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lara31 [8.8K]
2 years ago
12

In Drosophila, the autosomal recessive brown eye color mutation displays interactions with both the X-linked recessive vermilion

mutation and the autosomal recessive scarlet mutation. Flies homozygous for brown and simultaneously hemizygous or homozygous for vermilion have white eyes. Flies simultaneously homozygous for both brown and scarlet mutations also have white eyes. Wild-type flies have red eyes. Predict the F1 and F2 progeny phenotype(s) of crossing the following true-breeding parents. Include proportion of every possible F1 and F2 phenotype
Biology
1 answer:
Brrunno [24]2 years ago
3 0

Answer:

Explanation:

According to the given information,

b/b = brown, b+/_ = Wild type (WT);  

XvXv or XvY = vermillion, Xv+_=WT;  

sc/sc = scarlet, sc+/_ = WT

b/b; XvXv =white eyed female, b/b;XvY = white eyed male

b/b;sc/sc = white-eyed male or female

You should have at least one b+ and one v+ to be WT(e.g. XvXvb+b =  

vermillion female, not WT)

1. vermilion females × brown males  Cross is: XvXv b+/b+x Xv+Ybb

F1: XvY b+b – vermillion males; and XvXv+b+b – WT females

F2:  both males and females have the same ratios: 1/8 white, 3/8 WT, 1/8 brown and  

3/8 vermillion

Xvb+  Xvb  Yb+  Yb

Xvb+  vermillion  vermillion  vermillion  WT

Xvb  vermillion  white  vermillion  white

Xv+b+  WT  WT  WT  WT

Xv+b  WT  Brown  WT  brown

Cross 2 – reciprocal of #1:  brown females (v+v+bb) x vermilion males (vYb+b+)  F1 =  

wild type females (XvXv+b+b) and WT males (Xv+Yb+b)   

F2: females ¾ v+_b+_ (WT): ¼ v+_bb (brown)

Males 3/8 v+Yb+_ (WT): 3/8 vYb+_ (vermilion): 1/8 v+Ybb (brown): 1/8 vYbb(white)  

Cross 3: scarlet females x brown males  

sc/sc;b+/b+ x sc+/sc+; b/b

sc/sc;b+ is scarlet, that you need at least one sc+ and one b+ to be  

WT

F1: sc/sc+; b+/b – all are WT

F2: no x-linked alleles in this cross. 9 WT: 3 scarlet: 3 brown: 1 white

scb+  scb  sc+b+  sc+b

scb+  scarlet  scarlet  WT  WT

scb  scarlet  white  WT  brown

sc+b+  WT  WT  WT  WT

sc+b  WT  brown  WT  brown

Cross #4:  brown females (b/b; sc+/sc+) x scarlet males (b+/b+; sc/sc)  F1 b+/b ; sc+/sc

(all WT)  F2 is the same as cross #3.

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