Hello. Attached I have 3 genetic labs that I need assistance with. In some of the questions, it says you'll need to watch the video to answer the questions. This for the most part is not true. I usual

g oin g to r e tu rn to th e to pic to day. B efo re yo u st art, r e vi ew th e b ackg ro und in fo rm atio n a bout C RIS PR in th e L earn in g R eso urce s fo r

th is unit .

G o to th is lin k: CRIS PR -C as9 M ech anism a nd A pplica tio ns a nd la unch th e C RIS PR in te ra ct ive .

Work t h ro ugh t h e w hole in te ra ctiv e t h en r e tu rn h ere a n d a n sw er t h e f o llo w in g q uestio ns.

P art 1 : U sin g C RIS PR t o in activ ate g en es

1. D iscu ss what it m eans to in act iva te a g ene. C an yo u th in k of a w ay to d o th is? T hin k back to w hat yo u kn ow a bout e uka ryo tic gene

re gula tio n.

2. A s sh ow n in th e C lick & L earn , th e C RIS PR -C as9 to ol u se s a D NA-cu ttin g e nzym e, o r n ucl ease , ca lle d C as9 . C as9 w as disco vere d

in b act eria a s a w ay fo r b act eria to fig ht o ff vi ru se s. S ci entist s co mbin e C as9 w it h a n R NA m ole cu le ca lle d a g uid e R NA to fo rm a

C as9 -R NA co mple x. P art o f th e g uid e R NA m atch es a ta rg et D NA se quence w it h in th e g ene th at th e sci entist s want to in act iva te . Y ou

w ill n ow g o th ro ugh e ach st ep in th e p ro ce ss of u si ng C RIS PR -C as9 to in act iva te a g ene, u si ng a se quence fr o m a n a ct ual g ene a s an

exa mple .

F ir st , th e C as9 -R NA co mple x re co gnize s and b in ds to a th re e-n ucl eotid e se quence ca lle d P A M , w hich st ands fo r “ p ro to -sp ace r

adja ce nt m otif .” P A M se quence s are a bundant th ro ughout th e g enom e a nd ca n o ccu r o n e it h er st ra nd o f D NA. E ve ry PA M se quence

has th e fo rm 5 '- N G G-3 ', w here th e “ N ” ca n b e a ny DNA n ucl eotid e ( A , C , G , o r T ).

T he p artia l g ene ( D NA) se quence b elo w co nta in s m ult ip le P A M se quence s. H ig hlig ht si x PA M se quence s in th e to p ( 5 ’ to 3 ’) st ra nd.

O nce C as9 b in ds to a P A M se quence , it u nw in ds th e D NA. If th e g uid e R NA m atch es th e D NA se quence n ext to th e P A M , th e g uid e

R NA w ill b in d to th e co mple m enta ry DNA st ra nd. If n ot, th e D NA w ill zi p b ack to geth er a nd C as9 w ill ke ep b in din g to o th er P A M

se quence s until it fin ds th e m atch in g ta rg et D NA

5 '-G CAC G GCG GAG CG GTTC TTG GCAG CG GCCG CAC G ATC TC G TTG CCG CCG G -3 '

3' -C G TG CCG CCTC G CCAAG AAC CG TC G CCG GCG TG CTA G AG CAAC G GCG GCC- 5'

Once C as9 b in ds to a P A M se quence , it u nw in ds th e D NA. If th e g uid e R NA m atch es th e D NA se quence n ext to t h e P A M , th e g uid e

R NA w ill b in d to th e co mple m enta ry DNA st ra nd. If n ot, th e D NA w ill zi p b ack to geth er a nd C as9 w ill ke ep b in din g to o th er P A M

se quence s until it fin ds th e m atch in g ta rg et D NA.

3. B elo w is a p artia l se quence o f a g uid e R NA. T he u nderlin ed se ctio n o f th e R NA is desi gned to m atch a sp eci fic ta rg et D NA

se quence .

5 '-G GCG GAG CG GUUCUUG GCAG G UUUUAG AG CUAG AAAU AG C -3'

Revi ew th e p artia l g ene se quence r e sh ow n b elo w . It co nta in s a ta rg et D NA se quence th at m atch es th e g uid e R NA a bove . H ig hlig ht

th e o ne P A M se quence in th e to p ( 5 ’ to 3 ’) st ra nd th at is next to th is ta rg et D NA se quence . ( T he se quence u pst re am , to w ard th e 5 ’

end, o f th is PA M sh ould m atch th e u nderlin ed se quence in th e g uid e R NA,w hich m ake s th e o pposi te D NA st ra nd co mple m enta ry to

th e u nderlin ed se quence . R em em ber th at U ’s in R NA a re e quiva le nt to T ’s in D NA.)

5 '-G CAC G GCG GAG CG GTTC TTG GCAG CG GCCG CAC G ATC TC G TTG CCG CCG G -3 '

3' -C G TG CCG CCTC G CCAAG AAC CG TC G CCG GCG TG CTA G AG CAAC G GCG GCC- 5'

Ste p 2 : B in din g

O nce C as9 b in ds to th e co rre ct P A M , th e g uid e R NA b in ds to th e co mple m ent o f th e ta rg et D NA se quence .

Sie rra N etz -A pr 0 6, 2 021, 2 :1 0 P M M DT 4. W rit e d ow n th e g uid e R NA se quence th at b in ds to th e D NA, a nd th e D NA se quence th at it b in ds to ( th e co mple m ent o f th e ta rg et

D NA). L abel th e 5 ' a nd 3 ' e nds fo r b oth th e R NA a nd D NA st ra nds.

S te p 3 : C le avin g

O nce th e g uid e R NA b in ds to th e co mple m ent o f th e ta rg et D NA se quence , it a ct iva te s th e n ucl ease a ct ivi ty (D NA-cu ttin g a bilit y) o f th e

C as9 e nzym e. C uttin g D NA is also ca lle d “ cl eavi ng.” C as9 a lw ays cleave s both st ra nds of D NA. It cl eave s both th e ta rg et D NA a nd it s

co mple m ent th re e n ucl eotid es upst re am ( to w ard th e 5 ’ e nd) o f th e P A M se quence .

5.R ew rit e th e ta rg et D NA se quence a nd it s co mple m ent b elo w , in dica tin g w here C as9 w ould cu t b oth st ra nds of D NA w it h a la rg e

sp ace o r ve rtica l lin e ( |) .

S te p 4 : D NA R ep air

A fte r C as9 cl eave s th e D NA, ce llu la r e nzym es will a tte m pt to r e pair th e b re ak. M ost o f th e tim e, th ese e nzym es re pair th e D NA w it h out

erro rs. H ow eve r, C as9 w ill ke ep cu ttin g th e D NA a t th e sa me lo ca tio n u ntil a n e rro r is m ade.

6.D NA r e pair e rro rs in cl ude lo si ng o r in se rtin g r a ndom n ucl eotid es at th e cu t si te . E xp la in h ow th ese ch anges m ig ht in act iva te a g en e.