l)tt)\1,./t- -^-

in pet piir-r,ts, Ciie ¿;'1.'':;r'pcd gene (i;) is dsniÌ¡ar'i ar¡:Ìî ùe,v-âjii)-'\'i-!í'ri g+ra (¿)

,l¿ Twq pea plants with greeu pods, Gg x GG, were crossed' Complete the Punnett squale below to show

' the rãsulti of this cross. [I]

as

[a Wn"t p"r""niãg,5äf *e'tffsp-rüg produced by thls cross will rnost likeþ have green pods? [I]

ls} v,

4Lo* th" g"oeti" makeup of two parent pea plants whose offspriug would all have yellowpods' [l]

A q o^^.t"-Ð--Base your answeis to questions € through@on the Punaett square below, which shows a cross between

wo t"ll p"" pl"átr (rt x t'l). 3 - 3

TaII Male PlantTt

Tall Female Pla¡t

Kty

T = tall gene (dominant)

t = short geûe (recessive)

3. á/----.:'--^.-.r^rrÐ r.r la¿ 7ofuhat percentage of the offspring will grow tall? hl ( vÇ ( "'/

(""ordtg ao ihe Punaett square, what is the probability of ar: oßpriog bleriting two tall genes? W"/c- {-. ó/ I /-.g

(Express your answer as a hactioû or percentage') [r] ) Ò 1 o ¿^" / /-

(J ' -- ¡-- L^¡-L ^-^^¡I., +) ' es for heíght are not exactly tJre same [i]É*pUi" *ny U"tn parent plants are taÌl, even though their gen

-( "i) úfu.b ,\\l

''f-tu4Ui[l A _ <rg"xÐ_ _ø]t=rw, _W-**I4

+

rì:.

ifàe Por*,eit sqr.:ai= ieii,;=", s.ho:i,s a¡ ?i_ä ¡.¡a Èi"¿at crcssed-wiLl aa ÌÈ- pea pian....nrK"y

.R = firll mund pod sàaþe (dominant)

r = wri¡Hed pod sLape (recessive)

7,6 Complete the Pr:laet squere provided bely to show the erossing of two fir pareuts.

F--r

$@ tr roo oerrp'iog were produced from the crossing shown in.þ Rgnett square. b9row, approximatelyhow many would have a wrinlded pod shape? hl S-O ZO ñ /;

'

A(-\_

{-

.þ, ---'r

€ l{ãat percentage of tåe otrspring will have a firll, rouad pod shape? ftl / 6 Ò o"

KeE

.R = fidl rot-nd pod shape (dominant)

r = wri¡Ided pod sh¡fè (recessive)

ßB = chesl+nølL brown)

LrJk) = u:hìk- 3¡) = Dú,[o¡ninÒ

frtendi" X'JG*r'r@dt' ri

JtuL,*1tCro55

',1{)x tùt4-J

1ltDUJ

lo) cøss 3ø x Br^-r' fa L,) )flr 9'o'"

ß | øplL) 5o/' ¡'t"ww

*EM#,i,Y,"r*bB: rbw

'' nh/

UJ

t ß¿ ¿ii f;i v z .-ì,".:,,P'¡,gnc1¡ ;,";zi

ÇDate

skin after a cut. They also go through ¡t as

úrüÈisÈiË

@**t*tI *,o '.*r,--r-"ÆæJ''"-*'*'--Æffi.\r.ffilW

"/ \ ?dds{ù{eË

@ @--I I 4 dårrolT!âÊ

@@ @@*.

*,**,*@**

N

I *n**,r**,ËllEL

f@\Wlt:|w"7\v"1 \ ,r.os**n

c€llg

@@U.9, Nãt¡c.n,rl Librérf (l I'fÈdicir'È

Cell cycle/MitosisWord bank for cell cycle/mitosis:

.l¡tle+phase, M€tephase,-D'4+tosît

Bøenær€ph€i€rRep¡eates,--Teloplræ€-

cells go through mitosis when you need

Meiosis

Meiosis is similar tothrough

gametes.

much DNA as

both processes involve a cell dividing. Not all body cells go

cells (sperm and eggs). Sex cells are also called

only have half as Word bank forcrossrng over, , 8añeteTse¡(cperm,

cell. Your regular body cells

n^u. t4Ç \2 but your gametes

only have L9 & chromosomes. This is because

),bell plus an egg cell make a baby. lf you want your baby to have one set ofi

chromosomes like you do, you need the sperm to give half a set of DNA and the egg to give half

a set of DNA l% + y, = 1 set of chromosomes). To end up w¡th cells that have half a set ofchromosomes instead of a full set, you need a special process. That's where meiosis comes in!

Meiosis is different from mitos¡s because it t "r@h/^l cell divisions instead of just one

.hrorororn"r. rhis is carle¿ C?) Cr'ðSS t b 5\.-v' . and this mixes up your DNA.

when the cells divide and ..t7ã* *tlraoï*o *llrã alike because they each have

different genes on them (see picture below)!

w@

<ffi-@

No irvo rex cells arc alikc, ilueto c¡rssi¡iE ol,e¡'!

w-@,@@

Codominance Worksheet (Blood types)

Human blood t)?es are determined by genes that lollow the CODOMINANCE pThere are two dõminant aìleles (IA and IB) and one recessive allele (i).

l Wdte the genotFe for each person based on the des_cription:

a. Homozygous for the "n" alleleb. Heterozygous for the "4" allelec. Type Od. Type "4" and had a t,?e "O" parente. Type "AB"f. Blood can be donated to a.rìybody

Iß L "¿ÚL9rÞrß#.r¿+-*LTW

2.

g. Can only get blood fiom a type "O'' donor _-#< øL' )-Pretend that Brad Pitt is homozygous for the type B allele, and Angelina Jolie is þpe "O." What

) Iï.4 EÒ I- - --1"'-"

Two parents think their baby was switched at the hospital. Its 1968, so DNA fingerprintingtechnology does not exist yet, The mother has blood type "Q,-:the father has blood type "48,"and the baby has blood type "¡-l

a. Mother's genotype: CO

are all the possible blood types oftheir baby? Wlo,% bC ;ffi

Draw a Punnett square showing all the possible blood types for the Jffspring produced by a type

,1ffi"o"motherandanarlpe*AB"father

,,ffi -Vr

^, ts

*

b. Falher's genotype: /A J\ A tc. Baby's genotyp": 'gn o, t+ "d. Punnetl square showiirg,^all possible genotypes fór childlen produced by this couple,ìT1 Ao, hD cr"hÒr[] ' /e. wasthebabvs*,"*0,

NbÞ- Nù\ ¿ ,'lL*¡ Co,, 7u0,tvc\ \,r" b ¿L^lL

Genotype Can donate blood to: Can receive blood aom:

o ll A,B,AB and Olnnìversal donor'l

o

AB IAIB O, AB A,B,AB and O(universal receiver)

IoIo or Ioi AB, A O,A

B I"Iu o, # AB,B o,B

'-"-6. Two other parents tbink their baby vvas switched at the hospital. The mother hàÈ'61¡tod type "4,"the father has blood type "8," a4d tþe baby l¡as,.bÌood type "AB.''

a. Mother's genot¡p", lll 4 o, /\ Ub. Father's eãnotñe, ^-tR ,Þ o. F-¡-c. Babv's ";.t";, IFB--O. zu*"ttiquare t¡at\hows tne Uaby's genotype as a possibiliry:

e. Was the baby switched?

7. Based on the information in this table, which man could not be the father of the baby? Jusri$your answer with a }ìrnnett square. -. - +'

8. Based on the information in this table, which man could not be the father ofthe baby? Justifyyour answer with a Punnett square.

NIOa--"-

¡r}tù^Criro øl"tb

þ-0 to

r'ê Bo4ì \ 8,7*,_

S<r^^*6*Þ(+, ì'W"^.

Í'iøk/Vrl.

Ð\

Name Blood Type

Mother ryp" 9Baby Type AB

Bartender Type O

Guv at the club 'l ype A.B

Cabdriver Iype A

Flight attendant Iype B

9. Explain why blood type data cannotlather þ¡q¡.

È'nt+prove who the father ofa baby þ, and can only prove who the

Compare,/contrast Table t/"Comparing Processes #-tlti#:;{TÏ'#:,{#i#:,lif,iî;:mpletethecompare/contrcsttøbreberta.' Iitp*^t" tl,ot i¡/îpä."-"' "

uLc t'øbte t0 a¡ríte your attswøs, write tn* *T' \

ïÈc

(ô

MitosisMeiosie

urowrq regeneration ofnerrrt cells

Locationìn Bodyy cells

N-umber of Daughter

),...-- Fou¡

Change inChromosome Number

B$ftûNu¡rbe¡ of phases u' rz/

6

Cell Divisions One

Between parent Celland-Daughter CeIIs

8.

ï'.l)'N-q--_üomologous chromosomesassort independentlv, soeach gamefe has a ufouecombinatíon of alleìes '

139

In00mplele anû [0dOminanm lv0ilisneel(Non-mendelian monohybrid crosse s )

B aa{ the allele fo¡ white is W. The heterozygous phenotype is known as erminette.2.

'1.

a. What is the genotype for black chickens? Þßb. What is the genotype for white chickens? ¡¿]\)_^c. What is the genotype for erminette chickens? Þ14/

If two erminette chickens were c¡ossed, what is the probability that:a. They would have u 61u"¡

"¡¡s¡r 25 7,

b. They would have a white chick I ÉCo rz t \

Parents: P U x

4. A black chicken and a white chickeneminctta ¡hi¡l¿c? /).---r Ø-**-

&)r'"are crossed. What is the probability that they will have

$w

00

Parenrs: Vh x ûJ tA)

1\lBNlsrlø,1ñwM

5. In snapdragons, flower color is controlled by ,n"o*Oß8á**ncç-The two alleles are red (R)and white (W). The heterozygous genotype is expiessed as pink

a. What is the phenotype of a plant lvith the genotype RR?b. What is the phenotype of a plant r.vith the genotype W-W?c. What is the phenotype of a plant with the genotype RW?

{#-.w

eTName:Period:

Date:Answer the following questions. Províde a punnett square to support lour answers whereind.icated. Express probøbìlities as percentages. For instønce, a prohøbilíty of one chance ín tenwould be 707o.

Explain the difference between in::rnplerglþAlg19e and codominanc"'--V n, T; a*'-{e7Ns'''P '

þÞhi,,xIn some chickens, the gene for feather color is controìled by codimin^ance. The allele for black is

6. A pink-flowered plant is crossed with a white-flowered plant. lVhat is the probability ofproducingapink-floweredptantt

ÇA_v" "*"no,pil, tJ!

Llt¿úWWIutw

7. What c¡oss will produce the most pink-flowered plants? Show a punnett square to support your

8. Another type of non-mendelian trait: Multiple alleles. Human hair color is controlled by one genewith four alleles (with some incomplete dominance):

HB'= brown HBd = blonde hR = r9d hbk = black

The possible genotypes and phenotypes:

HBdHBd or HBdhbk = blonde HBdHBT - mousy brownIJBdhR - cfrcrr¡hc1.rr¡ hlnnde I{Bq{Br ñr lfBrhù*- l" -r.HB.ÌrR = auburn hRhR or hRhYk = red ' hbkhbk = black=

What do you think your parent's phenotypes and genotypes for hair color are?

¡ Ør11Þr?What a¡e yóu¡ phenotype and genotype fo¡ hair color? À i .

rr someone with aubu¡n hai¡ has "^+1,"!:':fl":"e wirh red nun,o", *liÍ".l "!n^?or*ohair), what are the genotype and phenôtype probabilities for their children?

'nv x. V- t blc-

h, ,1". ¡ fL ; aalaa,,t--

11 Þ', h-',,-,Lunr - Y,þirr_ w*

oftr)'h

. {-h'KO,

T {L

€Class -- Date

Acnvlry ¡ Generics

Shoutd This Dog Be Cailed Spot?Imagine this microscopic dTru'4 sex cerì from a mare dogjoins with â sex celr from

:"i,H,:*"T;3*1gi::..r.ã,ì-".-äs ü;.äi,å,å*äir,. ren'ized egg that resurts

,h;;y;;.;,ìä'Ëkîä:;:fiffi:?.ïrXåi_ä::ffiî*:i.ï"parént.suppose

From the female dog

From the male dog

CHAPTER

æ

Each chromosome of the pair contains genes for the same Faits" But one:5"i:T*.yt!1v¡ a

foniinan¡ g"r,. *-d t¡. ott ei u ."..ssive gene. Use the drawinsqrru rr¡ç L¡¡¿r¡ l, , i1¡tÐ\ ref tne questlofui.

1. Would the new puppy have a spotted coat?

@ Prentice-Hall, lnc.

3. Does the male dog have a çotted coat?

E".plui,,, ìl i) f¿C¡$rl-

4. Whar would tlre puppy's hair texrure be?

5. Would the puppy have curly or straight hair?

6. a. Does t̡e female dog have curly hair?

b. Does the male dog have curly hair?

â.a- - a-q

7. Would the puppy be a hybrid or a purebred for hair length?

trhe eåghter Sídegnh^'*{'

t^vþ Ub^Ðb-rr,"^)+- r:

"You idiot, I meant transfer their genes.',

Name

I. Puanett Square A:

Bb

'Eünlt4ryalø,rlt_t_-_______l

Frobability and Geneticsô ilnderstanding Main ldeasCoffiplete the two Punnett squaræ below, ønd ihett answer the questimts on ascparate sheet of paper.

2. Punaett Square B:

E-h__þ2h

3. In the cross between fwo black g,.rine1 pigs ghown in p¡,¡neä square .A,, what is i ,theprobabilityrtratanotrspringwillb;6¡*tlrrr,1t"i j-.*f ;"tsì"Á*r"";çd¿5al¡þ*

l:^111,,"-:.llt_: ":, rte cross between rwo btack guinea pigs in punnett Square Awould not þroduê a white guinea pig r^d"t;.

'

;;-: Ë ü-;'ïi:; ; ,5' what color are the guinea pig par.ot" i' th...ossl.ho"á-i' ioooll"tt sq*'.åi 0 b,lr rrw6. Which guinea pig parent(s) in punnett Square B is homozygo.r,.","1ig"*?;;!'kj"jqifr,*.-- et. tt -. ,', t}*n".e û*_ = _bL:iïff H,"ihäffi n#i;,*affini'p*y"3*

neærozygols3 Expki¡ hqw you know. el , -T.calcutatethìþ'rob$ffiÍår-",#-hrp,r"*Hrlf #rlìyl

Square B. What is the probabitity thuì u"ãrrprirrs *i[ b;;i;ä- .Ç,)

% B/ù"è Fuiidins vbcabutary ';,;)^"o¡^Ffr"/o øh"\sIt¡T.9cÀo

Yrf,!:!,r:::.ith ix dfinitíonby writingtheteuer of tlrr ron"n a"¡nírlín lnthe line besîàe the term.

t ".

n"rrro"rro.

-A- 9. Punnetr square

!-t'''.no*"Q n. codominance

S rz. probabitity

Cnu.oo^orrroo,

f) r¿. ph.oowp.

z.aía cha¡t that shows all the possible combinations of alleles thatcan result trom a genetic cross

-.b,:-. the likelihood that a particuJar eyent will occur

.;;,.ah organism that has two identical alleles for a trait,*ãn,organismt physica.l appearance

;e.án organism's genetic makeup, or allele combinations

.f<áh organism that has two different alleles for a_traitg, inheritance pattern in whích the alleles are neither dominant

nor recessive - 6o4t M^ Eyr-rs5- g. :"

ìil(llì

i!"

t:t I

Teaching Resources Cø17