ show dominant phenotype _______ show recessive phenotype r Rr ROUND Rr Rr 100% 4/4 0% 0/4

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

ALL Pp 100% PURPLE FLOWERS 0% WHITE FLOWERS 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

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

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

75% ______ PURPLE? ______ WHITE? _____ PP? _____ Pp? _____ pp?
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% 25% 50% 25% 1:2:1 3:1

POSSIBLE PARENT GAMETES?
__________________________ INDEPENDENT ASSORTMENT

TTRR ________________________ TR TR TR TR INDEPENDENT ASSORTMENT

ttrr ________________________ tr tr tr tr INDEPENDENT ASSORTMENT

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

TtRr ________________________ TR tr tR Tr INDEPENDENT ASSORTMENT

RrYy ________________________ RY ry rY Ry INDEPENDENT ASSORTMENT

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

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

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

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

INCOMPLETE DOMINANCE COMPLETE DOMINANCE NON-MENDELIAN INHERITANCE
F2 generation- Not a 3:1 ratio Heterozygote= blended intermediate phenotype

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

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

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

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

BLOOD TYPE MULTIPLE ALLELE TRAIT & CODOMINANT TRAIT A B O ALLELES: ____ ____ ____ A O _____ allele is dominant to _____ allele. _____ allele is dominant to _____ allele. _____ and _____ are CODOMINANT. BOTH SHOW TOGETHER 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 XCXc XCY PHENOTYPE 100% normal vision 50% normal males
Cross a colorblind male with a normal vision (non-carrier) female Xc Y PHENOTYPE 100% normal vision 50% normal males 50% carrier females XCXc XCY XC

X-LINKED XHXh XHY XhXh XhY Xh
Cross a hemophilia male with a carrier female Xh Y PHENOTYPE 25%- hemophilia female 25%- hemophilia male 25%- normal male 25% - normal (carrier) females XHXh XHY XhXh XhY XH Xh

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. TT Tt HOMOZYGOUS RECESSIVE

TEST CROSSES T T T t t t t Tt Tt Tt tt Tt tt 50% will be TALL 50% will be SHORT All offspring will be TALL 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
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 Dd Dd Dd Dd d d Dd dd Dd dd An actual test cross results in a litter with: 5 hearing puppies 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

f Ff Ff Ff Ff Ff ff Ff ff DRAGONS What can you conclude?
An actual test cross results in a litter with: 6 firebreathing dragons 1 NON-firebreather F F F f f Ff Ff Ff Ff 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 F1 offspring _______________________________ I would expect this pattern in the F2 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 + + p ++ +p +p pp F1 100% wild type F2 75% wild type % purple eyed 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: WT MALES purple eyed MALES WY FEMALES purple eyed FEMALES Calculate Chi-square DO YOU ACCEPT OR REJECT THE NULL HYPOTHESIS?

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

m m + +m +m + m + ++ +m m +m mm F1 All = +m (wildtype)
IF GENE is AUTOSOMAL and RECESSIVE TO + MALES:FEMALES 1:1 m m + +m +m F1 All = +m (wildtype) m + m F2 ¼ = % wildtype ½ = +m ¼ = mm % mutant 3:1 ++ +m +m mm

M M + +M +M + M + ++ +M M +M MM IF GENE is AUTOSOMAL and DOMINANT TO +
MALES:FEMALES 1:1 M M + +M +M F1 All = +M (mutant) M + M F2 ¼ = % wildtype ½ = +M ¼ = MM % mutant 1:3 ++ +M +M MM

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

RrYy ________________________ RY ry rY Ry INDEPENDENT ASSORTMENT

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

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

http://image. slidesharecdn
CROSSING OVER If far apart on homologous chromosomes, end up together 50% of time RECOMBINANTS- Put different maternal/paternal alleles together on different chromosomes

INDEPENDENT ASSORTMENT
INDEPENDENT ASSORTMENT If on different chromosomes, END UP TOGETHER 50% of time.

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

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!

TOTAL = 433 offspring IF 9:3:3:1 then expect:
H0- 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 = Purple & Sweet = 433 X 3/16 = Yellow & Starchy = 433 X 3/16 = 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 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 _______________________________

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

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

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

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

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 = 1/256 ½ X ¼ X ¼ X ¼ = 1/128 ½ X ¼ X ¼ X ½ = 1/64

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

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 = 1/32 ½ X ¼ X ¼ X ¼ = 1/128 ½ X ½ X ½ X ½ =1/16

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

1 2 3 4 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?

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?

Aa Aa Aa or AA aa 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? YES

XB Y XB Xb XB Y Xb Xb 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? NO

If dad passed to son it must be AUTOSOMAL
The inheritance of the disorder in II-3 from his father rules out what form of inheritance? Xb Y Can’t be X-linked recessive Males get their X-linked allele from their mother If dad passed to son it must be AUTOSOMAL Xb Y

Is this trait inherited as AUTOSOMAL RECESSIVE?

Is this trait inherited as AUTOSOMAL RECESSIVE?
AUTOSOMAL RECESSIVE POSSIBLE aa Aa aa a a Aa aa aa A a A a aa aa aa aa

Is this trait inherited as AUTOSOMAL DOMINANT ?
AUTOSOMAL DOMINANT IMPOSSIBLE Aa aa Aa A a aa Aa Aa a a a a Aa Aa Aa Aa

Is this trait inherited as X-LINKED RECESSIVE?
X-LINKED RECESSIVE IMPOSSIBLE XaXa XA Y XaY Xa Y Xa Xa

Is this trait inherited as X-LINKED DOMINANT?
X-LINKED DOMINANT IMPOSSIBLE XAXa Xa Y XaY XAY Xa Xa Xa Xa XAY Xa Xa Xa Y XA X?

PATTERNS ARE THE KEY Image modified from:

MAKING PUNNETT PREDICTIONS

USING CHI-SQUARE TO ANALYZE GENETICS DATA
CROSS BETWEEN A RED EYED (WILD TYPE) FEMALE AND PURPLE EYED MALE DATA: F WT females; 615 WT males F WT females; 147 purple eyed females; 451 WT males; 138 Purple eyed males IGNORE SEX : 911 WT flies; 285 purple eyed flies p + p pur pur + + p p + p p p + p p p

USING CHI-SQUARE TO ANALYZE GENETICS DATA
CROSS BETWEEN A RED EYED (WILD TYPE) FEMALE AND SEPIA EYED MALE DATA: F WT females; 497 WT males F2 - 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 se

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

P1 Male yellow body Female WT t+ t+ Xt+ Y t+ t+ Xt+ Xt+
F1 622 male WT 590 female WT F2 608 female WT 326 male WT 281 male Yellow body

y y + +y +y + y + ++ +y y +y yy F1 All = +y (wildtype)
IF GENE is AUTOSOMAL and RECESSIVE TO + MALES:FEMALES 1:1 +y +y F1 All = +y (wildtype) y + y F2 ¼ = % wildtype ½ = +y ¼ = yy % yellow body 3:1 ++ +y +y yy

IF GENE is AUTOSOMAL and DOMINANT TO +
MALES:FEMALES 1:1 Y Y + +Y +Y F1 All = +Y (yellow body) Y + Y F2 ¼ = % wildtype ½ = +Y ¼ = YY % mutant 1:3 Y +Y YY

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

Xt+ y Xt y Xt+ Xt+ Xt+Xt+ Xt+y Xt+Xt+ Xt+y Xt Xt+Xt Xt+y Xt+Xt Xty
IF GENE is X-linked recessive Different pattern if gene is inherited from mom or dad Wild type mom X yellow body dad F1 F2 Xt y Xt+ Xt Xt y Xt+ Xt+Xt+ Xt+y Xt+Xt Xt+y Xt+Xt+ Xt+y Xt+Xt Xty 50% normal females 50% normal males 50% Wild type FEMALE % tan MALES 25% Wild type males When using Virtual fly lab Choose ignore sex and see if it changes the ratios

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

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

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

http://image. slidesharecdn
CROSSING OVER If far apart on homologous chromosomes, end up together 50% of time RECOMBINANTS- Put different maternal/paternal alleles together on different chromosomes

INDEPENDENT ASSORTMENT
INDEPENDENT ASSORTMENT 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: 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: 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: 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: 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: 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: gray body/red eyes 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: gray body/red eyes black body/purple eyes 49- gray body/purple eyes 45- black body/red-eyes What is the recombination frequency between these genes? Recombinants = Total = 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: normal bristles/red eyes 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: normal bristles/red eyes 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-D 10% B-C 15% B-D 5%

A-D 30% B-C 24% B-D 16%

A-D 30% B-C 24% B-D 16% CABD

A-B 8% A-C 28% A-D 25% B-C 20% B-D 33%

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