Genetic Problems

1.

T= trotting Tt X Tt

t= pacing genotypic ratio - 1:2:1

phenotypic ratio - 3:1

	T	t
T	TT	Tt
t	Tt	tt

2.

B= black bb X Bb

b= chestnut 50% could be black

50% could be chestnut

	B	b
b	Bb	bb
b	Bb	bb

3.

B = brown eyes father = bb

b= blue eyes Bb X bb

50% children could be blue-eyed

	b	b
B	Bb	Bb
b	bb	bb

4.

Yes, they could have a brown-eyed child!

5.

If she is Mm then, 50% chance she would get children who have migraines

	m	m
M	Mm	Mm
m	mm	mm

If she is MM then, 100% chance she would get children who have migraines

	m	m
M	Mm	Mm
M	Mm	Mm

So, there is a 50-100% chance her children will have migraines!

6. B=black T = trotting

b=chestnut t = pacing

bbTt X BbTt

First, find the gametes

Only gametes possible for the female are: bT and bt

	T	t
b	bT	bt
b	bT	bt

Only gametes possible for the male are: BT, Bt, bT, bt

	T	t
B	BT	Bt
b	bT	bt

Then, cross the gametes to find the offspring:

	BT	Bt	bT	bt
bT	BbTT	BbTt	bbTT	bbTt
bt	BbTt	Bbtt	bbTt	bbtt

3 black trotters

1 black pacer

3 chestnut trotters

1 chestnut pacer

7.

L = long eyelashes A = unattached ears

l = short eyelashes a = attached earlobes

Llaa X llAa

First find the gametes

Only possible gametes for the female are: La and la

	a	a
L	La	La
l	la	la

Only possible gametes for the male are: lA and la

	A	a
l	lA	la
l	lA	la

Then, cross the gametes to find the offspring:

	lA	la
La	LlAa	Llaa
la	llAa	llaa

1 long eyelashed, unattached ears Phenotypic ratio - 1:1:1:1

1 long eyelashed, attached ears

1 short eyelashed, unattached ears

1 short eyelashed, attached ears

8.

RR - red

R’R’ - white

R’R - pink

a. RR x R’R

	R’	R
R	R’R	RR
R	R’R	RR

genotypic ratio - 1:1

phenotypic ratio - 1:1

b. R’R’ x R’R

	R’	R
R’	R’R’	R’R
R’	R’R’	R’R

genotypic ratio and phenotypic ratios are the same as above.

9. A = normal pigment, a = albinism

10.

This is incomplete dominance where,

TT - long tail

T’T - short tail

T’T’ - no tail

If you cross 2 short tailed cats T’T x T’T

	T’	T
T’	T’T’	T’T
T	T’T	TT

Then, there is 1 long-tailed to 2 short-tailed to 1 no tailed cat or 1:2:1 ratio.

This is the approximate ratio of the litter, 2:6:3 is about a 1:2:1 ratio.

11.

a.

	i	i
I^A	I^Ai	I^A i
I^B	I^B i	I^B i

b.

	I^B	i
I^A	I^A I^B	I^A i
I^B	I^B I^B	I^B i

c.

	I^B	i
I^A	I^A I^B	I^A i
i	I^B i	i i

12. X^H = normal gene X^h = hemophilia X^— = unknown

13.

Assume that there are linked genes:

If Yh are linked and yH are linked then,

	Yh	yH
Yh	YYhh	YyHh
yH	YyHh	yyHH

1 yellow smooth: 2 yellow hairy: 1 red hairy

You never see a red smooth plant if the genes are linked this way. If you want proof then

try linking them the other way (YH and yh, or all different ways where there is not linkage YH, Yh, yH, yh) and see if you ever get the same result.

14.

Dominant traits will commonly show up more frequently in a population unless by some random event (i.e. disease or other catastrophe like tornado or earthquake) reduces those individuals that carry this trait, so that the recessive trait becomes more frequent in a population and the dominant trait is more infrequent.